nyuuzyou/wikis huggingface dataset:
Hausdorff Paradox/Proof 1

== Theorem ==
== Proof ==
Let $R \subset \map {\mathbb {SO} } 3$ be the group generated by the $\pi$ and $\dfrac {2 \pi} 3$ rotations around different axes.
The elements:
:$\psi = \begin {pmatrix} -\tfrac 1 2 & \tfrac {\sqrt 3} 2 & 0 \\ -\tfrac {\sqrt 3} 2 & -\tfrac 1 2 & 0 \\ 0 & 0 & 1 \\ \end {pmatrix}$
:$\phi = \begin {pmatrix} -\map \cos \vartheta & 0 & \map \sin \vartheta \\ 0 & -1 & 0 \\ \map \sin \vartheta & 0 & \map \cos \vartheta \\ \end {pmatrix}$
form a basis for $R$ for some $\vartheta$.
We have:
:$\psi^3 = \phi^2 = \mathbf I_3$
where $\mathbf I_3$ is the identity mapping in $\mathbb R^3$.
Therefore $\forall r \in R: r \ne \mathbf I_3, \psi, \phi, \psi^2$ there exists some $n \in \N$ and some set of numbers $m_k \in \set {1, 2}, 1 \le k \le n$ such that $r$ can written as one of the following:
:$\text{(a)}: \quad \ds r = \prod_{k \mathop = 1}^n \phi \psi^{m_k}$
:$\text{(b)}: \quad \ds r = \map {\psi^{m_1} } {\prod_{k \mathop = 2}^n \phi \psi^{m_k} }\phi$
:$\text{(c)}: \quad \ds r = \paren {\prod_{k \mathop = 1}^n \phi \psi^{m_k} } \phi$
:$\text{(d)}: \quad \ds r = \map {\psi^{m_1} } {\prod_{k \mathop = 2}^n \phi \psi^{m_k} }$
Now we fix $\vartheta$ such that $\mathbf I_3$ cannot be written in any of the ways $\text{(a)}$, $\text{(b)}$, $\text{(c)}$, $\text{(d)}$. 
The action of $R$ on $\mathbb S^2$ will leave two points unchanged for each element of $R$ (the intersection of the axis of rotation and the sphere, to be exact). 
Since $R$ is finitely generated, it is a countable group.
Therefore the set of points of $\mathbb S^2$ which are unchanged by at least one element of $R$ is also countable. 
We call this set $Q \subset \mathbb S^2$, so that $R$ acts freely on $\mathbb S^2 \setminus Q$.
By Set of Orbits forms Partition, this partitions $\mathbb S^2 \setminus Q$ into orbits.
By the , there is a set $X$ containing one element of each orbit. 
For any $r \in R$, let $X_r$ be the right action of $r$ on $X$.
We have:
:$\ds \mathbb S^2 \setminus Q = \bigcup_{r \mathop \in R} X_r$
Define the sets $A, B, C$ to be the smallest sets satisfying 
:$X \subseteq A$
:If $X_r \subset A, B, C$, then $X_{r\phi} \subset B, A, A$, respectively.
:If $X_r \subset A, B, C$, then $X_{r\psi} \subset B, C, A$, respectively.
:If $X_r \subset A, B, C$, then $X_{r\psi^2} \subset C, A, B$, respectively.
These sets are defined due to the uniqueness of the properties $\text{(a)}$ to $\text{(d)}$
Also, $A, B, C, B \cup C$ are congruent since they are rotations of each other, namely:
:$A_\psi = B, B_{\psi^2} = C, A_\phi = B \cup C$
Hence we have constructed the sets $A, B, C, Q$ of the theorem.
Category:Hausdorff Paradox
Hausdorff Paradox/Proof 2

== Theorem ==
== Proof ==
=== Lemma $1$ ===
=== Lemma $2$ ===
Let $Q$ be the set of all fixed points of $\Bbb S^2$ of all rotations $\alpha \in G$.
Each such $\alpha$ has $2$ fixed points.
Hence $Q$ is countable.
The set $\Bbb S^2 \setminus Q$ is a disjoint union of all orbits $P_x$ of $G$:
:$P_x = \set {x \alpha: \alpha \in G}$
By the , there exists a set $M$ which contains exactly $1$ element in each $P_x$ where $x \in \Bbb S^2 \setminus Q$.
Let:
It then follows from that:
:$A$, $B$ and $C$ are disjoint
:$A$, $B$ and $C$ are congruent to each other
:$B \cup C$ is congruent to each of $A$, $B$ and $C$
and moreover:
:$S = A \cup B \cup C \cup Q$
== Sources ==
* : $1.$ Introduction: $1.3$ A paradoxical decomposition of the sphere
Category:Hausdorff Paradox
Hausdorff Space iff Diagonal Set on Product is Closed

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\Delta_S$ be the diagonal set on $S$:
:$\Delta_S = \set {\tuple {x, x} \in S \times S: x \in S}$
where $S \times S$ is the Cartesian product of $S$ with itself.
Let $T^2 = \struct {S \times S, \TT}$ be the product space with product topology $\TT$.
Then $T$ is a $T_2$ (Hausdorff) space $\Delta_S$ is closed in $T^2$.
== Proof ==
Suppose $T$ is a $T_2$ (Hausdorff) space.
Let $\Delta: S \to S \times S$ be the diagonal mapping on $S$.
We have that $\map \Delta S$ consists of all the elements of $S \times S$ whose coordinates are equal.
:$\map \Delta S = \set {\tuple {x, x} \in S \times S: x \in S} = \set {\tuple {x, y} \in S \times S: x = y}$
By definition, $\Delta_S$ is the same set as $\map \Delta S$.
Now consider $H := \relcomp {S \times S} {\map \Delta S}$, its complement in $S \times S$.
We have that $H$ consists of all the elements of $S \times S$ whose coordinates are different:
:$H = \set {\tuple {x, y} \in S \times S: x \ne y}$
Let $\tuple{x, y} \in H$.
Because $T$ is a $T_2$ space:
:$\exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$
Because $U \cap V = \O$ it follows that $U \times V \cap \map \Delta S = \O$.
That is:
:$U \times V \subseteq H$
Now from Natural Basis of Product Topology of Finite Product, we have that $U \times V$ is an element of the natural basis of $T^2 = \struct {S \times S, \TT}$.
Hence $H$ is open in $T^2$.
Therefore, by definition, $\Delta_S$ is closed in $T^2 = \struct {S \times S, \TT}$.
Now suppose $\Delta_S$ is closed in $T^2 = \struct {S \times S, \TT}$.
Then $H = \set {\tuple {x, y} \in S \times S: x \ne y}$ is open in $T^2$.
Let $\tuple {x, y} \in H$.
We have that $x \ne y$ by the nature of $H$.
Then there exists an element $U \times V$ in the natural basis of $T^2$, and therefore open in $T^2$.
As $U \times V \subseteq H$ it follows that $U \cap V = \O$.
From Natural Basis of Product Topology of Finite Product, as $U \times V$ is open in $T^2$, then $U$ and $V$ are both open in $T$.
So, for any $\tuple {x, y}$ we have $U, V \subseteq S$ such that $x \in U, y \in V, U \cap V = \O$.
So $T$ is a $T_2$ (Hausdorff) space.
== Also see ==
* Open Projection and Closed Graph Implies Quotient is Hausdorff, a generalization for quotient spaces
Category:Hausdorff Spaces
Category:Product Topology
Hausdorff Space is Hereditarily Compact iff Finite

== Theorem ==
Let $\left({S, \tau}\right)$ be a Hausdorff space.
Then $\left({S, \tau}\right)$ is hereditarily compact $S$ is finite.
== Proof ==
=== Necessary Condition ===
Let $\left({S, \tau}\right)$ be hereditarily compact.
That is, every subspace of $\left({S, \tau}\right)$ is compact.
Let $H \subseteq S$ be a subspace of $\left({S, \tau}\right)$.
From Compact Subspace of Hausdorff Space is Closed, $H$ is closed.
Thus for all $H \subseteq S$, we find that $H$ is closed in $S$.
It follows by definition of discrete topology that $\tau$ is discrete.
But since $S$ is compact and Discrete Space is Compact iff Finite, $S$ is also finite.
=== Sufficient Condition ===
Let $S$ be finite.
From $T_2$ (Hausdorff) Space is $T_1$ Space, $\left({S, \tau}\right)$ is a $T_1$ (Fréchet) space.
From Finite $T_1$ Space is Discrete, it follows that $\left({S, \tau}\right)$ has the discrete topology.
As $\left({S, \tau}\right)$ (and every subspace) is finite and discrete, and Discrete Space is Compact iff Finite, then $S$ (and every subspace) is compact.
Thus $\left({S, \tau}\right)$ is hereditarily compact.
== Also see ==
* Indiscrete Space is Hereditarily Compact, an example of a non-Hausdorff, infinite, non-discrete hereditarily compact space.
Category:Finite Topological Spaces
Category:Hausdorff Spaces
Category:Hereditarily Compact Spaces
Haversine Function is Even

== Theorem ==
The haversine is an even function:
:$\forall \theta \in \R: \map \hav {-\theta} = \hav \theta$
== Proof ==
== Sources ==
* : Chapter $\text I$. Spherical Trigonometry: $13$. The haversine formula.
Category:Haversines
Category:Examples of Even Functions
Heart Curve/Formulation 3

== Theorem ==
:$\begin{cases}
x = & \map \sin t \map \cos t \ln \size t \\
y = & \size t^{0.3} \sqrt {\cos t}
\end{cases}$
where $t \in \hointr {-1} 0 \cup \hointl 0 1$.
:
Category:Heart Curves
Heaviside Expansion Formula

== Theorem == 
Let $P, Q$ be polynomials with coefficients in $\C$. 
Let $\deg Q \ge \deg P + 1$.
Let $\map Q z$ have a simple zero for $z \in X$. 
Let $\map {\laptrans f} z = \dfrac {\map P z} {\map Q z}$. 
Then:
:$\ds \map f t = \sum_{z \mathop \in X} e^{z t} \frac {\map P z} {\map {Q'} z}$
== Proof == 
 
== Sources == 
* : $8.2.3$: Heaviside Expansion Formula
Category:Laplace Transforms
Heaviside Step Function is Locally Integrable

== Theorem ==
Let $u_c$ be a Heaviside step function.
Then $u_c$ is a locally integrable function.
== Proof ==
== Sources ==
* : Chapter $\S 6.1$: A glimpse of distribution theory. Test functions, distributions, and examples
Category:Heaviside Step Function
Category:Examples of Locally Integrable Functions
Heaviside Step Function is Piecewise Continuous

== Theorem ==
Let $c \ge 0$ be a constant real number.
The Heaviside step function:
:$\map {\mu_c} t = \begin {cases} 1 & : t > c \\ 0 & : t < c \end {cases}$
is piecewise continuous for any interval of the form:
:$\closedint {c - M} {c + M}$
where $M > 0$ is some arbitrarily large constant.
== Proof ==
Let the finite subdivision of $\mu_c$ be:
:$\set {c - M, c, c + M}$
Let $\epsilon > 0$.
For $t \to \paren {c - M}^+$, choose $\delta$ to be any number at all, because:
:$c - M < t < c - M + \delta \implies \size {\map {\mu_c} t - 0} = \size {0 - 0} < \epsilon$
holds, from True Statement is implied by Every Statement.
The case for $t \to c^-$ is proved similarly.
For $t \to c^+$, again, choose $\delta$ to be any number at all, because:
:$c - \delta < t < c \implies \size {\map {\mu_c} t - 1} = \size {1 - 1} < \epsilon$
holds, from True Statement is implied by Every Statement.
The case for $t \to \paren {c + M}^-$ is proved similarly.
For all other $t$, we invoke Continuity of Heaviside Step Function.
Category:Heaviside Step Function
Category:Continuous Functions
Heine-Borel Theorem

== Theorem ==
=== Real Line ===
=== Euclidean Space ===
=== Metric Space ===
=== Normed Vector Space ===
=== Dedekind-Complete Linearly Ordered Space ===
== Also see ==
* Compact Subspace of Linearly Ordered Space
* Heine-Borel Theorem/Real Line/Closed and Bounded Interval
* Heine-Borel Theorem/Real Line/Closed and Bounded Set
== Sources ==
* 
Category:Compact Spaces
Category:Heine-Borel Theorem
Heine-Borel Theorem/Dedekind Complete Space

== Theorem ==
Let $T = \struct {X, \preceq, \tau}$ be a Dedekind-complete linearly ordered space.
Let $Y$ be a non-empty subset of $X$.
Then $Y$ is compact $Y$ is closed and bounded in $T$.
== Proof ==
=== Sufficient Condition ===
Let $Y$ be a compact subspace of $T$.
From:
:Order Topology is Hausdorff
:Compact Subspace of Hausdorff Space is Closed
it follows that $Y$ is closed in $T$.
From Compact Subspace of Linearly Ordered Space: Lemma 1, $\struct {Y, \preceq {\restriction_Y} }$ is a complete lattice.
Hence $Y$ is closed and bounded in $T$.
=== Necessary Condition ===
Let $Y$ be a closed and bounded subspace of $T$.
Let $S$ be a non-empty subset of $Y$.
Since $Y$ is bounded and $S \subseteq Y$, $S$ is bounded.
Since $X$ is Dedekind complete, $S$ has a supremum and infimum in $X$.
We will show that $\sup S, \inf S \in Y$.
 $\sup S \notin Y$.
By Closed Set in Linearly Ordered Space, $b$ is not a supremum of $S$, a contradiction.
Thus $\sup S \in Y$.
A similar argument shows that $\inf S \in Y$.
Thus by Compact Subspace of Linearly Ordered Space, $Y$ is compact.
Category:Linearly Ordered Spaces
Category:Dedekind Completeness Property
Category:Heine-Borel Theorem
Heine-Borel Theorem/Euclidean Space

== Theorem ==
Let $n \in \N_{> 0}$.
Let $C$ be a subspace of the Euclidean space $\R^n$.
Then $C$ is closed and bounded it is compact.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
Let $C \subseteq \R^n$ be compact.
From Compact Subspace of Metric Space is Bounded, it follows that $C$ is bounded.
From Metric Space is Hausdorff, it follows that $\R^n$ is a Hausdorff space.
Then Compact Subspace of Hausdorff Space is Closed shows that $C$ is closed.
== Sources ==
* : $1$: Complex Numbers: Point Sets: $2$. Heine-Borel Theorem
* 
Category:Analysis
Category:Euclidean Space
Category:Heine-Borel Theorem
Heine-Borel Theorem/Euclidean Space/Necessary Condition/Proof 1

== Theorem ==
== Proof ==
Let $C \subseteq \R^n$ be closed and bounded.
Since $C$ is bounded, $C \subseteq \closedint a b^n = B$ for some $a, b \in \R$.
By the Heine-Borel Theorem: Real Line and by Topological Product of Compact Spaces, it follows that $B$ is compact.
From Euclidean Topology is Product Topology, it follows that $B$ is compact in the usual Euclidean topology induced by the Euclidean metric.
By the Corollary to Closed Set in Topological Subspace, we have that $C$ is closed in $B$.
By Closed Subspace of Compact Space is Compact, $C$ is compact.
Category:Heine-Borel Theorem
Heine-Borel Theorem/Euclidean Space/Necessary Condition/Proof 2

== Theorem ==
== Proof ==
The proof holds for $n = 1$, as follows.
Suppose $C$ is a closed, bounded subspace of $\R$.
Then $C \subseteq \closedint a b$ for some $a, b \in \R$.
Moreover, $C$ is closed in $\closedint a b$ by the Corollary to Closed Set in Topological Subspace.
Hence $C$ is compact, by Closed Subspace of Compact Space is Compact.
Now suppose $C \subseteq \R^n$ is closed and bounded.
Since $C$ is bounded, $C \subseteq \closedint a b \times \closedint a b \times \cdots \times \closedint a b = B$ for some $a, b \in \R$.
Now $B$ is compact by Topological Product of Compact Spaces.
Also, $C$ is closed in $B$ by the Corollary to Closed Set in Topological Subspace.
Hence $C$ is compact, by Closed Subspace of Compact Space is Compact.
== Sources ==
* : $5$: Compact spaces: $5.7$: Compact Subspaces of $\R^n$: Theorem $5.7.1$
* : Chapter $5$, Theorem $10$
Category:Heine-Borel Theorem
Heine-Borel Theorem/Metric Space

== Theorem == 
A metric space is compact it is both complete and totally bounded.
== Proof ==
=== Necessary Condition ===
This follows directly from:
* Compact Metric Space is Complete
* Compact Metric Space is Totally Bounded
=== Sufficient Condition ===
This follows directly from:
* Complete and Totally Bounded Metric Space is Sequentially Compact
* Sequentially Compact Metric Space is Compact
== Also see ==
* Heine-Borel Theorem
* Heine-Borel Theorem: Real Line
== Sources ==
* 
Category:Heine-Borel Theorem
Category:Totally Bounded Metric Spaces
Category:Complete Metric Spaces
Category:Compact Spaces
Heine-Borel Theorem/Normed Vector Space

== Theorem ==
Let $\struct {X, \norm {\,\cdot\,}}$ be a finite-dimensional normed vector space.
A subset $K \subseteq X$ is compact $K$ is closed and bounded.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Closed and Bounded Subset of Normed Vector Space is not necessarily Compact
Category:Normed Vector Spaces
Category:Compact Spaces
Category:Analysis
Category:Heine-Borel Theorem
Heine-Borel Theorem/Normed Vector Space/Necessary Condition

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a finite-dimensional normed vector space.
Let $K \subseteq X$ be closed and bounded.
Then $K$ is a compact subset.
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $K$.
$\sequence {x_n}_{n \mathop \in \N}$ is bounded.
We have that bounded sequence in finite-dimensional space has a convergent subsequence.
Hence, $\sequence {x_n}_{n \mathop \in \N}$ has a convergent subsequence $\sequence {x_{n_k} }_{k \mathop \in \N}$.
Denote the limit $\ds \lim_{k \mathop \to \infty} \sequence {x_{n_k} } = L$.
By assumption, $K$ is closed.
By equivalence of definitions, $L \in K$.
By definition, $K$ is compact.
== Sources ==
* : Chapter $\S 1.5$: Normed and Banach spaces. Compact sets
Category:Heine-Borel Theorem
Heine-Borel Theorem/Real Line

== Theorem ==
Let $\R$ be the real number line considered as a Euclidean space.
Let $C \subseteq \R$.
Then $C$ is closed and bounded in $\R$ $C$ is compact.
== Proof ==
=== Necessary Condition ===
Let $C$ be closed and bounded in $\R$.
Then, by Closed Bounded Subset of Real Numbers is Compact, $C$ is compact.
=== Sufficient Condition ===
Let $C$ be compact in $\R$.
Then, by Compact Subspace of Real Numbers is Closed and Bounded, $C$ is closed and bounded in $\R$.
== Note ==
This does not apply in the general metric space.
A trivial example is $\left({0 \,.\,.\, 1}\right)$ as a subspace of itself.
It is closed and bounded but not compact.
== Also see ==
* Bolzano-Weierstrass Theorem
* Heine-Borel Theorem for Metric Spaces
* Compact Subspace of Linearly Ordered Space
The theorem is sometimes called the Borel-Lebesgue Theorem, for and .
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
Category:Real Analysis
Category:Euclidean Space
Category:Compact Spaces
Category:Heine-Borel Theorem
Heine-Borel Theorem/Real Line/Closed and Bounded Set

== Theorem ==
Let $F$ be a closed and bounded real set.
Let $C$ be a set of open real sets.
Let $C$ be a cover of $F$.
Then there is a finite subset of $C$ that covers $F$.
== Proof ==
We are given that $C$ is a set of open real sets that covers $F$.
In other words, $C$ is an open cover of $F$.
We need to show that there is a finite subset of $C$ that covers $F$.
In other words, we need to show that $C$ has a finite subcover.
Let $F_o$ be the complement of $F$ in $\R$.
By definition of closed real set, $F_o$ is open as $F$ is closed.
=== Step 1: $C^*$ is an Open Cover of $\closedint a b$ ===
It is demonstrated that $C^*$ is an open cover of $\closedint a b$.
Since $F$ is bounded, $F$ is contained in a closed and bounded interval $\closedint a b$ where $a, b \in \R$.
Define $C^* = C \cup \set {F_o}$.
Like $C$, $C^*$ is a set of open real sets as $F_o$ is open.
$C^*$ covers $F \cup F_o$ as $C$ covers $F$ and $\set {F_o}$ covers $F_o$.
$F_o \cup F$ equals $\R$ as $F_o$ is the complement of $F$ in $\R$.
So $C^*$ covers $\R$.
Furthermore, $C^*$ is an open cover of $\closedint a b$ as $\closedint a b$ is a subset of $\R$.
=== Step 2: $C^*$ has a Finite Subcover ===
It is demonstrated that $C^*$ has a finite subcover $C^*_f$.
$C^*$ is an open cover of the closed and bounded interval $\closedint a b$.
Therefore, by Open Cover of Closed and Bounded Real Interval has Finite Subcover, $C^*$ has a finite subcover $C^*_f$.
=== Step 3: $C$ has a Finite Subcover ===
It is demonstrated that $C$ has a finite subcover $C_f$.
Note that $F_o$ is the only element in $C^*$ that is not an element of $C$.
Therefore, $F_o$ is the only possible element in $C^*_f$ that is not an element of $C$ as $C^*_f$ is a subset of $C^*$.
This means that $C^*_f \setminus \set {F_o}$ is a subset of $C$.
Define $C_f = C^*_f \setminus \set {F_o}$.
According to the reasoning above, $C_f \subseteq C$.
Also, $C_f$ is finite as $C^*_f$ is finite.
What remains is to show that $C_f$ covers $F$.
We have:
Thus, $C_f$ covers $F$.
This finishes the proof.
	
== Sources ==
* : Ch. $2$: Section $5$, theorem $15$
Category:Real Analysis
Heine-Borel iff Dedekind Complete

== Theorem ==
Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Then $X$ is Dedekind complete every closed, bounded subset of $X$ is compact.
== Proof ==
The forward implication follows from Heine-Borel Theorem: Dedekind-Complete Space.
Suppose that $X$ is not Dedekind complete.
Then $X$ has a non-empty subset $S$ with an upper bound $b$ in $X$ but no supremum in $X$.
Let $a \in S$ and let $Y = {\bar \downarrow} S \cap {\bar \uparrow} a$.
$Y$ is nonempty and bounded below by $a$ and above by $b$.
=== $Y$ is closed in $X$ ===
Let $x \in X \setminus Y$.
Then $x \prec a$ or $x$ strictly succeeds every element of $S$.
If $x \prec a$, then $x \in {\downarrow} a \subseteq X \setminus Y$.
If $x$ strictly succeeds each element of $S$, then it is an upper bound of $S$.
Since $S$ has no supremum in $X$, it has an upper bound $p \prec x$.
Then $x \in {\uparrow p} \subseteq X \setminus Y$.
=== $Y$ is not compact ===
Let $\AA = \set { {\downarrow} s: s \in S}$.
Then $\AA$ is an open cover of $Y$ with no finite subcover.
Category:Order Topology
Category:Compact Spaces
Heine-Cantor Theorem

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $M_1$ be compact.
Let $f: A_1 \to A_2$ be a continuous mapping.
Then $f$ is uniformly continuous.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Uniform Continuity on Metric Space does not imply Compactness
== Sources ==
* : $4.19$
Category:Metric Spaces
Category:Compact Spaces
Category:Continuous Mappings
Category:Uniformly Continuous Mappings
Category:Heine-Cantor Theorem
Heine-Cantor Theorem/Proof 1

== Theorem ==
== Proof ==
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
For all $x \in A_1$, define:
:$\map \Delta x = \set {\delta \in \R_{>0}: \forall y \in A_1: \map {d_1} {x, y} < 2 \delta \implies \map {d_2} {\map f x, \map f y} < \dfrac \epsilon 2}$
Define:
:$\CC = \set {\map {B_{\delta} } x: x \in A_1, \, \delta \in \map \Delta x}$
where $\map {B_{\delta} } x$ denotes the open $\delta$-ball of $x$ in $M_1$.
From the definition of continuity, it follows that $\CC$ is a cover for $A_1$.
From Open Ball of Metric Space is Open Set, it therefore follows that $\CC$ is an open cover for $A_1$.
By the definition of a compact metric space, there exists a finite subcover $\set {\map {B_{\delta_1} } {x_1}, \map {B_{\delta_2} } {x_2}, \ldots, \map {B_{\delta_n} } {x_n} }$ of $\CC$ for $A_1$.
Define:
:$\delta = \min \set {\delta_1, \delta_2, \ldots, \delta_n}$
Let $x, y \in A_1$ satisfy $\map {d_1} {x, y} < \delta$.
By the definition of a cover, there exists a $k \in \set{1, 2, \ldots, n}$ such that $\map {d_1} {x, x_k} < \delta_k$.
Then:
By the definition of $\map \Delta {x_k}$, it follows that:
:$\map {d_2} {\map f x, \map f {x_k} } < \dfrac \epsilon 2$
:$\map {d_2} {\map f y, \map f {x_k} } < \dfrac \epsilon 2$
Hence:
The result follows from the definition of uniform continuity.
== Sources ==
* : $5$: Compact spaces: $5.8$: Compactness and Uniform Continuity: Proposition $5.8.2$
Category:Heine-Cantor Theorem
Heine-Cantor Theorem/Proof 2

== Theorem ==
== Proof ==
Let $A_1 \times A_1$ and $A_2 \times A_2$ be considered with the product topology.
Let $F: A_1 \times A_1 \to A_2 \times A_2$ be the mapping defined as:
:$\map F {x, y} = \tuple {\map f x, \map f y}$
By Projection from Product Topology is Continuous, we have that the (first and second) projections on $A_1 \times A_1$ are continuous.
By Composite of Continuous Mappings is Continuous and Continuous Mapping to Product Space, it follows that $F$ is continuous.
By Distance Function of Metric Space is Continuous and Composite of Continuous Mappings is Continuous, it follows that $d_2 \circ F: A_1 \times A_1 \to \R$ is continuous.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
By Continuity Defined from Closed Sets, the set:
:$C = \paren {d_2 \circ F}^{-1} \sqbrk {\hointr \epsilon \to} = \set {\tuple {x, y} \in A_1 \times A_1: \map {d_2} {\map f x, \map f y} \ge \epsilon}$
is closed in $A_1 \times A_1$.
By Topological Product of Compact Spaces, $A_1 \times A_1$ is compact.
By Closed Subspace of Compact Space is Compact, $C$ is compact.
By Distance Function of Metric Space is Continuous and Continuous Image of Compact Space is Compact, $d_1 \sqbrk C$ is compact.
Therefore, $d_1 \sqbrk C$ has a smallest element $\delta$.
By and , we have that $\delta > 0$.
By construction, it follows that:
:$\forall x, y \in A_1: \map {d_1} {x, y} < \delta \implies \map {d_2} {\map f x, \map f y} < \epsilon$
Hence, $f$ is uniformly continuous.
Category:Heine-Cantor Theorem
Hellinger-Toeplitz Theorem

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $T : \HH \to \HH$ be a Hermitian operator.
That is: 
:$\innerprod {T x} y = \innerprod x {T y}$ for each $x, y \in \HH$.
Then: 
:$T$ is bounded.
== Proof 1 ==
== Proof 2 ==
Category:Hermitian Operators
Category:Hellinger-Toeplitz Theorem