avewright/proofwiki-math huggingface dataset:
Theorem: Normed Vector Space of Bounded Sequences is not Separable

Let $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ be the normed vector space of bounded sequences.
$\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ is not separable.

Proof:
{{AimForCont}} $\struct {\ell^\infty, \norm {\, \cdot \,}_\infty}$ is separable.
Let $\mathbf x := \sequence {x_i}_{i \mathop \in \N}$, $\mathbf a := \sequence {a_i}_{i \mathop \in \N}$, $\mathbf b := \sequence {b_i}_{i \mathop \in \N}$ be sequences.
Let $D := \set {\mathbf x_i \in \R^\N : i \in \N}$ be a dense countable subset of $\ell^\infty$.
Let $A := \set {\sequence {y_i}_{i \mathop \in \N} : y_i = 0 \text{ or } y_i = 1}$ be the set of all sequences with terms equal to either $0$ or $1$.
Suppose $\mathbf a, \mathbf b \in A$ are two distinct elements of $A$.
Then:
:$\exists n \in \N : a_n \ne b_n$
and:
{{begin-eqn}}
{{eqn | l = \norm {\mathbf a - \mathbf b}_\infty
      | r = \sup_{n \mathop \in \N} \size {a_n - b_n}
      | c = {{Defof|Supremum Norm}}
}}
{{eqn | r = 1
}}
{{end-eqn}}
Hence, their mutual distance is $1$.
Since $D$ is dense in $\ell^\infty$, we have that:
:$\ds \forall \mathbf a \in A : \exists \mathbf x_{\map i {\mathbf a} } \in \map {B_{\frac 1 3} } {\mathbf a}$
For all $\mathbf a \in A$, the open balls $\map {B_{\frac 1 3} } {\mathbf a}$ are mutually disjoint because $\ds \frac 1 3 + \frac 1 3 < 1$.
Hence:
:$\forall {\mathbf a} \in A : \forall n \in \N : \mathbf a \mapsto \map n {\mathbf a}$
is an injection.
By definition, $A$ is countable.
However, $A$ is in a one-to-one correspondence with all real numbers via binary expansion.
Therefore, $A$ is uncountable.
Hence the contradiction.
{{qed}}
Theorem: Metric formed by Arc Length on Circle is Lipschitz Equivalent to Euclidean Metric

Let $A \subseteq \R^2$ be the set defined as:
:$A = \set {\tuple {x_1, x_2}: x_1^2 + y_2^2 = 1}$
Thus from Equation of Unit Circle, $A$ is the unit circle embedded in the Cartesian plane.
Let $d: A^2 \to \R$ be the metric defined as:
:$\forall \tuple {x, y} \in A^2: \map d {x, y} = \begin {cases} 0 & : x = y \\ \pi & : x = -y \\ l & : \text {otherwise} \end {cases}$
where:
:$x$ and $y$ are of the form $\tuple {x_1, x_2}$
:$l$ denotes the length of the minor arc of $A$ between $x$ and $y$.
Then $d$ is Lipschitz equivalent to the Euclidean metric $d_2$ on $A$.

Proof:
That $d$ forms a metric is demonstrated in Arc Length on Circle forms Metric.
Let $p_1 = \tuple {x_1, y_1}$ and $p_2 = \tuple {x_2, y_2}$ be points in $A$.
We have that $\map {d_2} {p_1, p_2}$ is the length of a line segment between $p_1$ and $p_2$.
This can never be longer than the length of the arc between $p_1$ and $p_2$.
Hence:
:$\map {d_2} {p_1, p_2} \le \map d {p_1, p_2}$
Then we have that $\map d {p_1, p_2}$ is at most $\pi$ times the diameter of $A$
Hence:
:$\map d {p_1, p_2} \le \dfrac \pi 2 \map {d_2} {p_1, p_2}$
The result follows by definition of Lipschitz equivalence.
{{qed}}
Theorem: Standard Bounded Metric is Metric/Topological Equivalence

$\bar d$ is topologically equivalent to $d$.

Proof:
That $\bar d$ forms a metric on $M$ is demonstrated in Standard Bounded Metric is Metric.
We have that:
:$\forall x, y \in A^2: \map {\bar d} {x, y} \le \map d {x, y}$
Hence:
:$\map {B_\epsilon} {x; d} \subseteq \map {B_\epsilon} {x; \bar d}$
where $\map {B_\epsilon} {x; d}$ denotes the open $\epsilon$-ball of $x$ in $\struct {A, d}$.
Hence:
:if $U$ is $\bar d$-open, the $U$ is $d$-open
where $U$ is a subset of $A$.
Let $U$ be $d$-open.
Let $x \in U$.
Then $\map {B_\epsilon} {x; d} \subseteq U$ for some $\epsilon \in \R_{>0}$.
Let us take $\epsilon < 1$.
Then:
:$\map {B_\epsilon} {x; \bar d} = \map {B_\epsilon} {x; d} \subseteq U$
demonstrating that $U$ is $\bar d$-open.
The result follows.
{{qed}}
Theorem: Metric over 1 plus Metric forms Metric/Topological Equivalence

$d_3$ is topologically equivalent to $d$.

Proof:
That $d_3$ forms a metric on $M$ is demonstrated in Metric over 1 plus Metric forms Metric.
We have that:
:$\forall x, y \in A^2: \map {d_3} {x, y} \le \map d {x, y}$
Hence:
:$\map {B_\epsilon} {x; d} \subseteq \map {B_\epsilon} {x; d_3}$
where $\map {B_\epsilon} {x; d}$ denotes the open $\epsilon$-ball of $x$ in $\struct {A, d}$.
Hence:
:if $U \subseteq A$ is $d_3$-open, the $U$ is $d$-open
where $U$ is a subset of $A$.
Let $U \subseteq A$ be $d$-open.
Let $x \in U$.
Then $\map {B_\epsilon} {x; d} \subseteq U$ for some $\epsilon \in \r_{>0}$.
Let $\delta = \min \set {\dfrac \epsilon 2, \dfrac 1 2}$.
Then:
:$\map {d_3} {x, y} < \delta \implies \map {d_3} {x, y} < \dfrac 1 2$
and so:
{{begin-eqn}}
{{eqn | l = \map d {x, y}
      | r = \dfrac {\map {d_3} {x, y} } {1 - \map {d_3} {x, y} }
      | c = 
}}
{{eqn | o = \le
      | r = \map {d_3} {x, y}
      | c = 
}}
{{eqn | o = <
      | r = 2 \map {d_3} {x, y}
      | c = 
}}
{{eqn | o = <
      | r = \epsilon
      | c = 
}}
{{end-eqn}}
Hence:
:$\map {B_\epsilon} {x; d_2} = \map {B_\epsilon} {x; d} \subseteq U$
demonstrating that $U$ is $d_2$-open.
Note that $\delta = \dfrac \epsilon {1 + \epsilon}$ is just as good:
:$\map f a = \dfrac a {1 + a}$ is strictly increasing for $a \ge 0$
and so:
{{begin-eqn}}
{{eqn | l = \map {d_3} {x, y}
      | o = <
      | r = \dfrac \epsilon {1 + \epsilon}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \map d {x, y}
      | o = <
      | r = \epsilon
      | c = 
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Euclidean Metric on Real Number Space is Translation Invariant

Let $\tau_{\mathbf t}: \R^n \to \R^n$ denote the translation of the real Euclidean space of $n$ dimensions by the vector $\mathbf t = \tuple {t_1, t_2, \ldots, t_n}$.
Let $d_2$ denote the Euclidean metric on $\R^n$.
Then $d_2$ is unchanged by application of $\tau$:
:$\forall \mathbf x, \mathbf y \in \R^n: \map {d_2} {\map \tau {\mathbf x}, \map \tau {\mathbf y} } = \map {d_2} {\mathbf x, \mathbf y}$

Proof:
Let $\mathbf x = \tuple {x_1, x_2, \ldots, x_n}$ and $\mathbf y = \tuple {y_1, y_2, \ldots, y_n}$ be arbitrary points in $\R^n$.
Then:
{{begin-eqn}}
{{eqn | l = \map {d_2} {\map \tau {\mathbf x}, \map \tau {\mathbf y} }
      | r = \map {d_2} {\mathbf x - \mathbf t, \mathbf y - \mathbf t}
      | c = {{Defof|Translation in Euclidean Space}}
}}
{{eqn | r = \sqrt {\paren {\sum_{i \mathop = 1}^n \paren {\paren {x_i - t_i} - \paren {y_i - t_i} }^2} }
      | c = Definition of $\mathbf t$, {{Defof|Euclidean Metric on Real Vector Space}}
}}
{{eqn | r = \sqrt {\paren {\sum_{i \mathop = 1}^n \paren {x_i - y_i}^2} }
      | c = simplification
}}
{{eqn | r = \map {d_2} {\mathbf x, \mathbf y}
      | c = {{Defof|Euclidean Metric on Real Vector Space}}
}}
{{end-eqn}}
{{qed}}
Theorem: Euclidean Metric on Real Number Plane is Rotation Invariant

Let $r_\alpha: \R^2 \to \R^2$ denote the rotation of the Euclidean plane about the origin through an angle of $\alpha$.
Let $d_2$ denote the Euclidean metric on $\R^2$.
Then $d_2$ is unchanged by application of $r_\alpha$:
:$\forall x, y \in \R^2: \map {d_2} {\map {r_\alpha} x, \map {r_\alpha} y} = \map {d_2} {x, y}$

Proof:
Let $x = \tuple {x_1, x_2}$ and $y = \tuple {y_1, y_2}$ be arbitrary points in $\R^2$.
Note that $\paren {\map {d_2} {x, y} }^2$ can be expressed as:
:$\paren {\map {d_2} {x, y} }^2 = \paren {\mathbf x - \mathbf y}^\intercal \paren {\mathbf x - \mathbf y}$
where:
:$x$ and $y$ are expressed in vector form: $\mathbf x = \begin {pmatrix} x_1 \\ x_2 \end {pmatrix}, y = \begin {pmatrix} y_1 \\ y_2 \end {pmatrix}$
:$\paren {\mathbf x - \mathbf y}^\intercal$ denotes the transpose of $\paren {\mathbf x - \mathbf y}$
Then:
{{begin-eqn}}
{{eqn | l = \map {d_2} {\map {r_\alpha} x, \map {r_\alpha} y}
      | r = \map {d_2} {\begin {pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end {pmatrix} \begin {pmatrix} x_1 \\ x_2 \end {pmatrix}, \begin {pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end {pmatrix} \begin {pmatrix} y_1 \\ y_2 \end {pmatrix} }
      | c = Matrix Form of Plane Rotation
}}
{{eqn | r = \map {d_2} {\begin {pmatrix} x_1 \cos \alpha + x_2 \sin \alpha \\ -x_1 \sin \alpha + x_2 \cos \alpha \end {pmatrix}, \begin {pmatrix} y_1 \cos \alpha + y_2 \sin \alpha \\ -y_1 \sin \alpha + y_2 \cos \alpha \end {pmatrix} }
      | c = 
}}
{{eqn | r = \sqrt {\begin {pmatrix} \paren {x_1 - y_1} \cos \alpha + \paren {x_2 - y_2} \sin \alpha \\ -\paren {x_1 - y_1} \sin \alpha + \paren {x_2 - y_2} \cos \alpha \end {pmatrix}^\intercal \begin {pmatrix} \paren {x_1 - y_1} \cos \alpha + \paren {x_2 - y_2} \sin \alpha \\ -\paren {x_1 - y_1} \sin \alpha + \paren {x_2 - y_2} \cos \alpha \end {pmatrix} }
      | c = {{Defof|Euclidean Metric on Real Number Plane}}
}}
{{eqn | r = \sqrt {\paren {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2} \paren {\cos^2 \alpha + \sin^2 \alpha} + 2 \paren {x_1 - y_1} \paren {x_2 - y_2} \cos \alpha \sin \alpha - 2 \paren {x_1 - y_1} \paren {x_2 - y_2} \cos \alpha \sin \alpha}
      | c = multiplying out and gathering terms
}}
{{eqn | r = \sqrt {\paren {x_1 - y_1}^2 + \paren {x_2 - y_2}^2}
      | c = Sum of Squares of Sine and Cosine and simplifying
}}
{{eqn | r = \map {d_2} {x, y}
      | c = {{Defof|Euclidean Metric on Real Number Plane}}
}}
{{end-eqn}}
{{qed}}
Theorem: Taxicab Metric on Real Number Plane is Translation Invariant

Let $\tau_{\mathbf t}: \R^2 \to \R^2$ denote the translation of the Euclidean plane by the vector $\mathbf t = \begin {pmatrix} a \\ b \end {pmatrix}$.
Let $d_1$ denote the taxicab metric on $\R^2$.
Then $d_1$ is unchanged by application of $\tau$:
:$\forall x, y \in \R^2: \map {d_1} {\map \tau x, \map \tau y} = \map {d_1} {x, y}$

Proof:
Let $x = \tuple {x_1, x_2}$ and $y = \tuple {y_1, y_2}$ be arbitrary points in $\R^2$.
Then:
{{begin-eqn}}
{{eqn | l = \map {d_1} {\map \tau x, \map \tau y}
      | r = \map {d_1} {x - \mathbf t, y - \mathbf t}
      | c = {{Defof|Translation in Euclidean Space}}
}}
{{eqn | r = \size {\paren {x_1 - a} - \paren {y_1 - a} } + \size {\paren {x_2 - b} - \paren {y_2 - b} }
      | c = Definition of $\mathbf t$, {{Defof|Taxicab Metric on Real Number Plane}}
}}
{{eqn | r = \size {x_1 - y_1} + \size {x_2 - y_2}
      | c = simplification
}}
{{eqn | r = \map {d_1} {x, y}
      | c = {{Defof|Taxicab Metric on Real Number Plane}}
}}
{{end-eqn}}
{{qed}}
Theorem: Chebyshev Distance on Real Number Plane is Translation Invariant

Let $\tau_{\mathbf t}: \R^2 \to \R^2$ denote the translation of the Euclidean plane by the vector $\mathbf t = \begin {pmatrix} a \\ b \end {pmatrix}$.
Let $d_\infty$ denote the Chebyshev distance on $\R^2$.
Then $d_1$ is unchanged by application of $\tau$:
:$\forall x, y \in \R^2: \map {d_\infty} {\map \tau x, \map \tau y} = \map {d_\infty} {x, y}$

Proof:
Let $x = \tuple {x_1, x_2}$ and $y = \tuple {y_1, y_2}$ be arbitrary points in $\R^2$.
Then:
{{begin-eqn}}
{{eqn | l = \map {d_\infty} {\map \tau x, \map \tau y}
      | r = \map {d_\infty} {x - \mathbf t, y - \mathbf t}
      | c = {{Defof|Translation in Euclidean Space}}
}}
{{eqn | r = \max \set {\size {\paren {x_1 - a} - \paren {y_1 - a} }, \size {\paren {x_2 - b} - \paren {y_2 - b} } }
      | c = Definition of $\mathbf t$, {{Defof|Chebyshev Distance on Real Number Plane}}
}}
{{eqn | r = \max \set {\size {x_1 - y_1}, \size {x_2 - y_2} }
      | c = simplification
}}
{{eqn | r = \map {d_\infty} {x, y}
      | c = {{Defof|Chebyshev Distance on Real Number Plane}}
}}
{{end-eqn}}
{{qed}}
Theorem: Taxicab Metric on Real Number Plane is not Rotation Invariant

Let $r_\alpha: \R^2 \to \R^2$ denote the rotation of the Euclidean plane about the origin through an angle of $\alpha$.
Let $d_1$ denote the taxicab metric on $\R^2$.
Then it is not necessarily the case that:
:$\forall x, y \in \R^2: \map {d_1} {\map {r_\alpha} x, \map {r_\alpha} y} = \map {d_1} {x, y}$

Proof:
;Proof by Counterexample:
Let $x = \tuple {0, 0}$ and $y = \tuple {0, 1}$ be arbitrary points in $\R^2$.
Then:
{{begin-eqn}}
{{eqn | l = \map {d_1} {x, y}
      | r = \map {d_1} {\tuple {0, 0}, \tuple {0, 1} }
      | c = Definition of $x$ and $y$
}}
{{eqn | r = \size {0 - 0} + \size {0 - 1}
      | c = {{Defof|Taxicab Metric on Real Number Plane}}
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
Now let $\alpha = \dfrac \pi 4 = 45 \degrees$.
{{begin-eqn}}
{{eqn | l = \map {d_1} {\map {r_\alpha} x, \map {r_\alpha} y}
      | r = \map {d_1} {\tuple {0, 0}, \tuple {\dfrac {\sqrt 2} 2, \dfrac {\sqrt 2} 2} }
      | c = {{Defof|Plane Rotation}}
}}
{{eqn | r = \size {0 - \dfrac {\sqrt 2} 2} + \size {0 - \dfrac {\sqrt 2} 2}
      | c = {{Defof|Taxicab Metric on Real Number Plane}}
}}
{{eqn | r = \sqrt 2
      | c = simplification
}}
{{eqn | o = \ne
      | r = 1
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Chebyshev Distance on Real Number Plane is not Rotation Invariant

Let $r_\alpha: \R^2 \to \R^2$ denote the rotation of the Euclidean plane about the origin through an angle of $\alpha$.
Let $d_\infty$ denote the Chebyshev distance on $\R^2$.
Then it is not necessarily the case that:
:$\forall x, y \in \R^2: \map {d_\infty} {\map {r_\alpha} x, \map {r_\alpha} y} = \map {d_\infty} {x, y}$

Proof:
;Proof by Counterexample:
Let $x = \tuple {0, 0}$ and $y = \tuple {1, 1}$ be arbitrary points in $\R^2$.
Then:
{{begin-eqn}}
{{eqn | l = \map {d_\infty} {x, y}
      | r = \map {d_\infty} {\tuple {0, 0}, \tuple {1, 1} }
      | c = Definition of $x$ and $y$
}}
{{eqn | r = \max \set {\size {0 - 1}, \size {0 - 1} }
      | c = {{Defof|Chebyshev Distance on Real Number Plane}}
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
Now let $\alpha = \dfrac \pi 4 = 45 \degrees$.
{{begin-eqn}}
{{eqn | l = \map {d_\infty} {\map {r_\alpha} x, \map {r_\alpha} y}
      | r = \map {d_\infty} {\tuple {0, 0}, \tuple {0, \sqrt 2} }
      | c = {{Defof|Plane Rotation}}
}}
{{eqn | r = \max \set {\size {0 - 0}, \size {0 - \sqrt 2} }
      | c = {{Defof|Chebyshev Distance on Real Number Plane}}
}}
{{eqn | r = \sqrt 2
      | c = simplification
}}
{{eqn | o = \ne
      | r = 1
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Open Balls of Supremum Metric on Continuous Real Functions on Closed Interval

Let $\mathscr C \closedint a b$ be the set of all continuous functions $f: \closedint a b \to \R$.
Let $d: \mathscr C^2 \to \R$ be the supremum metric on $\mathscr C \closedint a b$ defined as:
:$\ds \forall f, g \in \mathscr C \closedint a b: \map d {f, g} := \sup_{x \mathop \in \closedint a b} \size {\map f x - \map g x}$
where $\sup$ denotes the supremum.
Let $f, g \in \mathscr C \closedint a b$ be such that:
:$\forall x \in \closedint a b: \map f x < \map g x$
Consider the set $S$, defined as:
:$S = \set {h \in \mathscr C \closedint a b: \forall x \in \closedint a b: \map f x < \map h x < \map g x}$
Then:
:$S$ is an open ball of $\struct {\mathscr C \closedint a b, d}$
{{iff}}:
:$\exists c \in \R_{>0}: \map g x := \map f x + c$
That is, {{iff}} $f$ and $g$ differ by a constant.

Proof:
Recall the definition of open ball:
The '''open $\epsilon$-ball of $a$ in $M = \struct {A, d}$''' is defined as:
:$\map {B_\epsilon} a := \set {x \in A: \map d {x, a} < \epsilon}$
In this context, the open $\epsilon$-ball of $\phi$ in $\mathscr C \closedint a b$ is defined as:
:$\ds \map {B_\epsilon} \phi := \set {\rho \in \mathscr C \closedint a b: \sup_{x \mathop \in \closedint a b} \size {\map \rho x - \map \phi x} < \epsilon}$
Theorem: Equality of Open Balls does not imply Equality of Centers

Let $M = \struct {A, d}$ be a metric space.
Let $x, y \in A$ and $r, s \in \R$ such that:
:$\map {B_r} x = \map {B_s} y$
Then it is not necessarily the case that their centers $x$ and $y$ are equal.

Proof:
Let $A$ be arbitrary.
Let $d$ be the (standard) discrete metric on $A$.
Let $r \ge 1$ and $s \ge 1$.
Then from Open Ball in Standard Discrete Metric Space:
:$\forall x, y \in A: \map {B_r} x = \map {B_s} y = A$
whether $x = y$ or not.
{{qed}}
Theorem: Equality of Open Balls does not imply Equality of Radii

Let $M = \struct {A, d}$ be a metric space.
Let $x, y \in A$ and $r, s \in \R$ such that:
:$\map {B_r} x = \map {B_s} y$
Then it is not necessarily the case that their radii $r$ and $s$ are equal.

Proof:
Let $A$ be arbitrary.
Let $d$ be the (standard) discrete metric on $A$.
Let $r \ge 1$ and $s \ge 1$ such that $r \ne s$.
Then from Open Ball in Standard Discrete Metric Space:
:$\forall x, y \in A: \map {B_r} x = \map {B_s} y = A$
{{qed}}
Theorem: Mapping between Subspaces of Homeomorphic Spaces is Homeomorphism

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $f: T_1 \to T_2$ be a homeomorphism.
Let $H \subseteq S_1$.
Let $T_H = \struct {H, \tau_H}$ be the topological subspace of $T_1$ under the subspace topology $\tau_H$ induced by $\tau_1$.
Let $K = f \sqbrk H$ be the image of $H$ under $f$.
Let $T_K = \struct {K, \tau_K}$ be the topological subspace of $T_2$ under the subspace topology $\tau_K$ induced by $\tau_2$.
Let $g: H \to K$ be the restriction of $f$ to $H$.
Then $g$ is a homeomorphism.

Proof:
Let $U \in \tau_K$ be open in $K$.
Then either:
:$U \in \tau_2$
or:
:$U = K \cap V$
where $V \in \tau_2$.
Suppose $U \in \tau_2$.
Because $f$ is continuous:
:$f^{-1} \sqbrk U \in \tau_1$
{{begin-eqn}}
{{eqn | l = f^{-1} \sqbrk U
      | o = \in
      | r = \tau_1
      | c = {{Defof|Continuous Mapping}}
}}
{{eqn | ll= \leadsto
      | l = g^{-1} \sqbrk U
      | o = \in
      | r = \tau_1
      | c = {{Defof|Restriction of Mapping}}
}}
{{end-eqn}}
As $U \subseteq K$ we have that $g^{-1} \sqbrk U \subseteq H$.
So by definition of subspace topology, $g^{-1} \sqbrk U$ is open in $H$.
{{qed|lemma}}
Now suppose $U = K \cap V$ for some $V \in \tau_2$.
Then:
{{begin-eqn}}
{{eqn | l = g^{-1} \sqbrk U
      | r = g^{-1} \sqbrk {K \cap V}
      | c = 
}}
{{eqn | r = g^{-1} \sqbrk K \cap g^{-1} \sqbrk V
      | c = Preimage of Intersection under Mapping
}}
{{eqn | r = f^{-1} \sqbrk K \cap f^{-1} \sqbrk V
      | c = {{Defof|Restriction of Mapping}}
}}
{{end-eqn}}
But as $f$ is continuous:
:$f^{-1} \sqbrk V \in \tau_1$
As $f^{-1} \sqbrk K$ it follows that:
:$f^{-1} \sqbrk U = H \cap f^{-1} \sqbrk U$
Hence:
:$g^{-1} \sqbrk U \in \tau_H$
and the result follows.
Theorem: Mapping between Subspaces of Homeomorphic Spaces is Homeomorphism/Corollary

Let $\overline H$ denote the complement $S_1 \setminus H$ of $H$ relative to $S_1$.
Let $T_{\overline H} = \struct {\overline H, \tau_{\overline H} }$ be the topological subspace of $T_1$ under the subspace topology $\tau_{\overline H}$ induced by $\tau_1$.
Let $\overline K = f \sqbrk {\overline H}$ be the image of $\overline H$ under $f$.
Let $T_{\overline K} = \struct {\overline K, \tau_{\overline K} }$ be the topological subspace of $T_2$ under the subspace topology $\tau_{\overline K}$ induced by $\tau_2$.
Let $h: \overline H \to \overline K$ be the restriction of $f$ to $\overline H$.
Then $h$ is a homeomorphism.

Proof:
By definition of relative complement, $\overline H \subseteq S_1$.
As $f$ is a homeomorphism, it is a fortiori a bijection.
Hence from Image of Relative Complement under Bijection is Relative Complement of Image:
:$f \sqbrk {\overline H} = \overline K$
Hence Mapping between Subspaces of Homeomorphic Spaces is Homeomorphism applies directly, using $\overline H$ for $H$.
{{qed}}
Theorem: Image of Relative Complement under Bijection is Relative Complement of Image

Let $S$ and $T$ be sets.
Let $f: S \to T$ be a bijection.
Let $H \subseteq S$.
Let $f \sqbrk H = K$ be the image of $H$ under $f$.
Let $\relcomp S H$ denote the relative complement of $H$ in $S$.
Then:
:$f \sqbrk {\relcomp S H} = \relcomp T K$

Proof:
From Set with Relative Complement forms Partition, $\set {H \mid \relcomp S H}$ forms a partition of $S$.
The result follows from Bijection Preserves Set Partition.
{{qed}}
Category:Relative Complement
Category:Bijections
ok4glh22j4a3wuzk94eewx7lqh50ya9
Theorem: Bijection Preserves Set Partition

Let $S$ and $T$ be sets.
Let $f: S \to T$ be a bijection
Let $I$ be an indexing set.
Let $\family {S_i}_{i \mathop \in I}$ be a partitioning of $S$ indexed by $I$.
Hence, let $\set {S_i: i \in I}$ be the resulting partition of $S$.
Then the image of $f$ is a partition of $T$ indexed by $I$ such that:
:$T = \set {f \sqbrk {S_i}: i \in I}$

Proof:
By definition of partitioning:
:$(1): \quad \forall i \in I: S_i \ne \O$, that is, none of $S_i$ is empty
:$(2): \quad \ds S = \bigcup_{i \mathop \in I} S_i$, that is, $S$ is the union of $\family {S_i}_{i \mathop \in I}$
:$(3): \quad \forall i, j \in I: i \ne j \implies S_i \cap S_j = \O$, that is, the elements of $\family {S_i}_{i \mathop \in I}$ are pairwise disjoint.
It is sufficient to check that:
:$(1): \quad \forall i \in I: f \sqbrk {S_i} \ne \O$, that is, none of $f \sqbrk {S_i}$ is empty
:$(2): \quad \ds T = \bigcup_{i \mathop \in I} f \sqbrk {S_i}$, that is, $T$ is the union of $\family {f \sqbrk {S_i} }_{i \mathop \in I}$
:$(3): \quad \forall i, j \in I: i \ne j \implies f \sqbrk {S_i} \cap f \sqbrk {S_j} = \O$, that is, the elements of $\family {f \sqbrk {S_i} }_{i \mathop \in I}$ are pairwise disjoint.
We have that:
:$f \sqbrk {S_i} = \set {t \in T: t = \map f s: s \in S_i}$
As $S_i \ne \O$ from $(1)$ it follows that $f \sqbrk {S_i} \ne \O$.
Hence criterion $(1)$ is fulfilled by $f$.
{{qed|lemma}}
Because $f$ is a bijection, it is a fortiori a surjection.
Hence:
:$\forall t \in T: \exists s \in S: \map f s = t$
But as $S = \set {S_i: i \in I}$ is partitioned by $I$, it follows that:
:$\exists i \in I: s \in S_i$
Hence:
:$\exists i \in I: t \in f \sqbrk {S_i}$
So every element of $T$ is an element of (at least) one of the $f \sqbrk {S_i}$.
That is:
:$\ds T = \bigcup_{i \mathop \in I} f \sqbrk {S_i}$
Hence criterion $(2)$ is fulfilled by $f$.
{{qed|lemma}}
Because $f$ is a bijection, it is a fortiori an injection.
From Image of Intersection under Injection:
:$\forall S_i, S_j \subseteq S: f \sqbrk {S_i \cap S_j} = f \sqbrk {S_i} \cap f \sqbrk {S_j}$
But for $i \ne j$ we have that $S_i \cap S_j = \O$.
Hence:
:$\forall i, j \in I: i \ne j \implies f \sqbrk {S_i} \cap f \sqbrk {S_j} = \O$
and it follows that criterion $(3)$ is fulfilled by $f$.
{{qed|lemma}}
Thus all $3$ criteria for $T = \set {f \sqbrk {S_i}: i \in I}$ being a partition are fulfilled.
{{qed}}
Category:Set Partitions
Category:Bijections
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Theorem: Product Space is Homeomorphic to Product Space with Factors Commuted

Let $T_1$ and $T_2$ be topological spaces.
Let $T_1 \times T_2$ denote the product space of $T_1$ and $T_2$.
Let $t: T_1 \times T_2 \to T_2 \times T_1$ be the mapping defined as:
:$\forall \tuple {x, y} \in T_1 \times T_2: \map t {x, y} = \tuple {y, x}$
Then $t$ is a homeomorphism.

Proof:
$t$ is trivially a bijection.
Let $U$ be open in $T_2 \times T_1$.
Then by definition of product space:
:$U = U_2 \times U_1$
where:
:$U_2$ is open in $T_2$
:$U_1$ is open in $T_1$.
Hence by definition of product space:
:$t^{-1} \sqbrk {U_2 \times U_1} = U_1 \times U_2$ is open in $T_1 \times T_2$.
Hence it has been shown that $t$ is continuous.
Similarly, let $V$ be open in $T_1 \times T_2$.
Then by definition of product space:
:$V = V_1 \times V_2$
where:
:$V_1$ is open in $T_1$
:$V_2$ is open in $T_2$.
Hence by definition of product space:
:$t \sqbrk {V_1 \times V_2} = V_2 \times V_1$ is open in $T_2 \times T_1$.
Hence it has been shown that $t^{-1}$ is continuous.
The result follows by definition of homeomorphism.
{{qed}}
Theorem: Cartesian Product of Homeomorphisms is Homeomorphism

Let $S_1, S_2, T_1, T_2$ be topological spaces.
Let $f_1: S_1 \to T_1$ and $f_2: S_2 \to T_2$ be mappings.
Let:
:$f_1 \times f_2: S_1 \times S_2 \to T_1 \times T_2$
be defined as:
:$\forall \tuple {x, y} \in S_1 \times S_2: \map {\paren {f_1 \times f_2} } {x, y} = \tuple {\map {f_1} x, \map {f_2} y}$
where $S_1 \times S_2$ denotes the product space of $S_1$ and $S_2$, and similarly for $T_1 \times T_2$.
Let $f_1$ and $f_2$ be homeomorphisms.
Then $f_1 \times f_2$ is also a homeomorphism.

Proof:
From Cartesian Product of Mappings is Continuous iff Factor Mappings are Continuous:
:$f_1 \times f_2$ is continuous.
From Cartesian Product of Bijections is Bijection:
:$f_1 \times f_2$ is a bijection.
From Cartesian Product of Mappings is Continuous iff Factor Mappings are Continuous:
:$\paren {f_1 \times f_2}^{-1} = f_1^{-1} \times f_2^{-1}$ is continuous.
So, by definition, $f_1 \times f_2$ is a homeomorphism.
{{qed}}