About "A function surjects the rational numbers onto the irrational numbers"

[Ross Finlayson]

A function surjects the rational numbers onto the irrational numbers

Ross A. Finlayson
December 28, 2006

Abstract

There exists a surjection from the rational numbers onto the irrational
numbers.


Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.

There is a distinct q e Q for each p e P.
Proof. Let Q_i+ be the set of positive rationals
less than p_i, p_i e P+.
For each p_h < p_i, p_h e P+: Q_h+ subset Q_i+, and Q_h+ neq Q_i+.

Let

Q_nlt i = Q_i+ \ ( U p_h < p_i Q_h+ ) = Int p_h< p_i (Q_i+ \ Q_h+)

It is so that Q_nlt i neq 0
else Eq_h > p_h in a Q_h+, or some p_h = p_i.
Thus, Eq_i in Q_nlt i. A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+. As Q_nlt i neq 0 and
Q_nlt_i subset Q+: A p_i e P+, E q_i in Q+, where q_i
is related to p_i and not related to p_j neq p_i
for any p_j in P.

Similarly for Q- and P- via symmetry, replace < with >,
nlt with ngt, + with -, "less than" with "greater than",
and positive with negative: A p_i e P-, E q_i e Q-,
where q_i is related to p_i and not related to p_j neq p_i,
for any p_j e P.

Box

There exists an injection phi: P -> Q.
Proof. As there exists a distinct rational
q e Q for each distinct irrational p e P,
there exists an injection q: P -> Q.

E q_i in Q, A p_i, p_j in P: phi(p_i) = q_i, p_i neq p_j -> phi(p_i) ne
phi(p_j),

a one-to-one function from P to a subset of Q.

Box

There exists a surjection rho: Q -> P.
Proof. As there exists an injection phi: P -> Q
there exists a surjection rho: Q -> P.

Box

1


^^^
That's sort of it, while, it's typeset in TeX,
fits on one page, uses these symbols.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i".



A ∀
E ∃

e ∈


neq ≠

nlt ≮
ngt ≯

0 ∅

subset ⊂
U ⋃
Int ⋂

P ℙ
Q ℚ

=> ⇒
-> →

phi ϕ
rho ρ

Box □


Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.

There is a distinct q ∈ Q for each p ∈ P.
Proof. Let Q_i+ be the set of positive
rationals less than p_i, p_i ∈ P+.
For each p_h < p_i, p_h ∈ P+: Q_h+ ⊂ Q_i+, and Q_h+ ≠ Q_i+.

Let

Q_{≮i} = Q_i+ \ ( U_{p_h < p_i} Q_h+ ) = ⋂_{p_h< p_i} (Q_i+ \ Q_h+)

It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i.
Thus, ∃q_i ∈ Q_{≮i}.
∀q_i ∈ Q_{≮i} ⇒ q_i ∈ Q_i+, Q_{≮i} ⊂ Q_i+ ⊂ Q+.
As Q_{≮i} ≠ 0 and Q_{≮i} ⊂ Q+: ∀p_i ∈ P+, ∃q_i ∈ Q+,
where q_i is related to p_i and not related to
p_j ≠ p_i for any p_j ∈ P.

Similarly for Q- and P- via symmetry,
replace < with >, ≮ with ≯, + with -,
"less than" with "greater than",
and positive with negative:
∀p_i ∈ P-, ∃q_i ∈ Q-, where q_i is related to p_i and not related to p_j
≠ p_i, for any p_j ∈ P.

□

There exists an injection ϕ: P → Q.
Proof. As there exists a distinct rational
q ∈ Q for each distinct irrational p ∈ P,
there exists an injection q: P → Q.

∃ q_i ∈ Q, ∀ p_i, p_j ∈ P: ϕ(p_i) = q_i, p_i ≠ p_j ⇒ ϕ(p_i) ≠ ϕ(p_j),

a one-to-one function from P to a subset of Q.

□

There exists a surjection ρ: Q → P.
Proof. As there exists an injection ϕ: P → Q
there exists a surjection ρ: Q → P.

□





I figure that if you enjoy "Dedekind cuts"
then you'll get a kick out of this.

There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i".

[Ross Finlayson]

Post by Ross Finlayson
A function surjects the rational numbers onto the irrational numbers
Ross A. Finlayson
December 28, 2006
Abstract
There exists a surjection from the rational numbers onto the irrational
numbers.
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q e Q for each p e P.
Proof. Let Q_i+ be the set of positive rationals
less than p_i, p_i e P+.
For each p_h < p_i, p_h e P+: Q_h+ subset Q_i+, and Q_h+ neq Q_i+.
Let
Q_nlt i = Q_i+ \ ( U p_h < p_i Q_h+ ) = Int p_h< p_i (Q_i+ \ Q_h+)
It is so that Q_nlt i neq 0
else Eq_h > p_h in a Q_h+, or some p_h = p_i.
Thus, Eq_i in Q_nlt i. A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+. As Q_nlt i neq 0 and
Q_nlt_i subset Q+: A p_i e P+, E q_i in Q+, where q_i
is related to p_i and not related to p_j neq p_i
for any p_j in P.
Similarly for Q- and P- via symmetry, replace < with >,
nlt with ngt, + with -, "less than" with "greater than",
and positive with negative: A p_i e P-, E q_i e Q-,
where q_i is related to p_i and not related to p_j neq p_i,
for any p_j e P.
Box
There exists an injection phi: P -> Q.
Proof. As there exists a distinct rational
q e Q for each distinct irrational p e P,
there exists an injection q: P -> Q.
E q_i in Q, A p_i, p_j in P: phi(p_i) = q_i, p_i neq p_j -> phi(p_i) ne
phi(p_j),
a one-to-one function from P to a subset of Q.
Box
There exists a surjection rho: Q -> P.
Proof. As there exists an injection phi: P -> Q
there exists a surjection rho: Q -> P.
Box
1
^^^
That's sort of it, while, it's typeset in TeX,
fits on one page, uses these symbols.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i".
A ∀
E ∃
e ∈
neq ≠
nlt ≮
ngt ≯
0 ∅
subset ⊂
U ⋃
Int ⋂
P ℙ
Q ℚ
=> ⇒
-> →
phi ϕ
rho ρ
Box □
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q ∈ Q for each p ∈ P.
Proof. Let Q_i+ be the set of positive
rationals less than p_i, p_i ∈ P+.
For each p_h < p_i, p_h ∈ P+: Q_h+ ⊂ Q_i+, and Q_h+ ≠ Q_i+.
Let
Q_{≮i} = Q_i+ \ ( U_{p_h < p_i} Q_h+ ) = ⋂_{p_h< p_i} (Q_i+ \ Q_h+)
It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i.
Thus, ∃q_i ∈ Q_{≮i}.
∀q_i ∈ Q_{≮i} ⇒ q_i ∈ Q_i+, Q_{≮i} ⊂ Q_i+ ⊂ Q+.
As Q_{≮i} ≠ 0 and Q_{≮i} ⊂ Q+: ∀p_i ∈ P+, ∃q_i ∈ Q+,
where q_i is related to p_i and not related to
p_j ≠ p_i for any p_j ∈ P.
Similarly for Q- and P- via symmetry,
replace < with >, ≮ with ≯, + with -,
"less than" with "greater than",
∀p_i ∈ P-, ∃q_i ∈ Q-, where q_i is related to p_i and not related to p_j
≠ p_i, for any p_j ∈ P.
□
There exists an injection ϕ: P → Q.
Proof. As there exists a distinct rational
q ∈ Q for each distinct irrational p ∈ P,
there exists an injection q: P → Q.
∃ q_i ∈ Q, ∀ p_i, p_j ∈ P: ϕ(p_i) = q_i, p_i ≠ p_j ⇒ ϕ(p_i) ≠ ϕ(p_j),
a one-to-one function from P to a subset of Q.
□
There exists a surjection ρ: Q → P.
Proof. As there exists an injection ϕ: P → Q
there exists a surjection ρ: Q → P.
□
I figure that if you enjoy "Dedekind cuts"
then you'll get a kick out of this.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i".

Everybody pretty much knows the rationals are countable,
they're also everywhere dense and discontinuous in
the reals, and some have, that, these sorts sets,
are HUGE.

Then, a first sort of idea is that the term Q_{nlt i}
is empty, but, that would contradict the "It is so that ...".

So it speaks to that there's a book-keeping involved,
that gets into why it must be some kind of continuum
limit, that the rationals that fill _out_ and again _in_,
_dispersion_, then _density_, really rather involve
the numerical resources, how so the resulting structure,
in terms of what it derives as a topological result,
is some continuum limit, as to why similarly to the
line-reals that the continuum limit is not a Cartesian
function, that this sort of "rational-reals" (what results
for signal-reals"), strikes a balance, as of out over
the complete ordered field.

This then gets into notions like "algebra's establishment
of the uniqueness up to isomorphism of the complete
ordered field", about that algebra and arithmetic,
align together in the complete ordered field, but can
build separately and only meeting there, about the
deconstruction of algebra and arithmetic, and, the
deconstruction of arithmetic, into increment and partition.

The most usual description of "the rationals" as a function,
is the Dirichlet function.

f(x) = { 1, x is rational, 0, x is not rational }

The usual idea of course is that that has measure zero,
while, its complement 1-f, has measure b-a for f on [a,b].

Of course, that's the most usual fundamental result
up after descriptive set theory's objects of real analysis
their measure theory, that the unit interval of f has
measure zero, because it's countable.

Then, next to Dirichlet function, then next most usual
idea is about "the measure problem", which arises after
point-set topology, about Lebesgue measure, and it's
askance neighbor, Jordan measure, and about other
issues involved with the measure problem, in the assignment
of "analytical character", as that real analytical character
is measure (of area, or, length).

The derivation of the Fourier-style analysis, has a lot
going on to arrive at that the infinitesimal width pieces
have the contributions according to the coefficients
of all the sine and cosine powers up out after the
Eulerian-Gaussian, which one might imagine is to
be left aside in an alternate derivation for only
real-valued terms, that the idea of signal-reals,
gets into the Dirichlet function, in Fourier analysis.


The Dirichlet problem then up to the Poincare sphere,
illustrates the space of the _dispersed_ and _densified_
adding back up, and gets into why the Laplacian is all
partials, and harmonicity has after an anharmonic,
what results higher order harmonics and resonances,
more about what's going on, the real analytical character,
of these combined concerns.


So, these days there are the notions of ultrafilter,
up after topology, it builds these things for the
real analytical character, regardless that they would
so model or be modeled, by these line-reals and signal-reals,
much, much lower.

Then, the re-Vitali-ization of measure theory, and
the great concepts that Hausdorff carried up from
Vitali into Banach, the square-integrable, and for
"fuller derivatives", really help identify where in
the mathematics' fields today, these very same
notions of that the "completion" of the "complete
ordered field" reflects both "a least last straw",
the above example , and "a final redoubled
combinatorial explosion", the usual direct inference
due the uncountability of the complete ordered field,
get into how and why that the rationals are HUGE,
and that in some continuum limit for example as
Fourier-style analysis displays, that it's another result
of the non-Cartesian of the functions so relating things,
making a clear accounting (or, un-counting) of the book-keeping,
helps explain how mathematics is consistent,
as with regards to, "It is so ...".

[Ross Finlayson]

Post by Ross Finlayson

Post by Ross Finlayson
A function surjects the rational numbers onto the irrational numbers
Ross A. Finlayson
December 28, 2006
Abstract
There exists a surjection from the rational numbers onto the irrational
numbers.
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q e Q for each p e P.
Proof. Let Q_i+ be the set of positive rationals
less than p_i, p_i e P+.
For each p_h < p_i, p_h e P+: Q_h+ subset Q_i+, and Q_h+ neq Q_i+.
Let
Q_nlt i = Q_i+ \ ( U p_h < p_i Q_h+ ) = Int p_h< p_i (Q_i+ \ Q_h+)
It is so that Q_nlt i neq 0
else Eq_h > p_h in a Q_h+, or some p_h = p_i.
Thus, Eq_i in Q_nlt i. A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+. As Q_nlt i neq 0 and
Q_nlt_i subset Q+: A p_i e P+, E q_i in Q+, where q_i
is related to p_i and not related to p_j neq p_i
for any p_j in P.
Similarly for Q- and P- via symmetry, replace < with >,
nlt with ngt, + with -, "less than" with "greater than",
and positive with negative: A p_i e P-, E q_i e Q-,
where q_i is related to p_i and not related to p_j neq p_i,
for any p_j e P.
Box
There exists an injection phi: P -> Q.
Proof. As there exists a distinct rational
q e Q for each distinct irrational p e P,
there exists an injection q: P -> Q.
E q_i in Q, A p_i, p_j in P: phi(p_i) = q_i, p_i neq p_j -> phi(p_i) ne
phi(p_j),
a one-to-one function from P to a subset of Q.
Box
There exists a surjection rho: Q -> P.
Proof. As there exists an injection phi: P -> Q
there exists a surjection rho: Q -> P.
Box
1
^^^
That's sort of it, while, it's typeset in TeX,
fits on one page, uses these symbols.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i".
A ∀
E ∃
e ∈
neq ≠
nlt ≮
ngt ≯
0 ∅
subset ⊂
U ⋃
Int ⋂
P ℙ
Q ℚ
=> ⇒
-> →
phi ϕ
rho ρ
Box □
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q ∈ Q for each p ∈ P.
Proof. Let Q_i+ be the set of positive
rationals less than p_i, p_i ∈ P+.
For each p_h < p_i, p_h ∈ P+: Q_h+ ⊂ Q_i+, and Q_h+ ≠ Q_i+.
Let
Q_{≮i} = Q_i+ \ ( U_{p_h < p_i} Q_h+ ) = ⋂_{p_h< p_i} (Q_i+ \ Q_h+)
It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i.
Thus, ∃q_i ∈ Q_{≮i}.
∀q_i ∈ Q_{≮i} ⇒ q_i ∈ Q_i+, Q_{≮i} ⊂ Q_i+ ⊂ Q+.
As Q_{≮i} ≠ 0 and Q_{≮i} ⊂ Q+: ∀p_i ∈ P+, ∃q_i ∈ Q+,
where q_i is related to p_i and not related to
p_j ≠ p_i for any p_j ∈ P.
Similarly for Q- and P- via symmetry,
replace < with >, ≮ with ≯, + with -,
"less than" with "greater than",
∀p_i ∈ P-, ∃q_i ∈ Q-, where q_i is related to p_i and not related to p_j
≠ p_i, for any p_j ∈ P.
□
There exists an injection ϕ: P → Q.
Proof. As there exists a distinct rational
q ∈ Q for each distinct irrational p ∈ P,
there exists an injection q: P → Q.
∃ q_i ∈ Q, ∀ p_i, p_j ∈ P: ϕ(p_i) = q_i, p_i ≠ p_j ⇒ ϕ(p_i) ≠ ϕ(p_j),
a one-to-one function from P to a subset of Q.
□
There exists a surjection ρ: Q → P.
Proof. As there exists an injection ϕ: P → Q
there exists a surjection ρ: Q → P.
□
I figure that if you enjoy "Dedekind cuts"
then you'll get a kick out of this.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i".

Everybody pretty much knows the rationals are countable,
they're also everywhere dense and discontinuous in
the reals, and some have, that, these sorts sets,
are HUGE.
Then, a first sort of idea is that the term Q_{nlt i}
is empty, but, that would contradict the "It is so that ...".
So it speaks to that there's a book-keeping involved,
that gets into why it must be some kind of continuum
limit, that the rationals that fill _out_ and again _in_,
_dispersion_, then _density_, really rather involve
the numerical resources, how so the resulting structure,
in terms of what it derives as a topological result,
is some continuum limit, as to why similarly to the
line-reals that the continuum limit is not a Cartesian
function, that this sort of "rational-reals" (what results
for signal-reals"), strikes a balance, as of out over
the complete ordered field.
This then gets into notions like "algebra's establishment
of the uniqueness up to isomorphism of the complete
ordered field", about that algebra and arithmetic,
align together in the complete ordered field, but can
build separately and only meeting there, about the
deconstruction of algebra and arithmetic, and, the
deconstruction of arithmetic, into increment and partition.
The most usual description of "the rationals" as a function,
is the Dirichlet function.
f(x) = { 1, x is rational, 0, x is not rational }
The usual idea of course is that that has measure zero,
while, its complement 1-f, has measure b-a for f on [a,b].
Of course, that's the most usual fundamental result
up after descriptive set theory's objects of real analysis
their measure theory, that the unit interval of f has
measure zero, because it's countable.
Then, next to Dirichlet function, then next most usual
idea is about "the measure problem", which arises after
point-set topology, about Lebesgue measure, and it's
askance neighbor, Jordan measure, and about other
issues involved with the measure problem, in the assignment
of "analytical character", as that real analytical character
is measure (of area, or, length).
The derivation of the Fourier-style analysis, has a lot
going on to arrive at that the infinitesimal width pieces
have the contributions according to the coefficients
of all the sine and cosine powers up out after the
Eulerian-Gaussian, which one might imagine is to
be left aside in an alternate derivation for only
real-valued terms, that the idea of signal-reals,
gets into the Dirichlet function, in Fourier analysis.
The Dirichlet problem then up to the Poincare sphere,
illustrates the space of the _dispersed_ and _densified_
adding back up, and gets into why the Laplacian is all
partials, and harmonicity has after an anharmonic,
what results higher order harmonics and resonances,
more about what's going on, the real analytical character,
of these combined concerns.
So, these days there are the notions of ultrafilter,
up after topology, it builds these things for the
real analytical character, regardless that they would
so model or be modeled, by these line-reals and signal-reals,
much, much lower.
Then, the re-Vitali-ization of measure theory, and
the great concepts that Hausdorff carried up from
Vitali into Banach, the square-integrable, and for
"fuller derivatives", really help identify where in
the mathematics' fields today, these very same
notions of that the "completion" of the "complete
ordered field" reflects both "a least last straw",
the above example , and "a final redoubled
combinatorial explosion", the usual direct inference
due the uncountability of the complete ordered field,
get into how and why that the rationals are HUGE,
and that in some continuum limit for example as
Fourier-style analysis displays, that it's another result
of the non-Cartesian of the functions so relating things,
making a clear accounting (or, un-counting) of the book-keeping,
helps explain how mathematics is consistent,
as with regards to, "It is so ...".

I wrote this in 2006 as I was finishing up my mathematics degree,
it's its own self-contained thing.

[Chris M. Thomasson]

Post by Ross Finlayson
A function surjects the rational numbers onto the irrational numbers
Ross A. Finlayson
December 28, 2006
Abstract
There exists a surjection from the rational numbers onto the irrational
numbers.

Its fun to think about how any irrational can be approximated with a
rational up to a given level of accuracy. So, with infinite convergents
of continued fractions, there are infinite rationals to cover any
(infinite) precision of any irrational? Fair enough?

Post by Ross Finlayson
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q e Q for each p e P.
Proof. Let Q_i+ be the set of positive rationals
less than p_i, p_i e P+.
For each p_h < p_i, p_h e P+: Q_h+ subset Q_i+, and Q_h+ neq Q_i+.
Let
Q_nlt i = Q_i+ \ (  U p_h < p_i  Q_h+  ) = Int p_h< p_i (Q_i+ \ Q_h+)
It is so that Q_nlt i neq 0
else Eq_h > p_h in a Q_h+, or some p_h = p_i.
Thus, Eq_i in Q_nlt i.  A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+.  As Q_nlt i neq 0 and
Q_nlt_i subset Q+:  A p_i e P+, E q_i in Q+, where q_i
is related to p_i and not related to p_j neq p_i
for any p_j in P.
Similarly for Q- and P- via symmetry, replace < with >,
nlt with ngt, + with -, "less than" with "greater than",
and positive with negative: A p_i e P-, E q_i e Q-,
where q_i is related to p_i and not related to p_j neq p_i,
for any p_j e P.
Box
There exists an injection phi: P -> Q.
Proof.  As there exists a distinct rational
q e Q for each distinct irrational p e P,
there exists an injection q: P -> Q.
E q_i in Q, A p_i, p_j in P: phi(p_i) = q_i, p_i neq p_j -> phi(p_i) ne
phi(p_j),
a one-to-one function from P to a subset of Q.
Box
There exists a surjection rho: Q -> P.
Proof.  As there exists an injection phi: P -> Q
there exists a surjection rho:  Q -> P.
Box
1
^^^
That's sort of it, while, it's typeset in TeX,
fits on one page, uses these symbols.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i".
A  ∀
E ∃
e ∈
neq ≠
nlt ≮
ngt ≯
0 ∅
subset ⊂
U ⋃
Int ⋂
P ℙ
Q ℚ
=> ⇒
-> →
phi ϕ
rho ρ
Box □
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q ∈ Q for each p ∈ P.
Proof.  Let Q_i+ be the set of positive
rationals less than p_i, p_i ∈ P+.
For each p_h < p_i, p_h ∈ P+:  Q_h+ ⊂ Q_i+, and Q_h+ ≠ Q_i+.
Let
Q_{≮i} = Q_i+ \ (  U_{p_h < p_i}  Q_h+  ) = ⋂_{p_h< p_i} (Q_i+ \ Q_h+)
It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i.
Thus, ∃q_i ∈ Q_{≮i}.
∀q_i ∈ Q_{≮i} ⇒ q_i ∈ Q_i+, Q_{≮i} ⊂ Q_i+ ⊂ Q+.
As Q_{≮i} ≠ 0 and Q_{≮i} ⊂ Q+:  ∀p_i ∈ P+, ∃q_i ∈ Q+,
where q_i is related to p_i and not related to
p_j ≠ p_i for any p_j ∈ P.
Similarly for Q- and P- via symmetry,
replace < with >, ≮ with ≯, + with -,
"less than" with "greater than",
∀p_i ∈ P-, ∃q_i ∈ Q-, where q_i is related to p_i and not related to p_j
≠ p_i, for any p_j ∈ P.
□
There exists an injection ϕ: P → Q.
Proof.  As there exists a distinct rational
q ∈ Q for each distinct irrational p ∈ P,
there exists an injection q: P → Q.
∃ q_i ∈ Q, ∀ p_i, p_j ∈ P: ϕ(p_i) = q_i, p_i ≠ p_j ⇒ ϕ(p_i) ≠ ϕ(p_j),
a one-to-one function from P to a subset of Q.
□
There exists a surjection ρ: Q → P.
Proof.  As there exists an injection ϕ: P → Q
there exists a surjection ρ:  Q → P.
□
I figure that if you enjoy "Dedekind cuts"
then you'll get a kick out of this.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i".

[Ross Finlayson]

Post by Chris M. Thomasson

Post by Ross Finlayson
A function surjects the rational numbers onto the irrational numbers
Ross A. Finlayson
December 28, 2006
Abstract
There exists a surjection from the rational numbers onto the irrational
numbers.

Its fun to think about how any irrational can be approximated with a
rational up to a given level of accuracy. So, with infinite convergents
of continued fractions, there are infinite rationals to cover any
(infinite) precision of any irrational? Fair enough?

Post by Ross Finlayson
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q e Q for each p e P.
Proof. Let Q_i+ be the set of positive rationals
less than p_i, p_i e P+.
For each p_h < p_i, p_h e P+: Q_h+ subset Q_i+, and Q_h+ neq Q_i+.
Let
Q_nlt i = Q_i+ \ ( U p_h < p_i Q_h+ ) = Int p_h< p_i (Q_i+ \ Q_h+)
It is so that Q_nlt i neq 0
else Eq_h > p_h in a Q_h+, or some p_h = p_i.
Thus, Eq_i in Q_nlt i. A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+. As Q_nlt i neq 0 and
Q_nlt_i subset Q+: A p_i e P+, E q_i in Q+, where q_i
is related to p_i and not related to p_j neq p_i
for any p_j in P.
Similarly for Q- and P- via symmetry, replace < with >,
nlt with ngt, + with -, "less than" with "greater than",
and positive with negative: A p_i e P-, E q_i e Q-,
where q_i is related to p_i and not related to p_j neq p_i,
for any p_j e P.
Box
There exists an injection phi: P -> Q.
Proof. As there exists a distinct rational
q e Q for each distinct irrational p e P,
there exists an injection q: P -> Q.
E q_i in Q, A p_i, p_j in P: phi(p_i) = q_i, p_i neq p_j -> phi(p_i)
ne phi(p_j),
a one-to-one function from P to a subset of Q.
Box
There exists a surjection rho: Q -> P.
Proof. As there exists an injection phi: P -> Q
there exists a surjection rho: Q -> P.
Box
1
^^^
That's sort of it, while, it's typeset in TeX,
fits on one page, uses these symbols.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i".
A ∀
E ∃
e ∈
neq ≠
nlt ≮
ngt ≯
0 ∅
subset ⊂
U ⋃
Int ⋂
P ℙ
Q ℚ
=> ⇒
-> →
phi ϕ
rho ρ
Box □
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q ∈ Q for each p ∈ P.
Proof. Let Q_i+ be the set of positive
rationals less than p_i, p_i ∈ P+.
For each p_h < p_i, p_h ∈ P+: Q_h+ ⊂ Q_i+, and Q_h+ ≠ Q_i+.
Let
Q_{≮i} = Q_i+ \ ( U_{p_h < p_i} Q_h+ ) = ⋂_{p_h< p_i} (Q_i+ \ Q_h+)
It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i.
Thus, ∃q_i ∈ Q_{≮i}.
∀q_i ∈ Q_{≮i} ⇒ q_i ∈ Q_i+, Q_{≮i} ⊂ Q_i+ ⊂ Q+.
As Q_{≮i} ≠ 0 and Q_{≮i} ⊂ Q+: ∀p_i ∈ P+, ∃q_i ∈ Q+,
where q_i is related to p_i and not related to
p_j ≠ p_i for any p_j ∈ P.
Similarly for Q- and P- via symmetry,
replace < with >, ≮ with ≯, + with -,
"less than" with "greater than",
∀p_i ∈ P-, ∃q_i ∈ Q-, where q_i is related to p_i and not related to
p_j ≠ p_i, for any p_j ∈ P.
□
There exists an injection ϕ: P → Q.
Proof. As there exists a distinct rational
q ∈ Q for each distinct irrational p ∈ P,
there exists an injection q: P → Q.
∃ q_i ∈ Q, ∀ p_i, p_j ∈ P: ϕ(p_i) = q_i, p_i ≠ p_j ⇒ ϕ(p_i) ≠ ϕ(p_j),
a one-to-one function from P to a subset of Q.
□
There exists a surjection ρ: Q → P.
Proof. As there exists an injection ϕ: P → Q
there exists a surjection ρ: Q → P.
□
I figure that if you enjoy "Dedekind cuts"
then you'll get a kick out of this.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i".

That was known since about the time of Eudoxus,
though it's post-Pythagorean there are irrational
numbers, at all.

Back then, if there were irrational numbers at all,
they were few and far between, then though that
since Eudoxus, and positional notation, and infinite
expressions, it's sort of known that there are rationals
and irrationals each dense in the reals and each others'
complement in a set of, reals. And, each has an infinite
expression, and, some have dual representations in
the usual notation, those that end with b-1 in base b,
and 0 in base b.

So, ..., when you get to Cauchy, then he's like, well also
there are any number of series that add up to any given
real number, in fact we say these series that have a limit
have the property of being called Cauchy, and, these days
it's that the equivalence classes of series with the property
of being Cauchy, with equal limits the equivalence, actually
are the set of real numbers, it so results they model the
real numbers. It's just so that that's just sort of Eudoxus
wrapped as new.

Then there's Dedekind, he basically notices that Eudoxus
and Dedekind have arrived at irrationals from rational,
an infinite expression or as whatever establishes the limit,
which sort of falls back to Eudoxus, because, there must
be at least one such series for each, and the particularly
contrived example are just an infinite expression in
coefficients the terms of a negative power series, in base b.

So, Dedekind would be kind of redundant, and Cauchy would
be kind of redundant, though Cauchy gets employed in
the derivation of the Fourier-style analysis, which builds
a sort of continuous domain after windowing and boxing,
or what looks like the signal-reals, then Dedekind gets
employed to eastablish least-upper-bound, axiomatizing
that any irrational is some least upper bound of some
initial terms of some sequence that is Cauchy, or Eudoxus.

Then, Eudoxus/Cauchy/Dedekind, is the usual sort of
way that not merely approximation yet actually expressions
of values, exist, if infinite expressions, for each real number,
however it is that a series with implicit or explicit infinitely-many
terms has a limit and it is the value.


This here then is a bit more about the "it is so that, ...",
about an infinite set of irrationals, there being an infinite
set of rationals, set meaning they're each distinct, that
there's a pairing of the irrationals and rationals, thus
that the sum of the diferences is no different than zero.

It's a bit stronger than that, also, and it's about the signal-continuity,
as what could be employed for a Fourier-style analysis,
that a region or bounded interval of rationals, has that
the discontinuities in it are only points, that a stronger property
than "no gaps" is "no point-wise discontinuities", in the
sense that gaps are any discontinuity at all, that "point-wise
discontinuities" doesn't necessarily mean "no interval gaps".

Then, it is also gets involved with uncountability,
because, the reals are uncountable and rationals countable,
thus the rationals' complement the irrationals uncountable,
so it seems to contradict "it is so that...", unless as with regards
to the establishment of the function as above, it's necessarily
an analysis over a bounded region, which results treating
each region as for all its neighborhoods, because "it is so that ...".

I.e., not having "it is so that..." would violate the usual
infinite-divisibility,
and having "it is so that..." would violate the usual uncountability.

So, it's written as just about the usual "laws of numbers" or
"laws of arithmetic", and it doesn't say anything about uncountability,
yet, we're big boys and as conscientious mathematicians are
not allowed to forget either.

That said, for itself, it's allowed to say what's so, here "it is so
that, ...".


Thusly, I frame it in a milieu where there are at three continuous
domains, line-reals as we talk about "naturals n and d, f(n) = n/d,
d -> \infty", f(d) = 1 e ran(f)", field-reals as Eudoxus/Cauchy/Dedekind,
and signal-reals as about Dirichlet problem, Fourier-style analysis,
in some cases the ultrafilters, here "A function surjects...", and so on.


The rationals in a sense, are huge. In antiquity there were only them
and these days there are mostly not, or almost everywhere, rationals.

Then, what this is about is the bridge results, in a bounded interval,
so, according to number theory and geometry, from being all
the rationals to covering all the irrationals with those being only
a shade or hair or minimal modicum or least latitude a filter,
either the other.


It's a thing. "It is so that, ...".


Anyways it's been known that Eudoxus has at least an infinite
expression for any real number since a couple thousand years, yeah.


We're conscientious mathematicians, though, so
the whole world of mathematics exists.

[Jim Burns]

Post by Ross Finlayson
A function surjects the rational numbers onto
the irrational numbers
Ross A. Finlayson
December 28, 2006
Abstract
There exists a surjection
from the rational numbers
onto the irrational numbers.
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q e Q for each p e P.
Proof.

De.ASCII.fied.
|
| For pᵢ ∈ ℙ⁺ = ℝ⁺\ℚ⁺
| Q_i+ =
| ℚ⁺ᵢ = {q ∈ ℚ⁺: q<pᵢ}
|
| ∀pₕ < pᵢ: ℚ⁺ₕ ≠⊂ ℚ⁺ᵢ
|
| Q_nlt i =
| ℚᐠᑉᵢ =
| ℚ⁺ᵢ\(⋃[pₕ<pᵢ] ℚ⁺ₕ) =
| ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ
|
| It is so that ℚᐠᑉ ᵢ ≠ ∅
| else
| ∃ℚ⁺ₕ: ∃qₕ ∈ ℚ⁺ₕ: qₕ > pₕ or ∃pₕ = pᵢ

What is X else Y ?
X Y X.else.Y
t t ?
f t ?
t f ?
f f ?

----
It is not so that ℚᐠᑉᵢ ≠ ∅
ℚᐠᑉᵢ = ∅

For each two rational.or.irrational r₁ r₂
there is an irrational p₁₂ between them.
∀r₁,r₂ ∈ ℝ, r₁<r₂:
∃p₁₂ ∈ ℙ: r₁ < p₁₂ < r₂

∀q ∈ ℚ⁺, ∀pᵢ ∈ ℙ⁺, q < pᵢ:
∃pₕ ∈ ℙ⁺: q < pₕ < pᵢ:
q ∈ ℚ⁺ₕ
q ∉ ℚ⁺ᵢ\ℚ⁺ₕ
q ∉ ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ
q ∉ ℚᐠᑉᵢ

∀q ∈ ℚ⁺, ∀pᵢ ∈ ℙ⁺, q < pᵢ:
q ∉ ℚᐠᑉᵢ

∀pᵢ ∈ ℙ⁺: ℚᐠᑉᵢ = ∅

Post by Ross Finlayson
Proof.
Let Q_i+ be the set of positive rationals
less than p_i, p_i e P+.
Q_h+ subset Q_i+, and Q_h+ neq Q_i+.
Let
Q_nlt i =
Q_i+ \ (  U p_h < p_i  Q_h+  ) =
Int p_h< p_i (Q_i+ \ Q_h+)
It is so that Q_nlt i neq 0
else
Eq_h > p_h in a Q_h+, or some p_h = p_i.
Thus, Eq_i in Q_nlt i.
A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+.
A p_i e P+, E q_i in Q+, where
q_i is related to p_i and
not related to p_j neq p_i
for any p_j in P.
Similarly for Q- and P- via symmetry, replace < with >,
nlt with ngt, + with -, "less than" with "greater than",
and positive with negative: A p_i e P-, E q_i e Q-,
where q_i is related to p_i and not related to p_j neq p_i,
for any p_j e P.

[Ross Finlayson]

Post by Jim Burns

Post by Ross Finlayson
A function surjects the rational numbers onto
the irrational numbers
Ross A. Finlayson
December 28, 2006
Abstract
There exists a surjection
from the rational numbers
onto the irrational numbers.
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q e Q for each p e P.
Proof.

De.ASCII.fied.
|
| For pᵢ ∈ ℙ⁺ = ℝ⁺\ℚ⁺
| Q_i+ =
| ℚ⁺ᵢ = {q ∈ ℚ⁺: q<pᵢ}
|
| ∀pₕ < pᵢ: ℚ⁺ₕ ≠⊂ ℚ⁺ᵢ
|
| Q_nlt i =
| ℚᐠᑉᵢ =
| ℚ⁺ᵢ\(⋃[pₕ<pᵢ] ℚ⁺ₕ) =
| ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ
|
| It is so that ℚᐠᑉ ᵢ ≠ ∅
| else
| ∃ℚ⁺ₕ: ∃qₕ ∈ ℚ⁺ₕ: qₕ > pₕ or ∃pₕ = pᵢ
What is X else Y ?
X Y X.else.Y
t t ?
f t ?
t f ?
f f ?
----
It is not so that ℚᐠᑉᵢ ≠ ∅
ℚᐠᑉᵢ = ∅
For each two rational.or.irrational r₁ r₂
there is an irrational p₁₂ between them.
∃p₁₂ ∈ ℙ: r₁ < p₁₂ < r₂
q ∈ ℚ⁺ₕ
q ∉ ℚ⁺ᵢ\ℚ⁺ₕ
q ∉ ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ
q ∉ ℚᐠᑉᵢ
q ∉ ℚᐠᑉᵢ
∀pᵢ ∈ ℙ⁺: ℚᐠᑉᵢ = ∅

Post by Ross Finlayson
Proof.
Let Q_i+ be the set of positive rationals
less than p_i, p_i e P+.
Q_h+ subset Q_i+, and Q_h+ neq Q_i+.
Let
Q_nlt i =
Q_i+ \ ( U p_h < p_i Q_h+ ) =
Int p_h< p_i (Q_i+ \ Q_h+)
It is so that Q_nlt i neq 0
else
Eq_h > p_h in a Q_h+, or some p_h = p_i.
Thus, Eq_i in Q_nlt i.
A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+.
A p_i e P+, E q_i in Q+, where
q_i is related to p_i and
not related to p_j neq p_i
for any p_j in P.
Similarly for Q- and P- via symmetry, replace < with >,
nlt with ngt, + with -, "less than" with "greater than",
and positive with negative: A p_i e P-, E q_i e Q-,
where q_i is related to p_i and not related to p_j neq p_i,
for any p_j e P.

Yeah, I get that a lot, but then there wouldn't be what it is there.

So, "it is so that ...".

[Ross Finlayson]

Post by Ross Finlayson

Post by Jim Burns

Post by Ross Finlayson
A function surjects the rational numbers onto
the irrational numbers
Ross A. Finlayson
December 28, 2006
Abstract
There exists a surjection
from the rational numbers
onto the irrational numbers.
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q e Q for each p e P.
Proof.

De.ASCII.fied.
|
| For pᵢ ∈ ℙ⁺ = ℝ⁺\ℚ⁺
| Q_i+ =
| ℚ⁺ᵢ = {q ∈ ℚ⁺: q<pᵢ}
|
| ∀pₕ < pᵢ: ℚ⁺ₕ ≠⊂ ℚ⁺ᵢ
|
| Q_nlt i =
| ℚᐠᑉᵢ =
| ℚ⁺ᵢ\(⋃[pₕ<pᵢ] ℚ⁺ₕ) =
| ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ
|
| It is so that ℚᐠᑉ ᵢ ≠ ∅
| else
| ∃ℚ⁺ₕ: ∃qₕ ∈ ℚ⁺ₕ: qₕ > pₕ or ∃pₕ = pᵢ
What is X else Y ?
X Y X.else.Y
t t ?
f t ?
t f ?
f f ?
----
It is not so that ℚᐠᑉᵢ ≠ ∅
ℚᐠᑉᵢ = ∅
For each two rational.or.irrational r₁ r₂
there is an irrational p₁₂ between them.
∃p₁₂ ∈ ℙ: r₁ < p₁₂ < r₂
q ∈ ℚ⁺ₕ
q ∉ ℚ⁺ᵢ\ℚ⁺ₕ
q ∉ ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ
q ∉ ℚᐠᑉᵢ
q ∉ ℚᐠᑉᵢ
∀pᵢ ∈ ℙ⁺: ℚᐠᑉᵢ = ∅

Post by Ross Finlayson
Proof.
Let Q_i+ be the set of positive rationals
less than p_i, p_i e P+.
Q_h+ subset Q_i+, and Q_h+ neq Q_i+.
Let
Q_nlt i =
Q_i+ \ ( U p_h < p_i Q_h+ ) =
Int p_h< p_i (Q_i+ \ Q_h+)
It is so that Q_nlt i neq 0
else
Eq_h > p_h in a Q_h+, or some p_h = p_i.
Thus, Eq_i in Q_nlt i.
A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+.
A p_i e P+, E q_i in Q+, where
q_i is related to p_i and
not related to p_j neq p_i
for any p_j in P.
Similarly for Q- and P- via symmetry, replace < with >,
nlt with ngt, + with -, "less than" with "greater than",
and positive with negative: A p_i e P-, E q_i e Q-,
where q_i is related to p_i and not related to p_j neq p_i,
for any p_j e P.

Yeah, I get that a lot, but then there wouldn't be what it is there.
So, "it is so that ...".

(Mostly the "it is so ... or else [one of] ... or ..." bit,
neither of those being a thing.)

[Jim Burns]

Post by Ross Finlayson

Post by Ross Finlayson

[...]

Yeah, I get that a lot,
but then there wouldn't be what it is there.
So, "it is so that ...".

(Mostly the
"it is so ... or else [one of] ... or ..."
bit,
neither of those being a thing.)

Then, by
| It is so that ℚᐠᑉᵢ ≠ ∅
| else
| ∃ℚ⁺ₕ: ∃qₕ ∈ ℚ⁺ₕ: qₕ > pₕ or ∃pₕ = pᵢ
|
you're saying [1]
| ℚᐠᑉᵢ ≠ ∅ or
| ∃ℚ⁺ₕ: ∃qₕ ∈ ℚ⁺ₕ: qₕ > pₕ or
| ∃pₕ = pᵢ
|
Is that right?

It would be great if
I didn't have to guess why you think [1] is true.

Until you find time to help me understand you,
I will guess that you think that because
for each ℚ⁺ₕ: ℚ⁺ᵢ\ℚ⁺ₕ ≠ ∅

This is an invalid quantifier swap:
∀ℚ⁺ₕ: ∃qᵢ: qᵢ ∈ ℚ⁺ᵢ\ℚ⁺ₕ
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
☠ ∃qᵢ: ∀ℚ⁺ₕ: qᵢ ∈ ℚ⁺ᵢ\ℚ⁺ₕ

If it were valid, it would follow that
☠ ℚᐠᑉᵢ = ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ ≠ ∅

That inference isn't valid.
That conclusion doesn't follow and isn't true.

----
For each two rational.or.irrational r₁ r₂
there is an irrational p₁₂ between them.
∀r₁,r₂ ∈ ℝ, r₁<r₂:
∃p₁₂ ∈ ℙ=ℝ\ℚ: r₁ < p₁₂ < r₂

| Assume r₁,r₂ ∈ ℝ, r₁<r₂
|
| d = 1+max{d′ ∈ ℕ: d′ ≤ 3/(r₂-r₁) }
| n = 1+max{n′ ∈ ℕ: n′ ≤ d⋅r₁ }
|
| r₁ < n/d < (n+1)/d < r₂
| n/d, (n+1)/d ∈ ℚ
|
| n/d < (n+⅟√​̅2)/d < (n+1)/d
| r₁ < (n+⅟√​̅2)/d < r₂
| (n+⅟√​̅2)/d ∈ ℙ
|
| p₁₂ = (n+⅟√​̅2)/d
| p₁₂ ∈ ℙ
| r₁ < p₁₂ < r₂

Therefore,
for each two rational.or.irrational r₁ r₂
there is an irrational p₁₂ between them.
∀r₁,r₂ ∈ ℝ, r₁<r₂:
∃p₁₂ ∈ ℙ=ℝ\ℚ: r₁ < p₁₂ < r₂

∀q ∈ ℚ⁺, ∀pᵢ ∈ ℙ⁺, q < pᵢ:
∃pₕ ∈ ℙ⁺: q < pₕ < pᵢ:
q ∈ ℚ⁺ₕ
q ∉ ℚ⁺ᵢ\ℚ⁺ₕ
q ∉ ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ
q ∉ ℚᐠᑉᵢ

∀q ∈ ℚ⁺, ∀pᵢ ∈ ℙ⁺, q < pᵢ:
q ∉ ℚᐠᑉᵢ

∀pᵢ ∈ ℙ⁺: ℚᐠᑉᵢ = ∅

[Moebius]

There exists a surjection from the rational numbers onto the irrational numbers.

Certainly not.

[Ross Finlayson]

Post by Moebius

Post by Ross Finlayson
There exists a surjection from the rational numbers onto the
irrational numbers.

Certainly not.

It's like if you found a time machine and went
and asked Leibniz about iota-values.

"Oh you mean differentials? Newton's got
his fluxions. Wait you mean that they're
integrable and not 1/2 instead exactly 1?
I didn't get that from Bradwardine's De Continuo,
you should go back another 300 years and
see what the Merton school says about it."


Then, it seems there's an impasse, and indeed,
more than one. Here though when we look at
the self-contained and internally consistent
definition, and it works out that it can't really
agree, then one would aver something has to
give, yet if it can't be one or the other, then
what would seem best would be a development
that arrives at both the major concerns together,
without being wrong either way.

Of course, that looks about the same from
either side, either the "it is so that ..." or
the "it is not necessarily so that ...", yet,
given iota-values and other models of
real-valued numbers among the line-reals
and field-reals as continuous domains,
at least one way seems put together
while the other seems kind of stuck.

"Is that all they got these days?
They should write Berkeley
and get their subscription back.
Even in my time we had more
open minds than that. Yeah,
run that past Cavalieri, then
I'm very interested to learn
about Vitali and Hausdorff,
what great analytic geometers."

[Ross Finlayson]

Post by Ross Finlayson

Post by Moebius

There exists a surjection from the rational numbers onto the irrational numbers.

Certainly not.

It's like if you found a time machine and went
and asked Leibniz about iota-values.
"Oh you mean differentials? Newton's got
his fluxions. Wait you mean that they're
integrable and not 1/2 instead exactly 1?
I didn't get that from Bradwardine's De Continuo,
you should go back another 300 years and
see what the Merton school says about it."
Then, it seems there's an impasse, and indeed,
more than one. Here though when we look at
the self-contained and internally consistent
definition, and it works out that it can't really
agree, then one would aver something has to
give, yet if it can't be one or the other, then
what would seem best would be a development
that arrives at both the major concerns together,
without being wrong either way.
Of course, that looks about the same from
either side, either the "it is so that ..." or
the "it is not necessarily so that ...", yet,
given iota-values and other models of
real-valued numbers among the line-reals
and field-reals as continuous domains,
at least one way seems put together
while the other seems kind of stuck.
"Is that all they got these days?
They should write Berkeley
and get their subscription back.
Even in my time we had more
open minds than that. Yeah,
run that past Cavalieri, then
I'm very interested to learn
about Vitali and Hausdorff,
what great analytic geometers."

Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.

There is a distinct q ∈ Q for each p ∈ P.
Proof. Let Q_i+ be the set of positive
rationals less than p_i, p_i ∈ P+.
For each p_h < p_i, p_h ∈ P+: Q_h+ ⊂ Q_i+, and Q_h+ ≠ Q_i+.

Let

Q_{≮i} = Q_i+ \ ( U_{p_h < p_i} Q_h+ ) = ⋂_{p_h< p_i} (Q_i+ \ Q_h+)

_It is so that Q_{≮i} ≠ ∅ else ∃q_h > p_h in a Q_h+, or some p_h = p_i._
Thus, ∃q_i ∈ Q_{≮i}.
∀q_i ∈ Q_{≮i} ⇒ q_i ∈ Q_i+, Q_{≮i} ⊂ Q_i+ ⊂ Q+.
As Q_{≮i} ≠ 0 and Q_{≮i} ⊂ Q+: ∀p_i ∈ P+, ∃q_i ∈ Q+,
where q_i is related to p_i and not related to
p_j ≠ p_i for any p_j ∈ P.

Similarly for Q- and P- via symmetry,
replace < with >, ≮ with ≯, + with -,
"less than" with "greater than",
and positive with negative:
∀p_i ∈ P-, ∃q_i ∈ Q-, where q_i is related to p_i and not related to p_j
≠ p_i, for any p_j ∈ P.

□

There exists an injection ϕ: P → Q.
Proof. As there exists a distinct rational
q ∈ Q for each distinct irrational p ∈ P,
there exists an injection q: P → Q.

∃ q_i ∈ Q, ∀ p_i, p_j ∈ P: ϕ(p_i) = q_i, p_i ≠ p_j ⇒ ϕ(p_i) ≠ ϕ(p_j),

a one-to-one function from P to a subset of Q.

□

There exists a surjection ρ: Q → P.
Proof. As there exists an injection ϕ: P → Q
there exists a surjection ρ: Q → P.

□

[Moebius]

Post by Moebius

There exists a surjection from the rational numbers onto the irrational numbers.

Certainly not.

There exists a surjection ρ: Q → P. [...]

"Just the place for a Snark! I have said it thrice:
What I tell you three times is true."

[Ross Finlayson]

Post by Ross Finlayson

Post by Ross Finlayson

Post by Moebius

There exists a surjection from the rational numbers onto the irrational numbers.

Certainly not.

It's like if you found a time machine and went
and asked Leibniz about iota-values.
"Oh you mean differentials? Newton's got
his fluxions. Wait you mean that they're
integrable and not 1/2 instead exactly 1?
I didn't get that from Bradwardine's De Continuo,
you should go back another 300 years and
see what the Merton school says about it."
Then, it seems there's an impasse, and indeed,
more than one. Here though when we look at
the self-contained and internally consistent
definition, and it works out that it can't really
agree, then one would aver something has to
give, yet if it can't be one or the other, then
what would seem best would be a development
that arrives at both the major concerns together,
without being wrong either way.
Of course, that looks about the same from
either side, either the "it is so that ..." or
the "it is not necessarily so that ...", yet,
given iota-values and other models of
real-valued numbers among the line-reals
and field-reals as continuous domains,
at least one way seems put together
while the other seems kind of stuck.
"Is that all they got these days?
They should write Berkeley
and get their subscription back.
Even in my time we had more
open minds than that. Yeah,
run that past Cavalieri, then
I'm very interested to learn
about Vitali and Hausdorff,
what great analytic geometers."

Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q ∈ Q for each p ∈ P.
Proof. Let Q_i+ be the set of positive
rationals less than p_i, p_i ∈ P+.
For each p_h < p_i, p_h ∈ P+: Q_h+ ⊂ Q_i+, and Q_h+ ≠ Q_i+.
Let
Q_{≮i} = Q_i+ \ ( U_{p_h < p_i} Q_h+ ) = ⋂_{p_h< p_i} (Q_i+ \ Q_h+)
_It is so that Q_{≮i} ≠ ∅ else ∃q_h > p_h in a Q_h+, or some p_h = p_i._
Thus, ∃q_i ∈ Q_{≮i}.
∀q_i ∈ Q_{≮i} ⇒ q_i ∈ Q_i+, Q_{≮i} ⊂ Q_i+ ⊂ Q+.
As Q_{≮i} ≠ 0 and Q_{≮i} ⊂ Q+: ∀p_i ∈ P+, ∃q_i ∈ Q+,
where q_i is related to p_i and not related to
p_j ≠ p_i for any p_j ∈ P.
Similarly for Q- and P- via symmetry,
replace < with >, ≮ with ≯, + with -,
"less than" with "greater than",
∀p_i ∈ P-, ∃q_i ∈ Q-, where q_i is related to p_i and not related to p_j
≠ p_i, for any p_j ∈ P.
□
There exists an injection ϕ: P → Q.
Proof. As there exists a distinct rational
q ∈ Q for each distinct irrational p ∈ P,
there exists an injection q: P → Q.
∃ q_i ∈ Q, ∀ p_i, p_j ∈ P: ϕ(p_i) = q_i, p_i ≠ p_j ⇒ ϕ(p_i) ≠ ϕ(p_j),
a one-to-one function from P to a subset of Q.
□
There exists a surjection ρ: Q → P.
Proof. As there exists an injection ϕ: P → Q
there exists a surjection ρ: Q → P.
□

Yeah, I was pretty excited to learn about "De Continuo",
Bradwardine and the Merton school's late Middle-Age
Scholastics' reflections on continuity.

Being a couple hundred years before Galileo, while
addressing the early algebraization after geometry,
and getting into continuum mechanics, really helps
advise where Newton and Leibniz came up with this
stuff, and one must wonder that Cantor and cie must
have been well aware of it also.

So you know....

[Ross Finlayson]

Post by Ross Finlayson

Post by Ross Finlayson

Post by Ross Finlayson

Post by Moebius

There exists a surjection from the rational numbers onto the irrational numbers.

Certainly not.

It's like if you found a time machine and went
and asked Leibniz about iota-values.
"Oh you mean differentials? Newton's got
his fluxions. Wait you mean that they're
integrable and not 1/2 instead exactly 1?
I didn't get that from Bradwardine's De Continuo,
you should go back another 300 years and
see what the Merton school says about it."
Then, it seems there's an impasse, and indeed,
more than one. Here though when we look at
the self-contained and internally consistent
definition, and it works out that it can't really
agree, then one would aver something has to
give, yet if it can't be one or the other, then
what would seem best would be a development
that arrives at both the major concerns together,
without being wrong either way.
Of course, that looks about the same from
either side, either the "it is so that ..." or
the "it is not necessarily so that ...", yet,
given iota-values and other models of
real-valued numbers among the line-reals
and field-reals as continuous domains,
at least one way seems put together
while the other seems kind of stuck.
"Is that all they got these days?
They should write Berkeley
and get their subscription back.
Even in my time we had more
open minds than that. Yeah,
run that past Cavalieri, then
I'm very interested to learn
about Vitali and Hausdorff,
what great analytic geometers."

Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q ∈ Q for each p ∈ P.
Proof. Let Q_i+ be the set of positive
rationals less than p_i, p_i ∈ P+.
For each p_h < p_i, p_h ∈ P+: Q_h+ ⊂ Q_i+, and Q_h+ ≠ Q_i+.
Let
Q_{≮i} = Q_i+ \ ( U_{p_h < p_i} Q_h+ ) = ⋂_{p_h< p_i} (Q_i+ \ Q_h+)
_It is so that Q_{≮i} ≠ ∅ else ∃q_h > p_h in a Q_h+, or some p_h = p_i._
Thus, ∃q_i ∈ Q_{≮i}.
∀q_i ∈ Q_{≮i} ⇒ q_i ∈ Q_i+, Q_{≮i} ⊂ Q_i+ ⊂ Q+.
As Q_{≮i} ≠ 0 and Q_{≮i} ⊂ Q+: ∀p_i ∈ P+, ∃q_i ∈ Q+,
where q_i is related to p_i and not related to
p_j ≠ p_i for any p_j ∈ P.
Similarly for Q- and P- via symmetry,
replace < with >, ≮ with ≯, + with -,
"less than" with "greater than",
∀p_i ∈ P-, ∃q_i ∈ Q-, where q_i is related to p_i and not related to p_j
≠ p_i, for any p_j ∈ P.
□
There exists an injection ϕ: P → Q.
Proof. As there exists a distinct rational
q ∈ Q for each distinct irrational p ∈ P,
there exists an injection q: P → Q.
∃ q_i ∈ Q, ∀ p_i, p_j ∈ P: ϕ(p_i) = q_i, p_i ≠ p_j ⇒ ϕ(p_i) ≠ ϕ(p_j),
a one-to-one function from P to a subset of Q.
□
There exists a surjection ρ: Q → P.
Proof. As there exists an injection ϕ: P → Q
there exists a surjection ρ: Q → P.
□

Yeah, I was pretty excited to learn about "De Continuo",
Bradwardine and the Merton school's late Middle-Age
Scholastics' reflections on continuity.
Being a couple hundred years before Galileo, while
addressing the early algebraization after geometry,
and getting into continuum mechanics, really helps
advise where Newton and Leibniz came up with this
stuff, and one must wonder that Cantor and cie must
have been well aware of it also.
So you know....

Of course, my mathematical conscience would demand that
there can be no duplicitousness, and as well my mathematical
formalist's pride would require being never wrong, so any
matters of contradiction in mathematics, make for quite
a thorough examination and test of the formalist's the
constructivist's and for the intuitionist's and duelling
constructivist's views for the complementary duals of
the inverses of mathematical stipulations and opinion,
that it's considered one of the higher and better modes
of thought and reason, the purely mathematical.

What I'm saying is, I wouldn't have to bring this up,
except that somebody must, because, mathematical
formalists with a mathematical conscience abstractly
have no opinion about any who don't.

So, here it's a thing, that the n/d with one free and
the other infinite make a [0,1] line-reals, and n/d with
both free make an ordered field, and the complement
in the linear continuum which completes them, is for
a bounded region about the doubling-space and
doubling-measure that these simplest mathematical
resources so construct.


It's a thing.

[WM]

Post by Moebius

There exists a surjection from the rational numbers onto the irrational numbers.

Certainly not.

*All* digits of Cantor's diagonal represent rational numbers.

Regards, WM

[Moebius]

Post by WM

Post by Moebius

There exists a surjection from the rational numbers onto the irrational numbers.

Certainly not.

*All* digits of Cantor's diagonal represent rational numbers.

@Ross Finlayson:

Rule of thumb: If Mückenheim agrees with you, there's something VERY
wrong with your argument!

[Chris M. Thomasson]

Post by Moebius

Post by WM

Post by Moebius

There exists a surjection from the rational numbers onto the irrational numbers.

Certainly not.

*All* digits of Cantor's diagonal represent rational numbers.

Rule of thumb: If Mückenheim agrees with you, there's something VERY
wrong with your argument!

YIKES!

[Ross Finlayson]

Post by Moebius

Post by WM

Post by Moebius

There exists a surjection from the rational numbers onto the irrational numbers.

Certainly not.

*All* digits of Cantor's diagonal represent rational numbers.

Rule of thumb: If Mückenheim agrees with you, there's something VERY
wrong with your argument!

It is its own thing.

I'm not talking about sets here at all, except insofar
as they're intervals of numbers.

Of course, I know set theory very well and besides this
thing about the rationals and other sets "dense in the
reals, nowhere continuous, whose complement is also
dense in the reals and nowhere continuous", or
nowhere continuous, dense, and equi-distributed,
about things like "the rationals are HUGE" and this
kind of thing, it is its own thing.

It's involved with other parts of the theory of
the continuous domains, and about Cantor space for
example, and "Square" Cantor space, after Borel versus
Combinatorics, and real analytical character.

For example line-reals are a continuous domain,
and field-reals are a continuous domain,
and signal-reals are a continuous domain.

Mein Hut hat drei Ecke....

Danke fur Lesern, keine Beleidigung beabsichticht, MfG, E.S.

[Jim Burns]

Now! With More Unicode!

Post by Ross Finlayson
A function surjects the rational numbers
onto the irrational numbers
Ross A. Finlayson
December 28, 2006
Abstract
There exists a surjection
from the rational numbers
onto the irrational numbers.
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q e Q for each p e P.
Proof.
Let Q_i+ be the set of
positive rationals less than p_i,
p_i e P+.
Q_h+ subset Q_i+, and Q_h+ neq Q_i+.
Let
Q_nlt i =
Q_i+ \ (  U p_h < p_i  Q_h+  ) =
Int p_h< p_i (Q_i+ \ Q_h+)
It is so that Q_nlt i neq 0
else
Eq_h > p_h in a Q_h+,
or
some p_h = p_i.

I am guessing that you are depending on
an unreliable quantifier swap:
∀ℚ⁺ₕ: ∃qᵢ: qᵢ ∈ ℚ⁺ᵢ\ℚ⁺ₕ
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
☠ ∃qᵢ: ∀ℚ⁺ₕ: qᵢ ∈ ℚ⁺ᵢ\ℚ⁺ₕ

If it were reliable, it would follow that
☠ ℚᐠᑉᵢ = ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ ≠ ∅

However, that doesn't follow and isn't true.

∀pᵢ ∈ ℙ⁺ = ℝ⁺\ℚ⁺:
∀qⱼ ∈ ℚ⁺ᵢ = ℚ⁺∩[0,pᵢ):
dⱼ = ⌊1+3/(pᵢ-qⱼ)⌋
nⱼ = ⌊1+dⱼ⋅qⱼ⌋
qⱼ < nⱼ/dⱼ < (nⱼ+1)/dⱼ < pᵢ
pₕ = (nⱼ+⅟√​̅2)/dⱼ
pₕ ∈ ℙ⁺: qⱼ < pₕ < pᵢ
qⱼ ∈ ℚ⁺ₕ = ℚ⁺∩[0,pₕ)
qⱼ ∉ ℚ⁺ᵢ\ℚ⁺ₕ
qⱼ ∉ ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ = ℚᐠᑉᵢ

∀pᵢ ∈ ℙ⁺: ∀qⱼ ∈ ℚ⁺ᵢ: qⱼ ∉ ℚᐠᑉᵢ

∀pᵢ ∈ ℙ⁺: ℚᐠᑉᵢ = ∅

Post by Ross Finlayson
Thus, Eq_i in Q_nlt i.
 A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+.
A p_i e P+, E q_i in Q+, where
q_i is related to p_i and
not related to p_j neq p_i
for any p_j in P.
Similarly for Q- and P- via symmetry,
replace < with >, nlt with ngt, + with -,
"less than" with "greater than",
A p_i e P-, E q_i e Q-, where
q_i is related to p_i and
not related to p_j neq p_i,
for any p_j e P.
Box
There exists an injection phi: P -> Q.
Proof.
As there exists a distinct rational q e Q
for each distinct irrational p e P,
there exists an injection q: P -> Q.
phi(p_i) = q_i,
p_i neq p_j -> phi(p_i) ne phi(p_j),
a one-to-one function from P to a subset of Q.
Box
There exists a surjection rho: Q -> P.
Proof.
As there exists an injection phi: P -> Q
there exists a surjection rho:  Q -> P.
Box
1
^^^
That's sort of it, while, it's typeset in TeX,
fits on one page, uses these symbols.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+,
or some p_h = p_i".

[...]

Post by Ross Finlayson
I figure that if you enjoy "Dedekind cuts"
then you'll get a kick out of this.

[Ross Finlayson]

Post by Jim Burns
Now! With More Unicode!

Post by Ross Finlayson
A function surjects the rational numbers
onto the irrational numbers
Ross A. Finlayson
December 28, 2006
Abstract
There exists a surjection
from the rational numbers
onto the irrational numbers.
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q e Q for each p e P.
Proof.
Let Q_i+ be the set of
positive rationals less than p_i,
p_i e P+.
Q_h+ subset Q_i+, and Q_h+ neq Q_i+.
Let
Q_nlt i =
Q_i+ \ ( U p_h < p_i Q_h+ ) =
Int p_h< p_i (Q_i+ \ Q_h+)
It is so that Q_nlt i neq 0
else
Eq_h > p_h in a Q_h+,
or
some p_h = p_i.

I am guessing that you are depending on
∀ℚ⁺ₕ: ∃qᵢ: qᵢ ∈ ℚ⁺ᵢ\ℚ⁺ₕ
⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
☠ ∃qᵢ: ∀ℚ⁺ₕ: qᵢ ∈ ℚ⁺ᵢ\ℚ⁺ₕ
If it were reliable, it would follow that
☠ ℚᐠᑉᵢ = ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ ≠ ∅
However, that doesn't follow and isn't true.
dⱼ = ⌊1+3/(pᵢ-qⱼ)⌋
nⱼ = ⌊1+dⱼ⋅qⱼ⌋
qⱼ < nⱼ/dⱼ < (nⱼ+1)/dⱼ < pᵢ
pₕ = (nⱼ+⅟√​̅2)/dⱼ
pₕ ∈ ℙ⁺: qⱼ < pₕ < pᵢ
qⱼ ∈ ℚ⁺ₕ = ℚ⁺∩[0,pₕ)
qⱼ ∉ ℚ⁺ᵢ\ℚ⁺ₕ
qⱼ ∉ ⋂[pₕ<pᵢ] ℚ⁺ᵢ\ℚ⁺ₕ = ℚᐠᑉᵢ
∀pᵢ ∈ ℙ⁺: ∀qⱼ ∈ ℚ⁺ᵢ: qⱼ ∉ ℚᐠᑉᵢ
∀pᵢ ∈ ℙ⁺: ℚᐠᑉᵢ = ∅

Post by Ross Finlayson
Thus, Eq_i in Q_nlt i.
A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+.
A p_i e P+, E q_i in Q+, where
q_i is related to p_i and
not related to p_j neq p_i
for any p_j in P.
Similarly for Q- and P- via symmetry,
replace < with >, nlt with ngt, + with -,
"less than" with "greater than",
A p_i e P-, E q_i e Q-, where
q_i is related to p_i and
not related to p_j neq p_i,
for any p_j e P.
Box
There exists an injection phi: P -> Q.
Proof.
As there exists a distinct rational q e Q
for each distinct irrational p e P,
there exists an injection q: P -> Q.
phi(p_i) = q_i,
p_i neq p_j -> phi(p_i) ne phi(p_j),
a one-to-one function from P to a subset of Q.
Box
There exists a surjection rho: Q -> P.
Proof.
As there exists an injection phi: P -> Q
there exists a surjection rho: Q -> P.
Box
1
^^^
That's sort of it, while, it's typeset in TeX,
fits on one page, uses these symbols.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+,
or some p_h = p_i".

[...]

Post by Ross Finlayson
I figure that if you enjoy "Dedekind cuts"
then you'll get a kick out of this.

Thanks Jim, I'll look to this.

Yeah, you noticed I just established the criteria
either way and didn't start induction.

[Ross Finlayson]

Post by Ross Finlayson
A function surjects the rational numbers onto the irrational numbers
Ross A. Finlayson
December 28, 2006
Abstract
There exists a surjection from the rational numbers onto the irrational
numbers.
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q e Q for each p e P.
Proof. Let Q_i+ be the set of positive rationals
less than p_i, p_i e P+.
For each p_h < p_i, p_h e P+: Q_h+ subset Q_i+, and Q_h+ neq Q_i+.
Let
Q_nlt i = Q_i+ \ ( U p_h < p_i Q_h+ ) = Int p_h< p_i (Q_i+ \ Q_h+)
It is so that Q_nlt i neq 0
else Eq_h > p_h in a Q_h+, or some p_h = p_i.
Thus, Eq_i in Q_nlt i. A q_i e Q_nlt i -> q_i e Q_i+,
Q_nlt i subset Q_i+ subset Q+. As Q_nlt i neq 0 and
Q_nlt_i subset Q+: A p_i e P+, E q_i in Q+, where q_i
is related to p_i and not related to p_j neq p_i
for any p_j in P.
Similarly for Q- and P- via symmetry, replace < with >,
nlt with ngt, + with -, "less than" with "greater than",
and positive with negative: A p_i e P-, E q_i e Q-,
where q_i is related to p_i and not related to p_j neq p_i,
for any p_j e P.
Box
There exists an injection phi: P -> Q.
Proof. As there exists a distinct rational
q e Q for each distinct irrational p e P,
there exists an injection q: P -> Q.
E q_i in Q, A p_i, p_j in P: phi(p_i) = q_i, p_i neq p_j -> phi(p_i) ne
phi(p_j),
a one-to-one function from P to a subset of Q.
Box
There exists a surjection rho: Q -> P.
Proof. As there exists an injection phi: P -> Q
there exists a surjection rho: Q -> P.
Box
1
^^^
That's sort of it, while, it's typeset in TeX,
fits on one page, uses these symbols.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i".
A ∀
E ∃
e ∈
neq ≠
nlt ≮
ngt ≯
0 ∅
subset ⊂
U ⋃
Int ⋂
P ℙ
Q ℚ
=> ⇒
-> →
phi ϕ
rho ρ
Box □
Let Q be the set of rational numbers,
and P be the set of irrational numbers,
Q+ the set of positive rational numbers,
P+ the set of positive irrational numbers,
Q- the set of negative rational numbers,
P- the set of negative irrational numbers.
There is a distinct q ∈ Q for each p ∈ P.
Proof. Let Q_i+ be the set of positive
rationals less than p_i, p_i ∈ P+.
For each p_h < p_i, p_h ∈ P+: Q_h+ ⊂ Q_i+, and Q_h+ ≠ Q_i+.
Let
Q_{≮i} = Q_i+ \ ( U_{p_h < p_i} Q_h+ ) = ⋂_{p_h< p_i} (Q_i+ \ Q_h+)
It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i.
Thus, ∃q_i ∈ Q_{≮i}.
∀q_i ∈ Q_{≮i} ⇒ q_i ∈ Q_i+, Q_{≮i} ⊂ Q_i+ ⊂ Q+.
As Q_{≮i} ≠ 0 and Q_{≮i} ⊂ Q+: ∀p_i ∈ P+, ∃q_i ∈ Q+,
where q_i is related to p_i and not related to
p_j ≠ p_i for any p_j ∈ P.
Similarly for Q- and P- via symmetry,
replace < with >, ≮ with ≯, + with -,
"less than" with "greater than",
∀p_i ∈ P-, ∃q_i ∈ Q-, where q_i is related to p_i and not related to p_j
≠ p_i, for any p_j ∈ P.
□
There exists an injection ϕ: P → Q.
Proof. As there exists a distinct rational
q ∈ Q for each distinct irrational p ∈ P,
there exists an injection q: P → Q.
∃ q_i ∈ Q, ∀ p_i, p_j ∈ P: ϕ(p_i) = q_i, p_i ≠ p_j ⇒ ϕ(p_i) ≠ ϕ(p_j),
a one-to-one function from P to a subset of Q.
□
There exists a surjection ρ: Q → P.
Proof. As there exists an injection ϕ: P → Q
there exists a surjection ρ: Q → P.
□
I figure that if you enjoy "Dedekind cuts"
then you'll get a kick out of this.
There's a particular emphasis on
"It is so that Q_{≮i} ≠ ∅
else ∃q_h > p_h in a Q_h+, or some p_h = p_i".