avewright/proofwiki-math huggingface dataset:
Theorem: Euclidean Metric on Real Number Line is Metric

The Euclidean metric on the real number line $\R$ is a metric.

Proof:
Consider the real number line under the Euclidean metric:
:$M = \struct {\R, d}$
where $d$ is the distance function given by:
:$\map d {x, y} = \size {x - y}$
=== Proof of {{Metric-space-axiom|1|nolink}} ===
{{begin-eqn}}
{{eqn | l = \map d {x, x}
      | r = \size {x - x}
      | c = Definition of $d$
}}
{{eqn | r = \size 0
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{end-eqn}}
So {{Metric-space-axiom|1}} holds for $d$.
{{qed|lemma}}
=== Proof of {{Metric-space-axiom|2|nolink}} ===
{{begin-eqn}}
{{eqn | l = \map d {x, y} + \map d {y, z}
      | r = \size {x - y} + \size {y - z}
      | c = Definition of $d$
}}
{{eqn | o = \ge
      | r = \size {\paren {x - y} + \paren {y - z} }
      | c = Triangle Inequality for Real Numbers
}}
{{eqn | r = \size {x - z}
      | c = 
}}
{{eqn | r = \map d {x, z}
      | c = Definition of $d$
}}
{{end-eqn}}
So {{Metric-space-axiom|2}} holds for $d$.
{{qed|lemma}}
=== Proof of {{Metric-space-axiom|3|nolink}} ===
{{begin-eqn}}
{{eqn | l = \map d {x, y}
      | r = \size {x - y}
      | c = Definition of $d$
}}
{{eqn | r = \size {y - x}
      | c = 
}}
{{eqn | r = \map d {y, x}
      | c = Definition of $d$
}}
{{end-eqn}}
So {{Metric-space-axiom|3}} holds for $d$.
{{qed|lemma}}
=== Proof of {{Metric-space-axiom|4|nolink}} ===
{{begin-eqn}}
{{eqn | l = x
      | o = \ne
      | r = y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x - y
      | o = \ne
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {x - y}
      | o = >
      | r = 0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map d {x, y}
      | o = >
      | r = 0
      | c = Definition of $d$
}}
{{end-eqn}}
So {{Metric-space-axiom|4}} holds for $d$.
{{qed}}
Theorem: Unit n-Cube under Chebyshev Distance is Subspace of Real Vector Space

Let $n \in \N$.
Let $I^n$ denote the unit $n$-cube:
:$I^n = \closedint 0 1^n$
that is, the Cartesian product of $n$ instances of the closed real interval $\set {x \in \R: 0 \le x \le 1}$.
Let $d_c: I^n \times I^n \to \R$ be defined as:
:$\ds \map {d_c} {x, y} = \max_{i \mathop = 1}^n \set {\size {x_i - y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {x_1, x_2, \ldots, x_n} \in I^n$.
Then $\struct {I_n, d_c}$ is a metric subspace of $\struct {\R^n, d_\infty}$, where $d_\infty$ is the Chebyshev distance on the real vector space $\R_n$.

Proof:
{{ProofWanted|Straightforward but a bit tedious.}}
Theorem: Unit n-Sphere under Euclidean Metric is Metric Subspace of Euclidean Real Vector Space

Let $\Bbb S^n$ be the unit $n$-sphere.
Let $d_S: \Bbb S^n \times \Bbb S^n \to \R$ be the real-valued function defined as:
:$\ds \forall x, y \in \Bbb S^n: \map {d_S} {x, y} = \sqrt {\sum_{i \mathop = 1}^{n + 1} \paren {x_i - y_i}^2}$
where $x = \tuple {x_1, x_2, \ldots, x_{n + 1} }, y = \tuple {y_1, y_2, \ldots, y_{n + 1} }$.
Then $\struct {\Bbb S^n, d_S}$ is a metric subspace of $\struct {\R^{n + 1}, d}$, where $d$ is the Euclidean metric on the real vector space $\R^{n + 1}$.

Proof:
The metric given is the Euclidean metric restricted to the subset $\Bbb S^n$ of the real vector space $\R^{n + 1}$.
The result follows from Subspace of Metric Space is Metric Space.
{{qed}}
Theorem: Vector Subspace of Real Vector Space under Chebyshev Metric is Metric Subspace

Let $n \in \N$.
Let $A$ be the set of all ordered $n+1$-tuples $\tuple {x_1, x_2, \ldots, x_{n + 1} }$ of real numbers such that $x_{n + 1} = 0$.
Let $d: A \times A \to \R$ be the function defined as:
:$\ds \forall x, y \in A: \map d {x, y} = \max_{i \mathop = 1}^n \set {\size {x_i - y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_{n + 1} }, y = \tuple {y_1, y_2, \ldots, y_{n + 1} }$.
Then $\struct {A, d}$ is a metric subspace of $\struct {\R^{n + 1}, d_\infty}$ where $d_\infty$ is the Chebyshev distance on the real vector space $\R^{n + 1}$.

Proof:
The metric given is the Chebyshev distance restricted to the subset $A$ of the real vector space $\R^{n + 1}$.
The result follows from Subspace of Metric Space is Metric Space.
{{qed}}
Theorem: Inclusion Mapping on Metric Space is Continuous

Let $M = \struct {A, d}$ be a metric space.
Let $\struct {H, d_H}$ be a metric subspace of $M$.
Then the inclusion mapping $i_H: H \to A$ is continuous from $\struct {H, d_H}$ to $\struct {A, d}$.

Proof:
Let $a \in H$.
Let $\epsilon \in \R_{>0}$.
Let $\delta = \epsilon$.
Then:
{{begin-eqn}}
{{eqn | l = \map {d_H} {a, y}
      | o = <
      | r = \delta
      | c = for some $y \in H$
}}
{{eqn | ll= \leadsto
      | l = \map d {\map {i_H} a, \map {i_H} y}
      | r = \map d {a, y}
      | c = {{Defof|Inclusion Mapping}}
}}
{{eqn | r = \map {d_H} {a, y}
      | c = Definition of $d_H$
}}
{{eqn | o = <
      | r = \delta
      | c = 
}}
{{eqn | r = \epsilon
      | c = {{hypothesis}}
}}
{{end-eqn}}
So by definition $i_H$ is continuous at $a$ {{WRT}} $d_H$ and $d$.
As $a \in H$ is arbitrary, it follows that $i_H$ is continuous from $\struct {H, d_H}$ to $\struct {A, d}$.
{{qed}}
Theorem: Bertrand-Chebyshev Theorem

For all $n \in \N_{>0}$, there exists a prime number $p$ with $n < p \le 2 n$.

Proof:
We will first prove the theorem for the case $n \le 2047$. 
Consider the following sequence of prime numbers:
:$2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503$
Each of these prime number is smaller than twice the previous one.
Hence every interval $\set {x: n < x \le 2 n}$, with $n \le 2047$, contains one of these prime numbers.
=== {{Lemma|Bertrand-Chebyshev Theorem|1}} ===
{{:Bertrand-Chebyshev Theorem/Lemma 1}}{{qed|lemma}}
=== {{Lemma|Bertrand-Chebyshev Theorem|2}} ===
{{:Bertrand-Chebyshev Theorem/Lemma 2}}{{qed|lemma}}
=== {{Lemma|Bertrand-Chebyshev Theorem|3}} ===
{{:Bertrand-Chebyshev Theorem/Lemma 3}}{{qed|lemma}}
From {{Lemma|Bertrand-Chebyshev Theorem|3}}:
:if $p > \sqrt {2 n}$, then $p$ appears at most once in $\dbinom {2 n} n$.
For $n \ge 3$, there is no prime factor $p$ with $\dfrac {2 n} 3 < p \le n$, because such a prime number divides $n!$ exactly once and $\paren {2 n}!$ exactly twice.
Therefore, by {{Lemma|Bertrand-Chebyshev Theorem|1}}:
:$\ds \frac {2^{2 n} } {2 n + 1} \le \dbinom {2 n} n \le \prod_{p \mathop \le \sqrt {2 n} } 2 n \prod_{\sqrt {2 n} \mathop < p \mathop \le \frac {2 n} 3} p \prod_{n \mathop < p \mathop \le 2 n} p$
for $n \ge 3$.
{{AimForCont}} there is no prime number $p$ with $n < p \le 2 n$. 
Then we have:
{{begin-eqn}}
{{eqn | l = \frac {2^{2 n} } {2 n + 1} 
      | r = \prod_{p \mathop \le \sqrt {2 n} } 2 n \prod_{\sqrt {2 n} \mathop < p \mathop \le \frac {2 n} 3} p 
      | o = \le
      | c = 
}}
{{eqn | r = \paren {2 n}^{\sqrt {2 n} } \prod_{p \mathop \le \frac {2 n} 3} p
      | o = \le
      | c = 
}}
{{eqn | r = \paren {2 n}^{\sqrt {2 n} } 2^{\frac {4 n} 3}
      | o = \le
      | c = {{Lemma|Bertrand-Chebyshev Theorem|2}}
}}
{{end-eqn}}
This is a contradiction for sufficiently large $n$.
Indeed, we have:
:$2^{2 n / 3} \le \paren {2 n + 1} \paren {2 n}^{\sqrt {2 n} }$
Now:
:$2 n + 1 \le \paren {2 n}^2 \le \paren {2 n}^{\sqrt {2 n} / 3}$
for $n \ge 18$.
So:
:$2^{2 n} \le \paren {2 n}^{4 \sqrt {2 n} }$
Put $r = \sqrt {2 n}$.
Then:
:$2^{r^2} \le r^{8 r}$
or equivalently:
:$2^r \le r^8$
This fails when $r = 2^6 = 64$.
It fails thereafter, since $2^r$ increases faster than $r^8$.
So our proof works if:
:$n \ge 2^{11} = 2048$
and the examples show it is true for smaller $n$.
{{qed}}
Theorem: Bertrand-Chebyshev Theorem

For all $n \in \N_{>0}$, there exists a prime number $p$ with $n < p \le 2 n$.

Proof:
We will first prove the theorem for the case $n \le 426$. 
Consider the following sequence of prime numbers:
:$2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631$
Each of these prime number is smaller than twice the previous one.
Hence every interval $\set {x: n < x \le 2 n}$, with $n \le 426$, contains one of these prime numbers.
=== {{Lemma|Bertrand-Chebyshev Theorem|2}} ===
{{:Bertrand-Chebyshev Theorem/Lemma 2}}{{qed|lemma}}
=== {{Lemma|Bertrand-Chebyshev Theorem|3}} ===
{{:Bertrand-Chebyshev Theorem/Lemma 3}}{{qed|lemma}}
=== {{Lemma|Bertrand-Chebyshev Theorem|4}} ===
{{:Bertrand-Chebyshev Theorem/Lemma 4}}{{qed|lemma}}
From {{Lemma|Bertrand-Chebyshev Theorem|3}}:
:if $p > \sqrt {2 n}$, then $p$ appears at most once in $\dbinom {2 n} n$.
For $n \ge 3$, there is no prime factor $p$ with $\dfrac 2 3 n < p \le n$, because such a prime number divides $n!$ exactly once and $\paren {2 n}!$ exactly twice.
For $n \ge 5$:
:$ \sqrt {2 n} < \dfrac {2 n} 3$
Therefore, by {{Lemma|Bertrand-Chebyshev Theorem|4}}:
:$\ds \frac {2^{2 n} } {2 n} \le \dbinom {2 n} n \le \paren{\prod_{p \mathop \le \sqrt {2 n} } 2 n} \paren{\prod_{\sqrt {2 n} \mathop < p \mathop \le \frac 2 3 n} p} \paren{\prod_{n \mathop < p \mathop \le 2 n} p}$
for $n \ge 3$.
Let $\map \pi n$ denote the prime-counting function.
Then:
:$\forall n \ge 1: \map \pi n \le n - 1$
as $1$ is neither prime nor composite.
{{AimForCont}} there is no prime number $p$ such that $n < p \le 2 n$.
Then we have:
{{begin-eqn}}
{{eqn | l = \frac {2^{2 n} } {2 n} 
      | r = \paren {\prod_{p \mathop \le \sqrt {2 n} } 2 n} \paren {\prod_{\sqrt {2 n} \mathop < p \mathop \le \frac {2 n} 3} p}
      | o = \le
      | c = 
}}
{{eqn | r = \paren {\prod_{p \mathop \le \sqrt {2 n} } 2 n} \paren {\prod_{p \mathop \le \frac {2 n} 3} p}
      | o = \le
      | c = 
}}
{{eqn | r = \paren {2 n}^{\sqrt {2 n} - 1} 2^{4 n / 3}
      | o = \le
      | c = {{Lemma|Bertrand-Chebyshev Theorem|2}}
}}
{{end-eqn}}
This is a contradiction for sufficiently large $n$.
Indeed, we have:
:$2^{2 n / 3} \le \paren {2 n}^{\sqrt {2 n} }$
So:
:$2^{2 n} \le \paren {2 n}^{3 \sqrt {2 n} }$
Let $r = \sqrt {2 n}$.
Then:
:$2^{r^2} \le r^{6 r}$
or equivalently:
:$2^r \le r^6$
This fails when $r = 30$.
It fails thereafter, since $2^r$ increases faster than $r^6$.
So our proof works if:
:$n \ge 427$
and the examples show it is true for smaller $n$.
{{qed}}
Theorem: Sequence of Binomial Coefficients is Strictly Increasing to Half Upper Index

Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $\dbinom n k$ be the binomial coefficient of $n$ over $k$ for a positive integer $k \in \Z_{\ge 0}$.
Let $S_n = \sequence {x_k}$ be the sequence defined as:
:$x_k = \dbinom n k$
Then $S_n$ is strictly increasing exactly where $0 \le k < \dfrac n 2$.

Proof:
When $k \ge 0$, we have:
{{begin-eqn}}
{{eqn | l = \binom n {k + 1}
      | r = \frac {n!} {\paren {k + 1}! \paren {n - k - 1}!}
      | c = {{Defof|Binomial Coefficient}}
}}
{{eqn | r = \frac {n - k} {n - k} \frac {n!} {\paren {k + 1}! \paren {n - k - 1}!}
      | c = 
}}
{{eqn | r = \frac {n - k} {\paren {k + 1} \paren {n - k} } \frac {n!} {k! \paren {n - k - 1}!}
      | c = extracting $k + 1$ from its factorial
}}
{{eqn | r = \frac {n - k} {k + 1} \frac {n!} {k! \paren {n - k}!}
      | c = inserting $n - k$ into its factorial
}}
{{eqn | r = \frac {n - k} {k + 1} \binom n k
      | c = {{Defof|Binomial Coefficient}}
}}
{{end-eqn}}
In order for $S_n$ to be strictly increasing, it is necessary for $\dfrac {n - k} {k + 1} > 1$.
So:
{{begin-eqn}}
{{eqn | l = \dfrac {n - k} {k + 1}
      | o = >
      | r = 1
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = n - k
      | o = >
      | r = k + 1
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = n
      | o = >
      | r = 2 k + 1
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = n
      | o = >
      | r = 2 \paren {k + 1} - 1
      | c = 
}}
{{end-eqn}}
Thus $\dbinom n {k + 1} > \dbinom n k$ {{iff}} $k + 1$ is less than half of $n$.
Hence the result.
{{Qed}}
Category:Binomial Coefficients
8q69d7aejhqoyayb0jhrvielrzv0gx1
Theorem: Isometry between Metric Spaces is Continuous

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be an isometry.
Then $\phi: M_1 \to M_2$ is a continuous mapping.

Proof:
Let $a \in A_1$.
Let $\epsilon \in \R_{>0}$.
Let $\delta = \epsilon$.
Then:
{{begin-eqn}}
{{eqn | l = \map {d_1} {a, y}
      | o = <
      | r = \delta
      | c = for some $y \in A_1$
}}
{{eqn | ll= \leadsto
      | l = \map {d_2} {\map \phi a, \map \phi y}
      | o = <
      | r = \delta
      | c = as $\map {d_2} {\map \phi a, \map \phi y} = \map {d_1} {a, y}$
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
So by definition $\phi$ is continuous at $a$.
As $a \in H$ is arbitrary, it follows that $d_H$ is continuous on $H$.
{{qed}}
Theorem: Equivalence of Definitions of Topologically Equivalent Metrics

{{TFAE|def = Topologically Equivalent Metrics}}
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.

Proof:
=== Definition 1 implies Definition 2 ===
Let $d_1$ and $d_2$ be topologically equivalent by definition 1.
Let $\struct {B, d}$ and $\struct {C, d'}$ be metric spaces:
Let $f: B \to A$ and $g: A \to C$ be mappings.
Then by definition:
:$f$ is $\tuple {d, d_1}$-continuous {{iff}} $f$ is $\tuple {d, d_2}$-continuous
:$g$ is $\tuple {d_1, d'}$-continuous {{iff}} $g$ is $\tuple {d_2, d'}$-continuous.
Suppose $U \subseteq A$ is $d_2$-open.
Let $h: A \to A$ be the identity mapping.
That is:
:$\forall a \in A: \map h a = a$
From Identity Mapping is Continuous, $h$ is $\tuple {d_2, d_2}$-continuous.
Hence because $h$ is $\tuple {d_1, d'}$-continuous {{iff}} $h$ is $\tuple {d_2, d'}$-continuous, $h$ is $\tuple {d_1, d_2}$-continuous.
From Metric Space Continuity by Open Set, it follows that $h^{-1} \sqbrk U$ is $d_1$-open.
But $h^{-1} \sqbrk U = U$, so $U$ is $d_1$-open.
A similar argument, still starting with the same proposition that $g$ is $\tuple {d_1, d'}$-continuous {{iff}} $g$ is $\tuple {d_2, d'}$-continuous, shows that if $U$ is $d_1$-open then it is $d_2$-open.
Thus $d_1$ and $d_2$ are topologically equivalent by definition 2.
{{qed|lemma}}
Theorem: Isometric Metric Spaces are Homeomorphic

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ and $M_2$ be isometric.
Then $M_1$ and $M_2$ are homeomorphic.

Proof:
By the definition of an isometry, there exists a bijection $f: A_1 \to A_2$ such that:
:$\forall a, b \in A_1: \map {d_1} {a, b} = \map {d_2} {\map \phi a, \map \phi b}$
From Isometry of Metric Spaces is Homeomorphism, $f$ is a homeomorphism from $M_1$ to $M_2$.
The result follows by definition of homeomorphic metric spaces.
{{qed}}
Theorem: Homeomorphic Metric Spaces are not necessarily Isometric

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ and $M_2$ be homeomorphic.
Then it is not necessarily the case that $M_1$ and $M_2$ are isometric.

Proof:
Consider the spaces $\struct {\openint 0 4, d}$ and $\struct {\openint 0 1, d}$, where $d$ is the Euclidean Metric.
By Open Real Intervals are Homeomorphic, they are homeomorphic.
Let $\phi: \openint 0 4 \to \openint 0 1$ be a mapping.
Then:
{{begin-eqn}}
{{eqn | l = \map d {\map \phi 1, \map \phi 3}
      | o = \le
      | r = \map d {0, 1}
}}
{{eqn | r = 1
}}
{{eqn | o = < 
      | r = 2
}}
{{eqn | r = \map d {1, 3}
}}
{{end-eqn}}
showing that $\phi$ cannot be an isometry.
Therefore the two spaces above are not isometric.
{{qed}}
Theorem: Identity Mapping between Metrics separated by Scale Factor is Continuous

Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $d_1$ and $d_2$ be such that:
:$\forall x, y \in A: \map {d_2} {x, y} \le K \map {d_2} {x, y}$
Let $I_A: A \to A$ be the identity mapping on $A$.
Then $I_A$ is continuous from $M_1$ to $M_2$.

Proof:
Let $\epsilon \in \R_{>0}$.
Let $a \in A$.
Set $\delta = \dfrac \epsilon K$.
Then:
{{begin-eqn}}
{{eqn | l = \map {d_1} {x, a}
      | o = <
      | r = \delta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map {d_2} {\map {I_A} x, \map {I_A} a}
      | o = \le
      | r = K \map {d_1} {x, a}
      | c = 
}}
{{eqn | o = <
      | r = K \delta
      | c = 
}}
{{eqn | r = \epsilon
      | c = 
}}
{{end-eqn}}
Hence by definition $I_A$ is continuous at $a$.
As $a$ is arbitrary, it follows that this is true for all $a \in A$.
Thus $I_A$ is continuous on the whole of $M_1$.
{{qed}}
Theorem: Metric Spaces on Topologically Equivalent Metrics on same Underlying Set are Homeomorphic

Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $d_1$ and $d_2$ be topologically equivalent.
Then $M_1$ and $M_2$ are homeomorphic.

Proof:
By definition, $d_1$ and $d_2$ are topologically equivalent on $A$ {{iff}}:
:For all metric spaces $\struct {B, d}$ and $\struct {C, d'}$:
:For all mappings $f: B \to A$ and $g: A \to C$:
::$(1): \quad f$ is $\tuple {d, d_1}$-continuous {{iff}} $f$ is $\tuple {d, d_2}$-continuous
::$(2): \quad g$ is $\tuple {d_1, d'}$-continuous {{iff}} $g$ is $\tuple {d_2, d'}$-continuous.
Set $B = C = A$.
The result follows by definition of homeomorphic metric spaces.
{{qed}}
Category:Topologically Equivalent Metrics
Category:Homeomorphisms (Metric Spaces)
jzhf142p0pw82no38pkivu1ycpsxmt3
Theorem: Real Vector Space under Chebyshev Distance is Homeomorphic to that under Euclidean Metric

Let $\R^n$ be an $n$-dimensional real vector space.
Let $d_\infty: \R^n \times \R^n \to \R$ be the Chebyshev distance on $\R^n$.
Let $d_2: \R^n \times \R^n \to \R$ be the Euclidean metric on $\R^n$.
Let $M_1 = \struct {\R^n, d_\infty}$ and $M_2 = \struct {\R^n, d_2}$ be the corresponding metric spaces.
Then $M_1$ and $M_2$ are homeomorphic.

Proof:
From Relation between $p$-Product Metric and Chebyshev Distance on Real Vector Space:
:$\forall x, y \in \R^n: \map {d_\infty} {x, y} \le \map {d_p} {x, y} \le n^{1 / p} \map {d_\infty} {x, y}$
The Euclidean metric $d_2$ is a special case of the $p$-product metric $d_p$ for $p = 2$.
It follows by definition that $d_\infty$ and $d_2$ are Lipschitz equivalent.
It follows from Lipschitz Equivalent Metrics are Topologically Equivalent that $d_\infty$ and $d_2$ are topologically equivalent.
The result follows from Metric Spaces on Topologically Equivalent Metrics on same Underlying Set are Homeomorphic.
{{qed}}
Theorem: Equivalence of Definitions of Homeomorphism between Metric Spaces

{{TFAE|def = Homeomorphism (Metric Spaces)}}
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: A_1 \to A_2$ be a bijection.

Proof:
In order to prove the assertion it is sufficient to prove that the conditions for homeomorphism in definitions $2$ to $4$ are necessary and sufficient conditions for $f$ and $f^{-1}$ to be continuous on $M_1$ and $M_2$ respectively.
Theorem: Liouville Numbers are Irrational

Liouville numbers are irrational.

Proof:
Let $x$ be a Liouville number. 
{{AimForCont}} $x$ were rational, that is:
: $x = \dfrac a b$
with $a, b \in \Z$ and $b > 0$.
By definition of a Liouville number, for all $n \in \N$, there exist $p,q \in \Z$ (which may depend on $n$) with $q > 1$ such that:
:$0 < \size {x - \dfrac p q} < \dfrac 1 {q^n}$
Let $n$ be a positive integer such that $2^{n - 1} > b$.
Let $p$ and $q$ be any integers with $q > 1$.
We have:
:$\size {x - \dfrac p q} = \dfrac {\size {a q - b p} } {b q}$
If $\size {a q - b p} = 0$, this would violate the first inequality.
If $\size {a q - b p} \ne 0$, then:
{{begin-eqn}}
{{eqn |l = \size {x - \frac p q}
      |o = \ge
      |r = \frac 1 {b q}
      |c = as $\size {a q - b p}$ is a positive integer
}}
{{eqn |o = >
      |r = \frac 1 {2^{n - 1} q}
      |c = by our choice of $n$
}}
{{eqn |o = \ge
      |r = \frac 1 {q^n}
      |c = as $q > 1$ by definition
}}
{{end-eqn}}
which would violate the second inequality.
Therefore, in any case, if $n$ is sufficiently large, there cannot exist integers $p$ and $q$ with $q > 1$ satisfying the two inequalities.
This is a contradiction.
Thus $x$ is irrational.
{{qed}}
Category:Number Theory
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Theorem: Cardinality of Power Set of Natural Numbers Equals Cardinality of Real Numbers

The cardinality of the power set of the natural numbers is equal to the cardinality of the real numbers.
:$\card {2^\N} = \card \R$

Proof:
{{refactor|level = basic|What is being stated is covered in passing by the below}}
{{mergeto|Power Set of Natural Numbers has Cardinality of Continuum}}
This is a direct corollary of:
:Power Set of Natural Numbers has Cardinality of Continuum
:Cardinality of Real Numbers is Cardinality of Continuum.
{{qed}}
Theorem: Liouville's Constant is Transcendental

Liouville's constant:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 1}^\infty \dfrac 1 {10^{n!} }
      | r = \frac 1 {10^1} + \frac 1 {10^2} + \frac 1 {10^6} +  \frac 1 {10^{24} } + \cdots
      | c = 
}}
{{eqn | r = 0.11000 \, 10000 \, 00000 \, 00000 \, 00010 \, 00 \ldots
      | c = 
}}
{{end-eqn}}
is transcendental.

Proof:
Let $q = 10^{n!}$ and write:
:$\ds L = \frac p q + \sum_{k \mathop = n + 1}^\infty \frac 1 {10^{k!} }$
for some suitable $p \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = \size {L - \frac p q}
      | r = \sum_{k \mathop = n + 1}^\infty \frac 1 {10^{k!} }
      | c =  
}}
{{eqn | r = \frac 2 {10^{\paren {n + 1}!} }
      | o = \le
      | c = 
}}
{{eqn | r = \frac 2 {q^{n + 1} }
      | c = 
}}
{{eqn | r = \frac 1 {q^n}
      | o = <
      | c = as $q \ge 10$ for all $n \in \N$
}}
{{end-eqn}}
Thus, by definition, $L$ is a Liouville number.
Therefore, by Liouville's Theorem, $L$ is transcendental.
{{qed}}
Theorem: Isometry of Metric Spaces is Equivalence Relation

Let $M_1$ and $M_2$ be metric spaces.
Let $M_1 \sim M_2$ denote that $M_1$ and $M_2$ are isometric.
The relation $\sim$ is an equivalence relation.

Proof:
Checking in turn each of the criteria for equivalence:
Theorem: Homeomorphism of Metric Spaces is Equivalence Relation

Let $M_1$ and $M_2$ be metric spaces.
Let $M_1 \sim M_2$ denote that $M_1$ and $M_2$ are homeomorphic.
The relation $\sim$ is an equivalence relation.

Proof:
Checking in turn each of the criteria for equivalence:
Theorem: Isometry of Metric Spaces is Homeomorphism

Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $f: M_1 \to M_2$ be an isometry.
Then $f$ is a homeomorphism from $M_1$ to $M_2$.

Proof:
By the definition of an isometry, $f$ is a bijection $f: A_1 \to A_2$ such that:
:$\forall a, b \in A_1: \map {d_1} {a, b} = \map {d_2} {\map \phi a, \map \phi b}$
By Isometry between Metric Spaces is Continuous, $f$ is a continuous mapping from $M_1$ to $M_2$.
By {{Corollary|Isometry between Metric Spaces is Continuous}}, $f^{-1}$ is a continuous mapping from $M_2$ to $M_1$.
Thus $f$ is a homeomorphism from $M_1$ to $M_2$.
Hence the result.
{{qed}}
Category:Isometries (Metric Spaces)
Category:Homeomorphisms (Metric Spaces)
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Theorem: Equivalence Class of Isometries is Subset of Equivalence Class of Homeomorphisms

Let $\CC_I$ be an equivalence class of isometries on the set of metric spaces.
Then $\CC_I$ is a subset of an equivalence class of homeomorphisms on the set of metric spaces.

Proof:
From Isometry of Metric Spaces is Equivalence Relation, all isometries can be partitioned into equivalence classes.
From Homeomorphism of Metric Spaces is Equivalence Relation, all homeomorphisms can be partitioned into equivalence classes.
Let $\CC_I$ be an equivalence class of isometries.
Let $f: M_1 \to M_2$ be an element of $\CC_I$ where $M_1$ and $M_2$ are some metric spaces.
From Isometry of Metric Spaces is Homeomorphism, $\CC_I$ is a homeomorphism.
Thus $\CC_I$ is an element of some equivalence class of homeomorphisms on the set of metric spaces.
{{qed}}