avewright/proofwiki-math huggingface dataset:
Theorem: Product of Closed and Half-Open Unit Intervals is Homeomorphic to Product of Half-Open Unit Intervals

Let $\closedint 0 1$ denote the closed unit interval $\set {x \in \R: 0 \le x \le 1}$.
Let $\hointr 0 1$ denote the half-open unit interval $\set {x \in \R: 0 \le x < 1}$.
Let both $\closedint 0 1$ and $\hointr 0 1$ have the Euclidean topology.
Then the product space:
:$\closedint 0 1 \times \hointr 0 1$
is homeomorphic to:
:$\hointr 0 1 \times \hointr 0 1$

Proof:
First we take the square $\Box ABCD$ embedded in the Cartesian plane such that $AD$ corresponds to $\closedint 0 1$ and $AB$ corresponds to $\hointr 0 1$:
:300px
This corresponds to the set $\closedint 0 1 \times \hointr 0 1$.
It is noted that the line segment $BC$ which corresponds to $\closedint 0 1 \times \set 1$ is not in the set $\closedint 0 1 \times \hointr 0 1$.
Then we apply a homeomorphism which maps the perimeter of $\closedint 0 1 \times \hointr 0 1$ to the circle whose center is $\tuple {\dfrac 1 2, \dfrac 1 2}$ and whose radius is $\dfrac {\sqrt 2} 2$.
The points $A$, $B$, $C$ and $D$ are fixed by this homeomorphism.
:320px
Note how the point $E$ is mapped to the point $E'$.
Then we apply a homeomorphism to the circle $\bigcirc ABE'CD$ which maps:
:the arc $BADC$ to the arc $BAD$
:the arc $BE'C$ to the arc $BE' 'C' '$
which in the process maps:
:$E'$ to $E' '$ which is the same as $C$.
:$C$ to $C' '$ which is the same point as $D$.
:320px
Then we apply a homeomorphism to the circle $\bigcirc ABE' 'C' '$ back to the square $\Box ABE' 'C' '$.
It is seen that the line segment $E' 'C$ is now identified with the $\set 1 \times \hointr 0 1$.
Hence $\closedint 0 1 \times \hointr 0 1$ has been transformed via $3$ homeomorphisms to $\hointr 0 1 \times \hointr 0 1$.
{{qed}}
Theorem: Closed Image of Closure of Set under Continuous Mapping equals Closure of Image

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $H \subseteq S_1$ be a subset of $S_1$.
Let $\map \cl H$ denote the closure of $H$.
Let $f: T_1 \to T_2$ be a continuous mapping.
Let $f \sqbrk {\map \cl H}$ be closed in $T_2$.
Then:
:$f \sqbrk {\map \cl H} = \map \cl {f \sqbrk H}$

Proof:
By Continuity Defined by Closure:
:$f \sqbrk {\map \cl H} \subseteq \map \cl {f \sqbrk H}$
{{qed|lemma}}
{{begin-eqn}}
{{eqn | l = H
      | o = \subseteq
      | r = \map \cl H
      | c = Set is Subset of its Topological Closure
}}
{{eqn | ll= \leadsto
      | l = f \sqbrk H
      | o = \subseteq
      | r = f \sqbrk {\map \cl H}
      | c = Image of Subset is Subset of Image
}}
{{eqn-intertext|We are given that $f \sqbrk {\map \cl H}$ is closed in $T_2$, hence:}}
{{eqn | ll= \leadsto
      | l = \map \cl {f \sqbrk H}
      | o = \subseteq
      | r = f \sqbrk {\map \cl H}
      | c = Closure of Subset of Closed Set of Topological Space is Subset
}}
{{end-eqn}}
{{qed|lemma}}
The proof follows by definition of set equality.
{{qed}}
Theorem: Closure of Image under Continuous Mapping is not necessarily Image of Closure

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $H \subseteq S_1$ be a subset of $S_1$.
Let $\map \cl H$ denote the closure of $H$.
Let $f: T_1 \to T_2$ be a continuous mapping.
Then it is not necessarily the case that:
:$f \sqbrk {\map \cl H} = \map \cl {f \sqbrk H}$

Proof:
Proof by Counterexample:
Let $\R$ be the real numbers under the usual (Euclidean) topology.
Let $f: \R \to \R$ be the (real) hyperbolic tangent function:
:$\forall x \in \R: \map f x = \tanh x$
It is accepted that $f$ is continuous.
Let $H \subseteq \R$ be the subset of $\R$ defined as:
{{begin-eqn}}
{{eqn | l = H
      | r = \openint 0 \to
      | c = which is an open set in $\R$.
}}
{{eqn | ll= \leadsto
      | l = \map \cl H
      | r = \hointr 0 \to
      | c = {{Defof|Closure (Topology)}}
}}
{{eqn | ll= \leadsto
      | l = f \sqbrk H
      | r = \openint 0 1
      | c = {{Defof|Real Hyperbolic Tangent}}
}}
{{eqn | l = f \sqbrk {\map \cl H}
      | r = \hointr 0 1
      | c = {{Defof|Real Hyperbolic Tangent}}
}}
{{eqn | l = \map \cl {f \sqbrk H}
      | r = \closedint 0 1
      | c = {{Defof|Closure (Topology)}}
}}
{{end-eqn}}
As can be seen:
:$f \sqbrk {\map \cl H} \ne \map \cl {f \sqbrk H}$
{{qed}}
Theorem: Closure of Preimage under Continuous Mapping is not necessarily Preimage of Closure

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $H \subseteq S_1$ be a subset of $S_1$.
Let $\map \cl H$ denote the closure of $H$.
Let $f: T_1 \to T_2$ be a continuous mapping.
Then it is not necessarily the case that:
:$f^{-1} \sqbrk {\map \cl H} = \map \cl {f^{-1} \sqbrk H}$

Proof:
Proof by Counterexample:
Let $\R$ be the real numbers under the usual (Euclidean) topology.
Let $f: \R \to \R$ be the real function:
:$\forall x \in \R: \map f x = \begin {cases} -1 & : x \le -1 \\ x & : -1 \le x \le 1 \\ 1 & : x \ge 1 \end {cases}$
It is accepted that $f$ is a continuous mapping.
Let $H \subseteq \R$ be the subset of $\R$ defined as:
{{begin-eqn}}
{{eqn | l = H
      | r = \openint {-1} 1
      | c = which is an open set in $\R$.
}}
{{eqn | ll= \leadsto
      | l = \map \cl H
      | r = \closedint {-1} 1
      | c = {{Defof|Closure (Topology)}}
}}
{{eqn | ll= \leadsto
      | l = f^{-1} \sqbrk H
      | r = \openint {-1} 1
      | c = Definition of $f$
}}
{{eqn | l = f^{-1} \sqbrk {\map \cl H}
      | r = \R
      | c = Definition of $f$
}}
{{eqn | l = \map \cl {f^{-1} \sqbrk H}
      | r = \closedint {-1} 1
      | c = {{Defof|Closure (Topology)}}
}}
{{end-eqn}}
As can be seen:
:$f^{-1} \sqbrk {\map \cl H} \ne \map \cl {f^{-1} \sqbrk H}$
{{qed}}
Theorem: Normed Vector Space of Rational Numbers is not Banach Space

Let $\struct {\Q, \size {\, \cdot \,}}$ be the normed vector space of rational numbers.
Then $\struct {\Q, \size {\, \cdot \,}}$ is not a Banach space.

Proof:
Let $\sequence {x_n}_{n \mathop \in \N}$ be a sequence in $\Q$ defined recursively in the following way:
:$\ds x_0 = \frac 3 2$
:$\ds \forall n \in \N_{> 0} : x_{n \mathop + 1} = \frac {4 + 3 x_n} {3 + 2 x_n}$
We have that:
:$\forall n \in \N : x_n \ge 0$
Note that:
{{begin-eqn}}
{{eqn | l = x_{n \mathop + 1}^2 - 2
      | r = \frac {\paren {4 + 3 x_n}^2 } {\paren {3 + 2 x_n}^2} - 2
}}
{{eqn | r = \frac {16 + 24 x_n + 9 x_n^2 - 18 - 24 x_n - 8 x_n^2} {\paren {3 + 2 x_n}^2}
}}
{{eqn | r = \frac {x_n^2 - 2} {\paren {3 + 2 x_n}^2}
}}
{{end-eqn}}
Hence:
:$\paren {x_n \ge \sqrt 2} \implies \paren {x_{n \mathop + 1} \ge \sqrt 2}$
For $n = 0$ we have $\ds \frac 9 4 \ge 2$, or $\ds \frac 3 2 \ge \sqrt 2$.
Therefore:
:$\forall n \in \N : x_n \ge \sqrt 2$
Furthermore:
{{begin-eqn}}
{{eqn | l = x_n - x_{n \mathop + 1}
      | r = x_n - \frac {4 + 3 x_n} {3 + 2 x_n}
}}
{{eqn | r = \frac {3 x_n + 2 x_n^2 - 4 - 3x_n} {3 + 2 x_n}
}}
{{eqn | r = \frac {2 \paren {x_n^2 - 2} } {3 + 2 x_n}
}}
{{eqn | o = \ge
      | r = 0
}}
{{end-eqn}}
Hence:
:$\forall n \in \N : x_n \ge x_{n \mathop + 1}$
By monotone convergence theorem, $\sequence {x_n}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$.
By Convergent Sequence in Normed Vector Space is Cauchy Sequence, $\sequence {x_n}_{n \mathop \in \N}$ is Cauchy in $\struct {\R, \size {\, \cdot \,}}$.
On the other hand:
:$\forall n \in \N : x_n \in \Q$
Thus, $\sequence {x_n}_{n \mathop \in \N}$ is also Cauchy in $\struct {\Q, \size {\, \cdot \,}}$.
Suppose $\sequence {x_n}_{n \mathop \in \N}$ converges to $L \in \Q$.
By combination theorem for sequences:
:$\ds L = \frac {4 + 3 L} {3 + 2 L}$
or $L^2 = 2$.
$L$ has to be positive, so $L = \sqrt 2$.
However, $\sqrt 2 \notin \Q$.
Hence, $\sequence {x_n}_{n \mathop \in \N}$ does not converge in $\struct {\Q, \size {\, \cdot \,} }$.
Altogether, we have that $\sequence {x_n}_{n \mathop \in \N} \in \Q$ is a Cauchy sequence which is not convergent in $\struct {\Q, \size {\, \cdot \,} }$.
By definition, $\struct {\Q, \size {\, \cdot \,}}$ is not a Banach space.
{{qed}}
Theorem: Intersection of Open Set with Closure of Set is Subset of Closure of Intersection

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \in \tau$ be an open set of $T$.
Let $K \subseteq S$ be an arbitrary subset of $S$.
Then:
:$H \cap \map \cl K \subseteq \map \cl {H \cap K}$
where $\cl$ denotes set closure.

Proof:
Let $x \in H \cap \map \cl K$.
Then:
:$x \in H$
and:
:$x \in \map \cl K$
Let $N_1$ be an arbitrary neighborbood of $x$.
Because $x \in H$, there exists a neighborbood $N_2$ of $x$ entirely within $U$.
Let $N_3 = N_1 \cap N_2$.
By Intersection of Neighborhoods in Topological Space is Neighborhood, $N_3$ is also a neighborbood of $x$.
Because $N_1 \cap N_2 \subseteq N_2$, $N_3$ is also contained entirely within $U$.
Because $x \in \map \cl K$, by definition of closure, there exists $y \in N_3$ which is also in $K$.
From $x \in N_3$ we know that $y \in H$.
Thus we have demonstrated the existence of $y \in N_1$ that is in $H \cap K$.
As $N_1$ is arbitrary, it follows that for every neighborbood of $x$ there exists $y \in N_1$ such that $y \in H$ and $y \in K$.
That is:
:$x \in \map \cl {H \cap K}$
Hence the result, by definition of subset.
{{qed}}
Theorem: Closure of Non-Empty Bounded Subset of Metric Space is Bounded

Let $M = \struct {A, d}$ be a metric space.
Let $S \subseteq A$ be bounded in $M$.
Then:
:$\map \cl S$ is also bounded in $M$.
where $\map \cl S$ denotes the closure of $S$ in $M$.

Proof:
By definition of bounded:
:$S$ is '''bounded''' {{iff}}:
:$\exists K \in \R: \forall x, y \in M': \map {d_B} {x, y} \le K$
That is, such that $S$ has a diameter $K$.
From Diameter of Closure of Subset is Diameter of Subset, if $S$ has a diameter $K$, then so does $\map \cl S$.
That is, $\map \cl S$ is also bounded.
{{qed}}
Theorem: Image under Projection from Closed Set of Product Topology is not necessarily Closed

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $\pr_1: T \to T_1$ and $\pr_2: T \to T_2$ be the first and second projections from $T$ onto its factors.
Let $K \subseteq S_1 \times S_2$ be closed in $T$.
Then $\pr_1 \sqbrk K$ is not necessarily closed in $T_1$.

Proof:
Proof by Counterexample:
Let $K = \set {\tuple {x, y} \in \R^2: x \ge 0, y \ge \dfrac 1 x}$
Then:
:$\pr_1 \sqbrk K = \openint 0 \to$
$K$ is closed in $T$, as follows:
Consider the mapping $f: \R^2 \to \R$ defined as:
:$\map f {x, y} = x y$
which is continuous on $\R^2$ (see Preimages of $\map f {x, y} = \tuple {x^2 + y^2, x y}$: Continuity).
Then we note that:
$K = f^{-1} \hointr 1 \to \cap p_1^{-1} \hointr 0 \to$
and the result follows.
{{qed}}
Theorem: Interior of Cartesian Product is Product of Interiors

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $H \subseteq T_1$ and $K \subseteq T_2$.
Then:
:$\Int {H \times K} = \Int H \times \Int K$
where $\Int H$, for example, denotes the interior of $H$.

Proof:
By definition of interior, both $\Int H$ and $\Int K$ are open in $T_1$ and $T_2$ respectively.
By Projection from Product Topology is Continuous, it follows that $\Int {H \times K}$ is an open set of $T$.
It remains to be shown that $\Int {H \times K}$ is the largest open subset of $H \times K$.
Let $H' \times K'$ be an open set of $T$ such that $H' \times K' \subseteq H \times K$.
Then from Projection from Product Topology is Open, $H'$ and $K'$ are open sets of $T_1$ and $T_2$ respectively such that $H' \subseteq H$ and $K' \subseteq K$.
By definition of interior, $\Int H$ and $\Int K$ are the largest open sets of $T_1$ and $T_2$ respectively such that $\Int H \subseteq H$ and $\Int K \subseteq K$.
Hence $H' \subseteq \Int H$ and $K' \subseteq \Int K$.
Hence by Cartesian Product of Subsets:
:$H' \times K' \subseteq \Int H \times \Int K$
That is, any open subset of $H \times K$ is also a subset of $\Int {H \times K}$.
Hence $\Int {H \times K}$ is the largest open subset of $H \times K$.
{{qed}}
Theorem: Closure of Cartesian Product is Product of Closures

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $H \subseteq T_1$ and $K \subseteq T_2$.
Then:
:$\map \cl {H \times K} = \map \cl H \times \map \cl K$
where $\map \cl H$, for example, denotes the closure of $H$.

Proof:
Consider the relative complements of $H$ and $K$ in $T_1$ and $T_2$ respectively:
:$\overline H = \relcomp {S_1} H$
:$\overline K = \relcomp {S_2} K$
Then from Interior of Cartesian Product is Product of Interiors:
:$\Int {\overline H \times \overline K} = \Int {\overline H} \times \Int {\overline K}$
From Complement of Interior equals Closure of Complement:
:$\map \cl H = \Int {\overline H}$
:$\map \cl K = \Int {\overline K}$
:$\map \cl {H \times K} = \Int {\overline H \times \overline K}$
{{questionable|The complement of $H \times K$ is ''not'' $\overline H \times \overline K$}}
Hence the result.
{{qed}}
Theorem: Boundary of Cartesian Product of Subsets

Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $T = \struct {T_1 \times T_2, \tau}$ be the product space of $T_1$ and $T_2$, where $\tau$ is the product topology on $S$.
Let $H \subseteq T_1$ and $K \subseteq T_2$.
Then:
:$\map \partial {H \times K} = \paren {\map \partial H \times \map \cl K} \cup \paren {\map \cl H \times \map \partial K}$
where:
:$\map \cl H$, for example, denotes the closure of $H$.
:$\map \partial H$, for example, denotes the boundary of $H$.

Proof:
{{begin-eqn}}
{{eqn | l = \map \partial {H \times K}
      | r = \map \cl {H \times K} \setminus \Int {H \times K}
      | c = {{Defof|Boundary (Topology)}}
}}
{{eqn | r = \paren {\map \cl H \times \map \cl K} \setminus \Int {H \times K}
      | c = Closure of Cartesian Product is Product of Closures
}}
{{eqn | r = \paren {\map \cl H \times \map \cl K} \setminus \paren {\Int H \times \Int K} 
      | c = Interior of Cartesian Product is Product of Interiors
}}
{{eqn | r = \paren {\map \cl H \times \paren {\map \cl K \setminus \Int K} } \cup \paren {\paren {\map \cl H \setminus \Int H} \times \map \cl K}
      | c = Set Difference of Cartesian Products
}}
{{eqn | r = \paren {\map \partial H \times \map \cl K} \cup \paren {\map \cl H \times \map \partial K}
      | c = {{Defof|Boundary (Topology)}}
}}
{{end-eqn}}
{{qed}}
Theorem: Composition of Identification Mappings is Identification Mapping

Let $T_1 = \struct {S_1, \tau_1}$ be a topological space.
Let $S_2$ and $S_3$ be sets.
Let $f_1: S_1 \to S_2$ and $f_2: S_2 \to S_3$ be mappings.
Let $\tau_2$ be the identification topology on $S_2$ with respect to $f_1$ and $\tau_1$.
Let $\tau_3$ be the identification topology on $S_3$ with respect to $f_2$ and $\tau_2$.
Let $\phi: S_1 \to S_3$ be the composition of $f_2$ with $f_1$:
:$\phi = f_2 \circ f_1$
Then $\phi$ is itself an identification mapping.

Proof:
Suppose $V \subseteq S_3$ is an arbitrary subset.
We have the following chain of equivalences:
{{begin-eqn}}
{{eqn | l = V
      | o = \in
      | r = \tau_3
      | c =
}}
{{eqn | ll= \leadstoandfrom
      | l = f_2^{-1} \sqbrk V
      | o = \in
      | r = \tau_2
      | c = $f_2$ is an identification mapping
}}
{{eqn | ll= \leadstoandfrom
      | l = f_1^{-1} \sqbrk {f_2^{-1} \sqbrk V}
      | o = \in
      | r = \tau_1
      | c = $f_1$ is an identification mapping
}}
{{eqn | ll= \leadstoandfrom
      | l = \phi^{-1} \sqbrk V
      | o = \in
      | r = \tau_1
      | c = Preimage of Subset under Composite Mapping
}}
{{end-eqn}}
Thus, $\phi$ is an identification mapping.
{{qed}}
Theorem: Graph of Continuous Mapping to Hausdorff Space is Closed in Product

Let $T_A = \struct {A, \tau_A}$ and $T_B = \struct {B, \tau_B}$ be topological spaces.
Let $T_B$ be a $T_2$ (Hausdorff) space.
Let $f: T_A \to T_B$ be a continuous mapping.
Then the graph of $f$ is a closed subset of $T_A \times T_B$ under the product topology.

Proof:
Let $G_f$ be the graph of $f$:
:$G_f = \set {\tuple {x, y} \in A \times B: \map f x = y}$
Let $I_B: T_B \to T_B$ be the identity mapping on $B$:
:$\forall y \in B: \map {I_B} y = y$
From Identity Mapping is Continuous, $I_B$ is continuous on $T_B$.
Let $f \times I_B: T_A \times T_B \to T_B \times T_B$ be the product map:
:$\map {f \times I_B} {x, y} = \tuple {\map f x, y}$
From Continuous Mapping to Product Space, $f \times I_B$ is continuous.
Let $\Delta_B$ be the diagonal relation on $B$.
From Hausdorff Space iff Diagonal Set on Product is Closed, $\Delta_B$ is a closed set of $T_B \times T_B$ under the product topology.
From Continuity Defined from Closed Sets:
{{begin-eqn}}
{{eqn | l = (\paren {f \times I_B}^{-1} \sqbrk {\Delta_B}
      | r = \set {\tuple {x, y} \in A \times B: \map {f \times I_B} {x, y} \in \Delta_B}
}}
{{eqn | r = \set {\tuple {x, y} \in A \times B : \tuple {\map f x, y} \in \Delta_B}
}}
{{eqn | r = \set {\tuple {x, y} \in A \times B : \map f x = y}
}}
{{eqn | r = G_f
}}
{{end-eqn}}
is closed in $T_A \times T_B$ endowed with the product topology.
{{qed}}
Theorem: Intersection of Open Sets of Hausdorff Space containing Point is Singleton

Let $T = \struct {S, \tau}$ be a Hausdorff space.
Let $x \in S$ be arbitrary.
Then the intersection of all open sets containing $x$ is $\set x$:
:$\ds \bigcap_{\substack {H \mathop \in \tau \\ x \mathop \in H} } = \set x$

Proof:
Let $x \in S$.
Let $K = \ds \bigcap_{\substack {H \mathop \in \tau \\ x \mathop \in H} }$, that is, the intersection of all open sets containing $x$.
{{AimForCont}} there exists $y \in S$ such that $y \in K$ but $y \ne x$.
By definition of Hausdorff space, there exist disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively.
We have that $y \in V$.
But we also have that $y \in K$.
By Intersection is Subset:
:$K \subseteq U$
and so:
:$y \in U$
So $y \in U$ and $y \in V$ and so $U \cap V \ne \O$.
This directly contradicts our assertion that $U, V \in \tau$ are disjoint.
In turn, this contradicts our supposition that $T$ is a Hausdorff space.
Hence by Proof by Contradiction there can be no such $y$ such that $y \in K$.
The result follows.
{{qed}}
Theorem: Space such that Intersection of Open Sets containing Point is Singleton may not be Hausdorff

Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$ be arbitrary.
Let $T$ be such that the intersection of all open sets containing $x$ is $\set x$:
:$\ds \bigcap_{\substack {H \mathop \in \tau \\ x \mathop \in H} } = \set x$
Then it is not necessarily the case that $T$ is a $T_2$ (Hausdorff) space.

Proof:
Let $T$ be the finite complement topology on the real numbers $\R$, for example.
The open sets of $T$ are subsets of $\R$ of the form $U$ such that $\R \setminus U$ is finite, together with $\O$.
Let $x \in \R$ be arbitrary.
Let $K = \ds \bigcap_{\substack {H \mathop \in \tau \\ x \mathop \in H} }$, that is, the intersection of all open sets of $T$ containing $x$.
Let $y \in \R$ such that $y \ne x$.
Note that the set $\set y$ is finite.
Thus $\R \setminus \set y$ is an open set of $T$ which contains $x$ but not $y$.
Hence $y \notin K$.
As $y$ is arbitrary, it follows that the only element of $\R$ which is in $K$ is $x$ itself.
That is:
:$\ds \bigcap_{\substack {H \mathop \in \tau \\ x \mathop \in H} } = \set x$
But from Finite Complement Space is not $T_2$, $T$ is not a $T_2$ (Hausdorff) space.
{{qed}}
Theorem: Point at Zero Distance from Subset of Metric Space is Limit Point or Element

Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be an arbitrary subset of $A$.
Let $x \in A$ be arbitrary.
Let $\map d {x, H}$ denote the distance between $x$ and $H$:
:$\ds \map d {x, H} = \inf_{y \mathop \in H} \paren {\map d {x, y} }$
Then:
:$\map d {x, H} = 0$
{{iff}}:
:either $x \in H$ or $x$ is a limit point of $H$.

Proof:
=== Necessary Condition ===
Let $x$ be such that either $x \in H$ or $x$ is a limit point of $H$.
If $x \in H$ then:
:$\map d {x, H} = 0$
from Distance from Subset to Element.
Otherwise $x$ is a limit point of $H$.
Then from Limit Point of Subset of Metric Space is at Zero Distance:
:$\map d {x, H} = 0$
{{qed|lemma}}
Theorem: Limit Point of Subset of Metric Space is at Zero Distance

Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be an arbitrary subset of $A$.
Let $x \in A$ be a limit point of $H$.
Let $\map d {x, H}$ denote the distance between $x$ and $H$:
:$\ds \map d {x, H} = \inf_{y \mathop \in H} \paren {\map d {x, y} }$
Then:
:$\map d {x, H} = 0$

Proof:
Let $x$ be a limit point of $H$.
{{AimForCont}} $\map d {x, H} \ne 0$.
By definition of metric, that means:
:$\map d {x, H} > 0$
Then:
:$\exists \epsilon \in \R_{>0}: \forall y \in H: \map d {x, y} > \epsilon$
That is:
:$\forall y \in H: y \notin \map {B_\epsilon} x \setminus \set x$
where $\map {B_\epsilon} x \setminus \set x$ denotes the deleted $\epsilon$-neighborhood of $x$.
That is:
:$\paren {\map {B_\epsilon} x \setminus \set x} \cap H = \O$
Hence, by definition, $x$ is not a limit point of $H$.
From this contradiction it follows that it cannot be the case that $\map d {x, H} \ne 0$
That is:
:$\map d {x, H} = 0$
{{qed}}
Theorem: Point not in Subset of Metric Space iff Distance from Elements is Greater than Zero

Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be an arbitrary subset of $A$.
Let $x \in A$ be arbitrary.
Then:
:$x \notin H$
{{iff}}:
:$\forall y \in H: \map d {x, y} > 0$

Proof:
{{begin-eqn}}
{{eqn | q = \forall y \in H
      | l = \map d {x, y}
      | o = >
      | r = 0
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall y \in H
      | l = \map d {x, y}
      | o = \ne
      | r = \map d {x, x}
      | c = {{Metric-space-axiom|1}}
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall y \in H
      | l = y
      | o = \ne
      | r = x
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | o = \notin
      | r = H
      | c = 
}}
{{end-eqn}}
{{qed}}
Category:Distance Function
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Theorem: Point at Distance Zero from Closed Set is Element

Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be an arbitrary subset of $A$.
Let $x \in A$ be arbitrary.
Let $\map d {x, H}$ denote the distance between $x$ and $H$:
:$\ds \map d {x, H} = \inf_{y \mathop \in H} \paren {\map d {x, y} }$
Let $H$ be closed in $M$.
Then:
:$\map d {x, H} = 0$ {{iff}} $x \in H$.

Proof:
=== Necessary Condition ===
Let $x \in H$.
Then from Distance from Subset to Element:
:$\map d {x, H} = 0$
whether $H$ is closed or not.
{{qed|lemma}}
Theorem: Square Root is not Lipschitz Continuous

Let $\sqrt {\size x} : \R \to \R_{\ge 0}$ be a real function.
$\sqrt {\size x}$ is not Lipschitz continuous.

Proof:
{{AimForCont}} $\sqrt {\size x}$ is Lipschitz continuous.
Then:
:$\exists L \in \R_{> 0} : \forall x, y \in \R : \size {\sqrt {\size x} - \sqrt {\size y} } \le L \size {x - y}$
Suppose $x = \dfrac 1 {n^2}$ with $n \in \N$ and $y = 0$.
Then:
:$\dfrac 1 n \le \dfrac L {n^2}$
In other words:
:$\forall n \in \N : n \le L$
However, $L$ was assumed to be finite.
Hence the contradiction.
{{qed}}
Theorem: Triangle Inequality on Distance from Point to Subset

Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$.
Then:
:$\forall x, y \in A: \map d {x, H} \le \map d {x, y} + \map d {y, H}$
where $\map d {x, H}$ denotes the distance between $x$ and $H$.

Proof:
{{begin-eqn}}
{{eqn | q = \forall z \in H
      | l = \map d {y, z}
      | o = \ge
      | r = \map d {x, z} - \map d {x, y}
      | c = {{Metric-space-axiom|2}}
}}
{{eqn | ll= \leadsto
      | q = \forall z \in H
      | l = \map d {y, z}
      | o = \ge
      | r = \map d {x, H} - \map d {x, y}
      | c = {{Defof|Distance between Element and Subset of Metric Space}}
}}
{{eqn | ll= \leadsto
      | l = \map d {y, H}
      | o = \ge
      | r = \map d {x, H} - \map d {x, y}
      | c = {{Defof|Distance between Element and Subset of Metric Space}}
}}
{{end-eqn}}
Hence the result.
{{qed}}