nyuuzyou/wikis huggingface dataset:
Metric Space is T4/Proof 1

== Theorem ==
== Proof ==
Let $H$ and $K$ be disjoint closed sets of $M$.
Let $g: A \to \R$ be defined as:
:$g = f_K - f_H$
where:
:$\forall x \in A: \map {f_K} x = \map d {x, K}$
:$\forall x \in A: \map {f_H} x = \map d {x, H}$
where $\map d {x, K}$, $\map d {x, H}$ denotes the distance from $x$ to $K$ and from $x$ to $H$ respectively.
From Distance from Point to Subset is Continuous Function, both $\map {f_K} x$ and $\map {f_H} x$ are continuous mappings.
Hence $g$ itself is a continuous mapping.
Consider the disjoint open intervals $\openint 0 \to$ and $\openint \gets 0$.
Because $g$ is continuous, it follows that $g^{-1} \sqbrk {\openint 0 \to}$ and $g^{-1} \sqbrk {\openint \gets 0}$ are open in $M$ and disjoint.
Let $x \in H$.
Then:
:$\map {f_H} x = 0$
Because $H$ and $K$ are disjoint:
:$H \cap K = \O$
and so:
:$x \notin K$
Because $K$ is closed sets in $M$, it follows from Point at Distance Zero from Closed Set is Element that:
:$\map {f_K} x > 0$
It follows that:
:$\map g x > 0$
that is:
:$H \subseteq g^{-1} \sqbrk {\openint 0 \to}$
Similarly:
:$K \subseteq g^{-1} \sqbrk {\openint \gets 0}$
The result follows.
== Sources ==
* : $4$: The Hausdorff condition: Exercise $4.3: 5$
Category:Metric Space is T4
Metric Space is T5

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $M$ is a $T_5$ space.
== Proof ==
Let $S, T \subseteq A$ such that $S$ and $T$ are separated in $A$.
Then:
:each point $x \in S$ has an open $\epsilon$-ball $\map {B_{\epsilon_x} } x$ which is disjoint from $T$
:each point $y \in T$ has an open $\epsilon$-ball $\map {B_{\epsilon_y} } y$ which is disjoint from $S$.
Then:
:$U_S = \ds \bigcup_{x \mathop \in S} \map {B_{\epsilon_x / 2} } x$
:$U_T = \ds \bigcup_{y \mathop \in T} \map {B_{\epsilon_y / 2} } y$
are disjoint open neighborhoods of $S$ and $T$ respectively.
Hence the result by the definition of $T_5$ space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:T5 Spaces
Metric Space is Weakly Countably Compact iff Countably Compact

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $M$ is weakly countably compact $M$ is countably compact.
== Proof ==
From Metric Space fulfils all Separation Axioms, a metric space is a $T_1$ (Fréchet) space.
The result follows from $T_1$ Space is Weakly Countably Compact iff Countably Compact.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Countably Compact Spaces
Category:Weakly Countably Compact Spaces
Metric Space is Weakly Locally Compact iff Strongly Locally Compact

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Then $M$ is weakly locally compact $M$ is strongly locally compact.
== Proof ==
We have from Strongly Locally Compact Space is Weakly Locally Compact that strong local compactness implies weak local compactness in all topological spaces, regardless of whether they are metric spaces or not.
So all we need to do is demonstrate that if $M$ is weakly locally compact then it is strongly locally compact.
We have that a metric space is a $T_2$ (Hausdorff) space.
The result follows from Weakly Locally Compact Hausdorff Space is Strongly Locally Compact.
== Also see ==
* Definition:Locally Compact Hausdorff Space
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces
Category:Metric Spaces
Category:Weakly Locally Compact Spaces
Category:Strongly Locally Compact Spaces
Category:Locally Compact Hausdorff Spaces
Metric Spaces on Topologically Equivalent Metrics on same Underlying Set are Homeomorphic

== Theorem ==
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 $A$ :
: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 $f$ is $\tuple {d, d_2}$-continuous
::$(2): \quad g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous.
Set $B = C = A$.
The result follows by definition of homeomorphic metric spaces.
Category:Topologically Equivalent Metrics
Category:Homeomorphisms (Metric Spaces)
Metric Subspace Induces Subspace Topology

== Theorem ==
Let $M = \struct {A,d}$ be a metric space.
Let $H \subseteq A$.
Let $\tau$ be the topology induced by the metric $d$.
Let $\tau_H$ be the subspace topology induced by $\tau$ on $H$.
Let $d_H$ be the subspace metric induced by $d$ on $H$.
Let $\tau_{d_H}$ be the topology induced by the metric $d_H$.
Then:
:$\tau_{d_H} = \tau_H$
== Proof ==
Let $\BB$ be the set of open $\epsilon$-balls in $M$.
Let $\BB_H$ be the set of open $\epsilon$-balls in $\struct {H, d_H}$.
Let $U \in \tau_{d_H}$.
By the definition of the topology induced by the metric $d_H$:
:$\exists \AA_H \subseteq \BB_H: U = \bigcup \AA_H$
Let $\AA = \set {B': B' \in \BB, B' \cap H \in \AA_H}$.
Let $V = \bigcup \set {B' : B' \in \AA}$.
By the definition of the topology induced by the metric $d$:
:$V \in \tau$
By the definition of the subspace metric:
:$\forall B \in \BB_H: \exists B' \in \BB: B = B' \cap H$
Hence:
By the definition of the subspace topology induced by $\tau$ on $H$:
:$U \in \tau_H$
Hence:
:$\tau_{d_H} \subseteq \tau_H$.
Let $U \in \tau_H$.
By the definition of the subspace topology induced by $\tau$ on $H$:
:$\exists V \in \tau : U = V \cap H$
By the definition of the topology induced by the metric $d$:
:$\exists \AA \subseteq \BB: V = \bigcup \set {B' : B' \in \AA}$
Hence:
By the definition of the subspace metric:
:$\forall B' \in \BB: B' \cap H \in \BB_H$
By the Definition of the topology induced by the metric $d_H$:
:$U \in \tau_{d_H}$
Hence:
:$\tau_H \subseteq \tau_{d_H}$
By Equivalence of Definitions of Set Equality:
:$\tau_H = \tau_{d_H}$
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 13$
* : $\S 3$: New Spaces From Old: Subspaces
Category:Metric Subspaces
Category:Topological Subspaces
Metric formed by Arc Length on Circle is Lipschitz Equivalent to Euclidean Metric

== Theorem ==
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.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 19$
Category:Examples of Metric Spaces
Category:Lipschitz Equivalence
Metric is Continous Mapping

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Consider the distance function:
:$d: A \times A \to \R$
Then $\R$ is a continuous function.
== Proof ==
Let $\epsilon > 0$.
Let $\tuple {x_1, x_2} \in A \times A$.
Let $\delta = \dfrac \epsilon 2$.
Then $U = \map {B_\delta} {x_1} \times \map {B_\delta} {x_2}$ is a neighborhood of $\tuple {x_1, x_2}$ in $X \times X$ such that:
:$d \sqbrk U \subseteq \openint {\map d {x_1, x_2} - \epsilon} {\map d {x_1, x_2} + \epsilon}$
Therefore $d: X \times X \to \mathbb R$ is a continuous function.
Category:Metric Spaces
Category:Continuous Mappings
Metric on P-adic Numbers Extends Metric on Rationals

== Theorem ==
Let $p$ be any prime number.
Let $\struct{\Q, \norm {\,\cdot\,}^\Q_p}$ be the rational numbers $\Q$ with the $p$-adic norm $\norm {\,\cdot\,}^\Q_p$.
Let $d^{\Q}_p$ be the $p$-adic metric on the rational numbers.
Let $\struct{\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $d_p$ be the $p$-adic metric on the $p$-adic numbers.
Then $d_p$ is an extension of $d^{\Q}_p$.
== Proof ==
From Rational Numbers are Dense Subfield of P-adic Numbers:
:$\norm {\,\cdot\,}_p$ on $\Q_p$ is an extension of $\norm {\,\cdot\,}^Q_p$ on $\Q$.
By definition of $p$-adic metric on the rational numbers:
:$\forall x, y \in \Q : \map {d^\Q_p} {x, y} = \norm{x - y}^\Q_p$
By definition of $p$-adic metric on the $p$-adic numbers:
:$\forall x, y \in \Q : \map {d_p} {x, y} = \norm{x - y}_p$
It follows that the $p$-adic metric on $\Q_p$ is an extension of the $p$-adic metric on $\Q$.
Category:P-adic Metrics
Metric on Shift of Finite Type is Metric

== Theorem ==
Let $\struct {X _\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.
Let $\theta \in \openint 0 1$.
Then the metric $d_\theta$ is indeed a metric on $X_\mathbf A$.
That is, $\struct {X _\mathbf A, d _\theta} $ is a metric space.
== Proof ==
=== M1 ===
Let $x\in X _\mathbf A$
Obviously, $x_i = x_i$ for all $i \in \openint {-\infty} \infty$.
Therefore:
:$\map {d _\theta} {x,x} = \theta ^\infty = 0$.
=== M3 ===
This is clear from the definition.
=== M4 ===
Let $x,y\in X _\mathbf A$ such that $x\ne y$.
That is:
:$\exists k\in\Z : x_k\ne y_k$.
Thus:
:$N:=\sup \set {n \in \N \cup \set \infty : x_i = y_i \text { for all } i \in \openint {-n} n}\leq \size k$
Therefore:
:$\map {d _\theta} {x,x} = \theta ^N > 0$.
=== M2 ===
Let $x,y,z\in X _\mathbf A$.
Let:
:$N_1 := \sup \set {n \in \N \cup \set \infty : x_i = y_i \text { for all } i \in \openint {-n} n}$
and:
:$N_2 := \sup \set {n \in \N \cup \set \infty : y_i = z_i \text { for all } i \in \openint {-n} n}$
so that:
:$\map {d _\theta} {x,y} = \theta ^{N_1}$
and:
:$\map {d _\theta} {y,z} = \theta ^{N_2}$
, say $N_1 \leq N_2$.
Then for all $i \in \openint {-N_1} {N_1}$:
:$x_i = y_i = z_i$
Hence:
:$\sup \set {n \in \N \cup \set \infty : x_i = z_i \text { for all } i \in \openint {-n} n} \geq N_1$
so that:
:$\map {d _\theta} {x,z} \leq \theta ^{N_1}$
Therefore:
:$\map {d _\theta} {x,z} \leq \theta ^{N_1} \leq \theta ^{N_1} + \theta ^{N_2} = \map {d _\theta} {x,y} + \map {d _\theta} {y,z}$
Category:Metric Spaces
Metric on Shift of Finite Type is Non-Archimedean

== Theorem ==
Let $\struct {X_\mathbf A, \sigma_\mathbf A}$ be a shift of finite type.
Let $\theta \in \openint 0 1$.
Then the metric $d_\theta$ on $X_\mathbf A$ is non-archimedean metric.
That is, $\struct {X_\mathbf A, d _\theta} $ is an ultrametric space.
== Proof ==
Let $x, y, z \in X_\mathbf A$.
Let:
:$N_1 := \sup \set {n \in \N \cup \set \infty : x_i = y_i \text { for all } i \in \openint {-n} n}$
and:
:$N_2 := \sup \set {n \in \N \cup \set \infty : y_i = z_i \text { for all } i \in \openint {-n} n}$
so that:
:$\map {d_\theta} {x, y} = \theta^{N_1}$
and:
:$\map {d_\theta} {y, z} = \theta^{N_2}$
Let $M := \min \set {N_1, N_2}$.
Then for all $i \in \openint {-M} M$:
:$x_i = y_i = z_i$
Hence:
:$\sup \set {n \in \N \cup \set \infty : x_i = z_i \text { for all } i \in \openint {-n} n} \ge M$
so that:
:$\map {d_\theta} {x, z} \le \theta ^M$
On the other hand:
:$\theta^M = \theta^{\min \set {N_1, N_2} } = \max \set {\theta^{N_1}, \theta^{N_2} }$
Therefore:
:$\map {d_\theta} {x, z} \le \max \set {\theta^{N_1}, \theta^{N_2} } = \max \set {\map {d_\theta} {x, y}, \map {d_\theta} {y, z} }$
Category:Metric Spaces
Metric over 1 plus Metric forms Metric

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $d_3: A^2 \to \R$ be the mapping defined as:
:$\forall \tuple {x, y} \in A^2: \map {d_3} {x, y} = \dfrac {\map d {x, y} } {1 + \map d {x, y} }$
Then $d_3$ is a metric for $A$.
=== Topological Equivalence ===
== Proof ==
It is to be demonstrated that $d_3$ satisfies all the metric space axioms.
=== Proof of ===
So holds for $d_3$.
=== Proof of ===
Note that $\map f x = \dfrac x {1 + x}$ is an increasing function of $x$ for $x = 0$.
In order to simplify the algebra, let $a = \map d {x, y}$, $b = \map d {y, z}$ and $c = \map d {x, z}$.
Then:
But $a + b + 2 a b > c$.
Because $\map f x = \dfrac x {1 + x}$ is an increasing function:
:$\dfrac {a + b + 2 a b} {1 + a + b + 2 a b} > \dfrac c {1 + c}$
and it follows by definition of $d_3$ that:
:$\map {d_3} {x, y} + \map {d_3} {y, z} \ge \map {d_3} {x, z}$
So holds for $d_3$.
=== Proof of ===
So holds for $d_3$.
=== Proof of ===
So holds for $d_3$.
Thus $d_3$ satisfies all the metric space axioms and so is a metric.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 6$
Category:Examples of Metric Spaces
Metric over 1 plus Metric forms Metric/Topological Equivalence

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $d_3: A^2 \to \R$ be the metric on $M$ defined as:
:$\forall \tuple {x, y} \in A^2: \map {d_3} {x, y} = \dfrac {\map d {x, y} } {1 + \map d {x, y} }$
$d_3$ 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:
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:
The result follows.
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 20$
Category:Examples of Metric Spaces
Metrics on Space are Lipschitz Equivalent iff Identity Mapping is Lipschitz Equivalence

== Theorem ==
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $I_A$ denote the identity mapping on $A$.
Then:
:$d_1$ and $d_2$ are Lipschitz equivalent
:
:$I_A: M_1 \to M_2$ is a Lipschitz equivalence.
== Proof ==
By definition of identity mapping:
:$\forall x \in A: \map {I_A} x = x$
=== Sufficient Condition ===
Let $d_1$ and $d_2$ be Lipschitz equivalent.
Then:
That is, by definition:
:$I_A: M_1 \to M_2$ is a Lipschitz equivalence.
=== Necessary Condition ===
Let $I_A$ be a Lipschitz equivalence.
Then:
That is, by definition:
:$d_1$ and $d_2$ are Lipschitz equivalent metrics.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics
Category:Lipschitz Equivalence
Metrics on Space are Topologically Equivalent iff Identity Mapping is Homemorphism

== Theorem ==
Let $M_1 = \struct {A, d_1}$ and $M_2 = \struct {A, d_2}$ be metric spaces on the same underlying set $A$.
Let $I_A$ denote the identity mapping on $A$.
Then:
:$d_1$ and $d_2$ are topologically equivalent
:
:$I_A: M_1 \to M_2$ is a homeomorphism.
== Proof ==
First we establish the following:
From Identity Mapping is Bijection, we have that $I_A$ is a bijection.
From Inverse of Identity Mapping, $\paren {I_A}^{-1} = I_A$.
By Identity Mapping on Metric Space is Continuous:
:$(1): \quad I_A$ is a $\tuple {d_1, d_1}$-continuous mapping.
:$(2): \quad I_A^{-1} = I_A$ is a $\tuple {d_2, d_2}$-continuous mapping.
=== Sufficient Condition ===
Let $d_1$ and $d_2$ be topologically equivalent.
Hence, by definition:
: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 $f$ is $\tuple {d, d_2}$-continuous
::$(2): \quad g$ is $\tuple {d_1, d'}$-continuous $g$ is $\tuple {d_2, d'}$-continuous.
In particular, for all mappings $f: A \to A$ and $g: A \to A$:
::$(1): \quad f$ is $\tuple {d_2, d_1}$-continuous $f$ is $\tuple {d_2, d_2}$-continuous
::$(2): \quad g$ is $\tuple {d_1, d_1}$-continuous $g$ is $\tuple {d_2, d_1}$-continuous.
By definition of topological equivalence:
:$I_A$ is a $\tuple {d_1, d_2}$-continuous mapping from $(1)$
:$I_A$ is a $\tuple {d_2, d_1}$-continuous mapping from $(2)$.
Hence, $I_A$ is a bijection such that:
:$I_A$ is continuous from $M_1$ to $M_2$
:$I_A^{-1}$ is continuous from $M_2$ to $M_1$.
That is, $I_A$ is a homeomorphism.
=== Necessary Condition ===
Let $I_A$ be a homeomorphism.
That is:
:$I_A$ is continuous from $M_1$ to $M_2$
:$I_A^{-1}$ is continuous from $M_2$ to $M_1$.
it having been established that $I_A$ is a bijection such that $I_A^{-1} = I_A$.
Let $U \subseteq A$ be open in $M_2$.
Because $I_A$ is a homeomorphism:
:$I_A^{-1} \sqbrk U$ is open in $M_1$.
Let $U \subseteq A$ be open in $M_1$.
Because $I_A^{-1} = I_A$ is a homeomorphism:
:$I_A \sqbrk U$ is open in $M_2$.
That is:
:$U \subseteq A$ is $d_1$-open $\iff$ $U \subseteq A$ is $d_2$-open.
That is:
:$d_1$ and $d_2$ are topologically equivalent.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics
Category:Topologically Equivalent Metrics
Metrizable Space is Hausdorff

== Theorem ==
Let $T$ be a metrizable topological space.
Then $T$ is a $T_2$ (Hausdorff) space.
== Proof ==
By definition, $T$ is homeomorphic to a topological space $\struct {S, \tau_d}$ such that $\tau_d$ is the topology induced by a metric $d$ on $S$.
From Metric Space is Hausdorff, $\struct {S, d}$ is a $T_2$ (Hausdorff) space.
As the open sets of $\struct {S, d}$ are the same as the open sets of $\struct {S, \tau_d}$, it follows that $\struct {S, \tau_d}$ is a $T_2$ (Hausdorff) space.
From Hausdorff Condition is Preserved under Homeomorphism it follows that $T$ is also a $T_2$ (Hausdorff) space.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Exercise $1$
Category:Hausdorff Spaces
Category:Metrizable Topologies
Metrizable Space is not necessarily Second-Countable

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is metrizable.
Then it is not necessarily the case that $T$ is second-countable.
== Proof ==
Let $T$ be an uncountable discrete space.
From Standard Discrete Metric induces Discrete Topology, $T$ is metrizable.
From Uncountable Discrete Space is not Second-Countable, $T$ is not second-countable.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $5$: Metric Spaces: Metrizability
Category:Metrizable Topologies
Category:Second-Countable Spaces
Metrization Theorems

== Theorem ==
=== Nagata-Smirnov Metrization Theorem ===
=== Bing's Metrization Theorem ===
=== Smirnov Metrization Theorem ===
=== Urysohn's Metrization Theorem ===
=== Frink's Metrization Theorem ===
Category:Metrization Theorems
Category:Metrizable Topologies
Metrization of Regular Second Countable Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_1$ space.
:$(1): \quad T$ is regular and second-countable
:$(2): \quad T$ is homeomorphic to a metric subspace of the Hilbert cube $I^\omega$
:$(3): \quad T$ is metrizable and separable
== Proof ==
=== Condition $(1)$ implies Condition $(2)$ ===
Follows immediately from Regular Second-Countable Space is Homeomorphic to Subspace of Hilbert Cube.
=== Condition $(2)$ implies Condition $(3)$ ===
Let $T$ be homeomorphic to a subspace of the Hilbert cube $\struct{I^\omega, d_2}$ 
where $d_2$ is the metric: $\ds \forall x = \sequence {x_i}, y = \sequence {y_i} \in I^\omega: \map {d_2} {x, y} := \paren {\sum_{k \mathop \in \N_{>0} } \paren {x_k - y_k}^2}^{\frac 1 2}$
By definition of metrizable topology:
:$\struct{I^\omega, \tau_{d_2}}$ is metrizable
where $\tau_{d_2}$ is the topology induced by $d_2$.
From Subspace of Metrizable Space is Metrizable Space:
:$T$ is homeomorphic to a metrizable space
It follows that $T$ is metrizable by definition.
From Hilbert Cube is Separable:
:the Hilbert cube $\struct{I^\omega, d_2}$ is separable
From Subspace of Separable Metric Space is Separable:
:$T$ is homeomorphic to a separable space
From Continuous Image of Separable Space is Separable:
:$T$ is separable
=== Condition $(3)$ implies Condition $(1)$ ===
Let $T$ be metrizable and separable.
By definition of metrizable space:
:there exists a metric $d$ on $S$ such that the topology $\tau$ on $S$ is the topology induced by $d$
From Metric Space is Separable iff Second-Countable
:$\struct{S, d}$ is a second-countable space
From Metric Space is Fully Normal:
:$\struct{S, \tau}$ is a fully normal space
From Fully Normal Space is Normal Space:
:$\struct{S, \tau}$ is a normal space
From Normal Space is Regular Space:
:$\struct{S, \tau}$ is a regular space
== Also known as ==
In some sources this theorem is known as Urysohn's Metrization Theorem as it includes the sufficient condition to the necessary conditions of metrization that is the subject of what is more commonly known as Urysohn's Metrization Theorem.
== Also see ==
* Urysohn's Metrization Theorem
== Sources ==
* : Chapter $4$: Embedding and Metrization
* : Chapter $7$: Metrizable Spaces: $\S23$: Metrization: Definition $23.1$
Category:Metrizable Topologies
Category:Regular Spaces
Category:Separable Spaces
Category:Second-Countable Spaces
Category:T1 Spaces
Mid-Interval Value/Examples/Heights of People

== Examples of Mid-Interval Values ==
For $200$ people, their exact height $x$ in centimetres is measured.
The observations are grouped as follows:
$\begin {array} {r|l} \hline
\text {Height $x$ (cm)} & \text {Count} \\
\hline
x < 165 & 6 \\
165 \le x < 170 & 39 \\
170 \le x < 175 & 93 \\
175 \le x < 180 & 44 \\
180 \le x < 185 & 15 \\
x \ge 185 & 3 \\
\hline \end {array}$
The mid-interval values for the above data are:
:$162.5, 167.5, 172.5, 177.5, 182.5, 187.5$
== Sources ==
* 
* 
Category:Examples of Mid-Interval Values
Middle Term of Valid Categorical Syllogism is Distributed at least Once

== Theorem ==
Let $M$ be the middle term of a valid categorical syllogism $Q$.
Then $M$ is distributed in at least one of the premises of $Q$ in which it appears.
== Proof ==
== Also see ==
To violate this rule is to commit the Fallacy of Undistributed Middle.
* Rules of Quantity
* Distributed Term of Conclusion of Valid Categorical Syllogism is Distributed in Premise
== Historical Note ==
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism: Exercises: $\text{A}$: Rules of Quantity: $2$
Category:Rules of Quantity
Category:Distributed Terms of Categorical Syllogisms
Midline Theorem

== Theorem ==
The midline of a triangle is parallel to the third side of that triangle and half its length.
== Proof ==
Let $\triangle ABC$ be a triangle.
Let $DE$ be the midline of $\triangle ABC$ through $AB$ and $AC$.
Extend $DE$ to $DF$ so $DE = EF$.
As $E$ is the midpoint of $AC$, the diagonals of the quadrilateral $ADCF$ bisect each other.
From Quadrilateral with Bisecting Diagonals is Parallelogram, $ADCF$ is a parallelogram.
By definition of a parallelogram, $AB \parallel CF$.
From Opposite Sides and Angles of Parallelogram are Equal, $AD = CF$.
But $AD = DB$ as $D$ is the midpoint of $AB$.
So $DB = CF$ and $DB \parallel CF$.
From Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel, $BCFD$ is a parallelogram.
Thus also by Quadrilateral is Parallelogram iff One Pair of Opposite Sides is Equal and Parallel $DF = BC$ and $DF \parallel BC$.
As $DE = EF$, $DE$ is the midpoint of $DF$ and so $DE = \dfrac 1 2 DF$.
Thus $DE = \dfrac 1 2 BC$ and $DE \parallel BC$.
Hence the result.
== Also known as ==
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.26$: Corollary $1$
* 
* : $(4)$
* 
* : $(4)$
* 
Category:Midline Theorem
Category:Triangles
Category:Named Theorems