avewright/proofwiki-math huggingface dataset:
Theorem: Locally Euclidean Space is Locally Compact

Let $M$ be a locally Euclidean space of some dimension $d$.
Then $M$ is locally compact.

Proof:
Let $m \in M$ be arbitrary.
By definition of locally Euclidean space, there exists an open neighborhood $U$ of $m$, homeomorphic to an open subset of $\R^d$.
By the definition of an open set, there is some open ball:
:$B = \map {B_\delta} {\map \phi m} = \set {x \in \R^d: \size {x - \map \phi m} < \delta}$
of radius $\delta$ containing $\map \phi m$, contained in $U$.
By Closure of Open Ball in Metric Space and Topological Closure is Closed, the set:
:$C = \set {x \in \R^d: \size {x - \map \phi m} \le \dfrac \delta 2}$
is closed, and $C \subseteq B \subseteq U$.
Moreover, $C$ is trivially bounded, hence compact by the Heine-Borel Theorem.
Now if $\phi$ is a homeomorphism $U \to \R^d$, then by definition $\phi^{-1}$ is continuous.
Therefore by Continuous Image of Compact Space is Compact, $\phi^{-1} \sqbrk C \subseteq M$ is compact.
Furthermore $m \in \phi^{-1} \sqbrk C$ because $\map \phi m \in C$.
Thus every point of $M$ has a compact neighborhood.
{{Explain|Why does the existence of one '''compact neighborhood''' imply the existence of a '''neighborhood basis of compact sets'''?}}
{{qed}}
Theorem: Locally Euclidean Space is Locally Compact

Let $M$ be a locally Euclidean space of some dimension $d$.
Then $M$ is locally compact.

Proof:
Let $m \in M$ be arbitrary.
From Locally Euclidean Space has Countable Neighborhood Basis Homeomorphic to Closed Balls:
:there exists a countable neighborhood basis $\family{K_n}_{n \in \N}$ of $m$ where each $N_n$ is the homeomorphic image of a closed ball of $\R^d$
From Closed Ball in Euclidean Space is Compact:
:$\forall n \in \N: K_n$ is the homeomorphic image of a compact subspace
From Continuous Image of Compact Space is Compact:
:$\forall n \in \N: K_n$ is a compact subspace
Hence:
:there exists a countable neighborhood basis $\family{K_n}_{n \in \N}$ of $m$ consisting of compact subspaces
It follows that $M$ is locally compact by definition.
{{qed}}
Theorem: Unique Factorization Domain is Integrally Closed

Let $A$ be a unique factorization domain (UFD).
Then $A$ is integrally closed.

Proof:
Let $K$ be the field of quotients of $A$.
Let $x \in K$ be integral over $A$.
Let:
$x = a / b$
for $a, b \in A$ with $\gcd \set {a, b} \in A^\times$.
This makes sense because a UFD is GCD Domain.
There is an equation:
:$\paren {\dfrac a b}^n + a_{n - 1} \paren {\dfrac a b}^{n - 1} + \dotsb + a_0$
{{explain|This is not actually an equation, it has no equals sign in it.}}
with $a_i \in A$, $i = 0, \dotsc, n - 1$.
Multiplying by $b^n$, we obtain:
:$a^n + b c = 0$
with $c \in A$.
Therefore:
:$b \divides a^n$
{{AimForCont}} $b$ is not a unit.
Then:
:$\gcd \set {a, b} \notin A^\times$
which is a contradiction.
So $b$ is a unit, and:
:$x = a b^{-1} \in A$
By definition of integrally closed integral domain:
:$A$ is '''integrally closed''' {{iff}} it is integrally closed in its field of fractions $K$.
Since $x$ is an arbitrary element in the integral closure of $A$ in $K$:
:$A$ is integrally closed.
{{qed}}
Theorem: Totally Disconnected Space is T1

Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected.
Then $T$ is a $T_1$ space.

Proof:
Let $T = \struct {S, \tau}$ be totally disconnected.
Then as its components are singletons, it follows that each of its points is closed.
From Equivalence of Definitions of $T_1$ Space, it follows that $T$ is a $T_1$ space.
{{qed}}
Theorem: Determinant of Linear Operator is Well Defined

Let $V$ be a nontrivial finite dimensional vector space over a field $K$.
Let $A: V \to V$ be a linear operator of $V$.
Then the determinant $\det A$ of $A$ is well defined, that is, does not depend on the choice of a basis of $V$.

Proof:
Let $A_\BB$ and $A_\CC$ be the matrices of $A$ relative to $\BB$ and $\CC$ respectively.
Let $\det$ also denote the determinant of a matrix.
We are required to show that $\det A_\BB = \det A_\CC$.
Let $P$ be the change of basis matrix from $\BB$ to $\CC$.
By Change of Coordinate Vectors Under Linear Mapping and since $A_\BB$ and $A_\CC$ represent the same linear operator {{WRT}} different bases, the following diagram commutes:
:250px
{{improve|Replace the above diagram with an xyplot}}
where $u \in V$, and $\sqbrk u_\BB$ indicates the coordinate vector of $u$ with respect to $\BB$, and similarly for $\sqbrk u_\CC$.
That is, $P A_\BB = A_\CC P$.
From Change of Basis Matrix is Nonsingular:
:$A_\BB = P^{-1} A_\CC P$
So:
{{begin-eqn}}
{{eqn | l = \map \det {A_\BB}
      | r = \map \det {P^{-1} A_\CC P}
}}
{{eqn | r = \map \det {P^{-1} } \map \det {A_\CC} \map \det P
      | c = Determinant of Matrix Product
}}
{{eqn | r = \map \det P^{-1} \map \det {A_\CC} \map \det P
      | c = Determinant of Inverse Matrix
}}
{{eqn | r = \map \det {A_\CC}
}}
{{end-eqn}}
Hence the result.
{{qed}}
Category:Linear Operators
Category:Determinants
p21z2zv952cttnrams307p79xrdp6yd
Theorem: Localization Preserves Integral Closure

Let $A \subseteq B$ be an extension of commutative rings with unity.
Let $C$ be the integral closure of $A$ in $B$.
Let $S \subseteq A$ be a multiplicatively closed subset.
Then $C_S$ is the integral closure of $A_S$ in $B_S$, where subscript $S$ indicates the localization at $S$.

Proof:
First we show that $C_S$ is an integral extension of $A_S$.
Let $x \in C_S$, and $\iota$ be the canonical inclusion from a ring to its localization.
There exists $s \in S$ such that $sx \in \iota(C)$, say $sx = \iota(c)$.
Since $c \in C$ is integral, there is an equation:
:$c^n + a_{n-1}c^{n-1} + \cdots + a_1 c + a_0 = 0$
with the $a_i \in A$. Since $\iota$ is a homomorphism, we have:
:$\left({s x}\right)^n + \iota \left({a_{n-1} }\right) \left({s x}\right)^{n-1} + \cdots + \iota \left({a_1}\right) \left({s x}\right) + \iota \left({a_0}\right) = 0$
{{explain|Just a link, probably, to indicate the nature of $\iota$ - so as to confirm that it is a (ring) homomorphism.}}
Multiplying by $s^{-n}$ this gives:
:$x^n + s^{-1}\iota \left({a_{n-1} }\right)x^{n-1} + \cdots + x^{1-n} \iota \left({a_1}\right) x + s^{-n}\iota \left({a_0}\right) = 0$
This is a monic polynomial with coefficients in $A_S$, so $x$ is integral over $A_S$.
Now let $x \in B_S$ be integral over $A_S$.
So we have an equation
:$x^n + r_{n-1}x^{n-1} + \cdots + r_1 x + r_0 = 0$
Let:
:$t \in S$ be such that $t x \in \iota \left({B}\right)$
:$u_i$ be such that $u_i r_i \in \iota \left({A}\right)$ for $i = 0, \ldots, n-1$
:$s = t u_0 \cdots u_{n-1}$.
Then we have:
:$\left({s x}\right)^n + r_{n-1} s \left({s x}\right)^{n-1} + \cdots + r_1 s^{n-1} \left({s x}\right) + r_0 s^n = 0$
Now let:
: $s x = \iota \left({y}\right)$ for $y \in B$
: $r_{n-i} s^i = \iota \left({q_i}\right)$ for $i = 1, \ldots, n$
Then:
:$\iota \left({y^n + q_1 y^{n-1} + \cdots + q_{n-1} y + q_n}\right) = 0$
Now:
: $\ker \left({\iota: B \to B_S}\right) = \left\{{x \in B : s x = 0 \text{ for some } s \in S}\right\}$
So there exists $v \in S$ such that:
:$\left({v y}\right)^n + q_1 v (\left({v y}\right)^{n-1} + \cdots + q_{n-1} v^{n-1} \left({v y}\right) + q_n v^n = 0$
Therefore $v y \in B$ is integral over $A$, so $vy \in C$.
Therefore:
:$\iota \left({v y}\right) \in C_S$
and:
:$\iota \left({y}\right) \in C_S$
Thus:
:$x = s^{-1} \iota \left({y}\right) \in C_S$
{{qed}}
{{MissingLinks}}
Category:Algebraic Number Theory
fn2hcx9gs0fpovp7mrjceq789a5r8vh
Theorem: Integrally Closed is Local Property

Let $A$ be an integral domain.
For a prime ideal $\mathfrak p$ of $A$, let $A_{\mathfrak p}$ denote the localization at $S = A \divides \mathfrak p$.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$A$ is integrally closed}}
{{item|(2):|$A_{\mathfrak p}$ is integrally closed for all prime ideals $\mathfrak p$}}
{{item|(3):|$A_{\mathfrak m}$ is integrally closed for all maximal ideals $\mathfrak m$}}
{{end-itemize}}

Proof:
=== $(1)$ implies $(2)$ ===
Let $\map Q R$ denote the field of quotients of an integral domain $R$.
We have by Localization Preserves Integral Closure that:
:$\map Q {A_{\mathfrak p}} = \map Q A$
Hence $A_{\mathfrak p}$ is integrally closed for all prime ideals $\mathfrak p$.
{{questionable|What is the point of considering field of quotients? The claim seems exactly Localization Preserves Integral Closure}}
{{qed|lemma}}
Theorem: Totally Disconnected and Locally Connected Space is Discrete

Let $T = \struct {S, \tau}$ be a topological space which is both totally disconnected and locally connected.
Then $T$ is the discrete space on $S$.

Proof:
So, let $T = \struct {S, \tau}$ be that topological space which is both totally disconnected and locally connected.
As $T$ is totally disconnected, every point is a component and therefore closed.
As $T$ is locally connected, there exists a basis $\BB$ of $T$ such that every element of $\BB$ is a component of $T$.
In order for $T$ to be covered by $\BB$, every singleton subset of $T$ must be in $\BB$.
From the definition of topology, the union of any number of these singletons is an open set of $T$.
That is, every subset of $S$ is open in $T$.
That is, every element of the power set of $S$ is open in $T$.
This is precisely the definition of the discrete space on $S$.
{{qed}}
Theorem: Totally Disconnected but Connected Set must be Singleton

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be both totally disconnected and connected.
Then $H$ is a singleton.

Proof:
If $H$ is totally disconnected then its individual points are separated.
If $H$ is connected it can not be represented as the union of two (or more) separated sets.
So $H$ can have only one point in it.
{{qed}}
Theorem: Identity Morphism is Unique

Let $\mathbf C$ be a category.
Let $X$ be an object of $\mathbf C$.
Then the identity morphism $\operatorname{id}_X : X \to X$ is unique.

Proof:
Let $\operatorname{id}_X^1$, $\operatorname{id}_X^2$ be two identity morphisms for $X$.
By definition, for any morphism $f : Y \to X$, we have:
:$\operatorname{id}_X^1 \circ f = f$
In particular, taking $Y = X$ and $f = \operatorname{id}_X^2$, we have:
:$\operatorname{id}_X^1 \circ \operatorname{id}_X^2 = \operatorname{id}_X^2$
Similarly, for any morphism $g : X \to Y$, we have:
:$g \circ \operatorname{id}_X^2 = g$
So taking $Y = X$ and $g = \operatorname{id}_X^1$ we have:
:$\operatorname{id}_X^1 \circ \operatorname{id}_X^2 = \operatorname{id}_X^1$
Putting this together we have:
:$\operatorname{id}_X^2 = \operatorname{id}_X^1 \circ \operatorname{id}_X^2 = \operatorname{id}_X^1$
as required.
{{qed}}
Category:Morphisms
4bevsbsoq13y7hzmmyvl3uieljurdnq
Theorem: Product Category is Category

Let $\mathbf C$ and $\mathbf D$ be metacategories.
Then the product category $\mathbf C \times \mathbf D$ is a metacategory.

Proof:
Let $\tuple {X, Y}$ and $\tuple {X', Y'}$ be objects of $\mathbf C \times \mathbf D$.
Let $\tuple {f, g}: \tuple {X, Y} \to \tuple {X', Y'}$ and $\tuple {h, k}: \tuple {X', Y'} \to \tuple {X, Y}$ be morphisms of $\mathbf C \times \mathbf D$.
Let $\operatorname {id}_X, \operatorname {id}_Y$ be the identity morphisms for the objects $X$ and $Y$ respectively.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {f, g} \circ \tuple {\operatorname {id}_X, \operatorname {id}_Y}
      | r = \tuple {f \circ \operatorname {id}_X, g \circ \operatorname {id}_Y}
      | c = {{Defof|Product Category}}: Composition of Morphisms
}}
{{eqn | r = \tuple {f, g}
      | c = {{Defof|Identity Morphism}}
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \tuple {\operatorname {id}_X, \operatorname {id}_Y} \circ \tuple {h, k}
      | r = \tuple {\operatorname {id}_X \circ h, \operatorname {id}_Y \circ k}
      | c = {{Defof|Product Category}}: Composition of Morphisms
}}
{{eqn | r = \tuple {h, k}
      | c = {{Defof|Identity Morphism}}
}}
{{end-eqn}}
Therefore, $\tuple {\operatorname {id}_X, \operatorname {id}_Y}$ satisfies the property of an identity morphism of $\mathbf C \times \mathbf D$.
Let $\tuple {f, g}$, $\tuple {h, k}$ and $\tuple {\ell, m}$ be composable morphisms of $\mathbf C \times \mathbf D$.
We have:
{{begin-eqn}}
{{eqn | l = \paren {\tuple {f, g} \circ \tuple {h, k} } \circ \tuple {\ell, m}
      | r = \tuple {f \circ h, g \circ k} \circ \tuple {\ell, m}
      | c = {{Defof|Product Category}}: Composition of Morphisms
}}
{{eqn | r = \tuple {\paren {f \circ h} \circ \ell, \paren {g \circ k} \circ m}
      | c = {{Defof|Product Category}}: Composition of Morphisms
}}
{{eqn | r = \tuple {f \circ \paren {h \circ \ell}, g \circ \paren {k \circ m} }
      | c = {{Defof|Composition of Morphisms}}: Associativity
}}
{{eqn | r = \tuple {f, g} \circ \tuple {h \circ \ell, k \circ m}
      | c = {{Defof|Product Category}}: Composition of Morphisms
}}
{{eqn | r = \tuple {f, g} \circ \paren {\tuple {h, k} \circ \tuple {\ell, m} }
      | c = {{Defof|Product Category}}: Composition of Morphisms
}}
{{end-eqn}}
Therefore, composition of morphisms in $\mathbf C \times \mathbf D$ is also associative.
{{qed}}
Category:Product Categories
27sgx3w3pneq21qxk0fnrjve7hmmdtb
Theorem: Category of Sets is Category

Let $\mathbf{Set}$ be the category of sets.
Then $\mathbf{Set}$ is a metacategory.

Proof:
Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a metacategory.
For any two mappings their composition (in the usual set theoretic sense) is again a mapping by Composite Mapping is Mapping.
For any set $X$, we have the identity mapping $\operatorname{id}_X$.
By Identity Mapping is Left Identity and Identity Mapping is Right Identity, this is the identity morphism for $X$.
Finally by Composition of Mappings is Associative, the associative property is satisfied.
Hence $\mathbf{Set}$ is a metacategory.
{{qed}}
Category:Examples of Categories
Category:Category of Sets
rpd0zesyfpvfk4le8e9ow3t0xrsnjyl
Theorem: Projection from Product Category

Let $\mathbf C$ and $\mathbf D$ be categories.
Let $\mathbf C \times \mathbf D$ be the product category.
Then the projection functors:
:$\pi_{\mathbf C}: \mathbf C \times \mathbf D \to \mathbf C$
:$\pi_{\mathbf D}: \mathbf C \times \mathbf D \to \mathbf D$
are indeed functors.
Moreover, $\pi_{\mathbf C}$ and $\pi_{\mathbf D}$ satisfy the following universal property:
{{Improve|To satisfy the universal property means here that they are uniquely determined by the property. Definition of universal property is necessary.<br/>
Also see Product Category is Product in Category of Categories.}}
:''For any category $\mathbf E$ and any functors $F : \mathbf E \to \mathbf C$, $G : \mathbf E \to \mathbf D$, there exists a unique functor $H : \mathbf  \to \mathbf C \times \mathbf D$ such that $F = \pi_{\mathbf C} H$ and $G = \pi_{\mathbf D} H$''
That is, the following diagram commutes:
:<nowiki>$\xymatrix {
\mathbf C & \mathbf C \times \mathbf D \ar[l]_{\pi_\mathbf C} \ar[r]^{\pi_\mathbf D} & \mathbf D \\
& \mathbf E \ar[lu]^F \ar[u]_{\exists ! H} \ar[ru]_G}$</nowiki>

Proof:
{{ProofWanted}}
Category:Product Categories
Category:Projections
laq3yrt19kqib748s6ez9aq5igaw3z6
Theorem: Extremally Disconnected Hausdorff Space is Totally Separated

Let $T = \struct {S, \tau}$ be an extremally disconnected Hausdorff space.
Then $T$ is totally separated.

Proof:
{{Recall|Totally Separated Space|totally separated topological space}}
{{:Definition:Totally Separated Space/Definition 1}}
Let $T = \struct {S, \tau}$ be an extremally disconnected Hausdorff space.
{{Recall|Extremally Disconnected Space|extremally disconnected topological space}}
{{:Definition:Extremally Disconnected Space/Definition 1}}
Let $x, y \in S$ be arbitrary such that $x \ne y$.
As $T$ is a $T_2$ (Hausdorff) space, there exist disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively.
Let $U^-$ denote the closure of $U$.
From Topological Closure is Closed, $U^-$ is closed.
But $U$ is open, so from the definition of extremally disconnected space, then $U^-$ is also open.
So $U^-$ is clopen.
From Open Set Disjoint from Set is Disjoint from Closure:
:$U^- \cap V = \O$
so:
:$y \notin U^-$
Thus $\set {U^- \mid S \setminus U^-}$ is a partition where:
:$x \in U^-$
and:
:$y \in X \setminus U^-$
As $x$ and $y$ are arbitrary, this applies to all $x, y \in S$ such that $x \ne y$.
Thus, by definition, $T$ is totally separated.
{{qed}}
Theorem: Zero Dimensional Space is T3

Let $T = \struct {S, \tau}$ be a zero dimensional topological space.
Then $T$ is a $T_3$ space.

Proof:
{{Recall|T3 Space|$T_3$ space}}
{{:Definition:T3 Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a zero dimensional space.
{{Recall|Zero Dimensional Space|zero dimensional space}}
{{:Definition:Zero Dimensional Space}}
Let $F \subseteq S$ be closed in $T$.
Let also $y \notin F$.
Then by definition of closed, $\relcomp S F$ is open in $T$, where $\relcomp S F$ is the complement of $F$ in $S$.
As $T$ is zero dimensional, it has a basis $\BB$ which consists entirely of clopen sets.
As $\BB$ is a basis for $T$, it follows that:
:$\ds \exists \UU \subseteq \BB: \relcomp S F = \bigcup \UU$
that is, $\relcomp S F$ is the union of a subset of elements of $\BB$.
Thus, there is a set $U \in \UU$ such that $y \in U$ and $F \subseteq \relcomp S U$.
But the elements of $\UU$ are clopen sets, so $U$ is itself clopen.
Thus, by definition, $\relcomp S U$  is also clopen.
So we have that $U$ and $\relcomp S U$ are open sets in $T$ such that:
:$\exists W, R \in \tau: F \subseteq W, y \in R: R \cap W = \O$
by setting $R = U$ and $W = \relcomp S U$.
That is, $T$ is a $T_3$ space.
{{qed}}
Theorem: Zero Dimensional T0 Space is Totally Separated

Let $T = \struct {S, \tau}$ be a zero dimensional topological space which is also a $T_0$ space.
Then $T$ is totally separated.

Proof:
{{Recall|Totally Separated Space|totally separated space}}
{{:Definition:Totally Separated Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a zero dimensional space which is also a $T_0$ space.
{{Recall|Zero Dimensional Space|zero dimensional space}}
{{:Definition:Zero Dimensional Space}}
{{Recall|T0 Space|$T_0$ space}}
{{:Definition:T0 Space/Definition 1}}
Let $x, y \in S$ be arbitrary such that $x \ne y$.
{{WLOG}}, suppose that $\exists U \in \tau: x \in U, y \notin U$.
Then:
:$\exists V \in \BB: x \in V$ and $V \subseteq U$
by definition of basis.
The set $V$ is clopen by the definition of $\BB$.
But then $x \in V$ which is open and $y \in S \setminus V$ which is also open.
From Clopen Set and Complement form Separation, $\set {V \mid S \setminus V}$ is a separation of $T$.
As $x$ and $y$ are arbitrary, this applies to all $x$ and $y$ in $S$.
Hence by definition $T$ is totally separated.
{{qed}}
Theorem: Scattered T1 Space is Totally Disconnected

Let $T = \struct {S, \tau}$ be a scattered topological space which is also a $T_1$ space.
Then $T$ is totally disconnected.

Proof:
Let $T = \struct {S, \tau}$ be a scattered space which is also a $T_1$ space.
We have that every Non-Trivial Connected Set in $T_1$ Space is Dense-in-itself.
As $T$ is scattered, every $H \subseteq S$ contains at least one point which is isolated in $H$.
So $H$ is not dense-in-itself and so if $H$ has more than one element it can not be connected.
As $H$ is arbitrary, it follows that $T$ is totally disconnected.
{{qed}}
Theorem: Non-Trivial Connected Set in T1 Space is Dense-in-itself

Let $T = \struct {S, \tau}$ be a $T_1$ space.
Let $H \subseteq S$ be connected in $T$.
If $H$ has more than one element, then $H$ is dense-in-itself.

Proof:
{{AimForCont}} $H$ is not dense-in-itself.
Then $\exists x \in H$ such that $x$ is isolated in $H$.
That is, $\exists U \in \tau: U \cap H = \set x$.
Since $H$ has more than one element, we can find $y \in H$ with $y \ne x$.
Since $T$ is a $T_1$ space:
:$\forall y \in H: y \ne x \implies \paren {\exists V_y \in \tau: y \in V_y, x \notin V_y}$
Define:
:$V = \ds \bigcup_{y \mathop \in H \\ y \mathop \ne x} V_y$
Then $\forall z \in H$:
{{begin-eqn}}
{{eqn | l = z = x
      | o = \implies
      | r = z \in U
}}
{{eqn | l = z \ne x
      | o = \implies
      | r = z \in V_z \subseteq V
}}
{{end-eqn}}
showing that:
:$H \subseteq U \cup V$
This also shows that both $U \cap H$ and $V \cap H$ are non-empty.
Since $x \notin V_y$ for any $y \in H$:
:$x \notin V$
This implies:
:$\set x \cap V = \O$
So:
:$H \cap U \cap V = \O$
Thus we have:
{{begin-eqn}}
{{eqn | l = H
      | o = \subseteq
      | r = U \cup V
}}
{{eqn | l = H \cap U \cap V
      | r = \O
}}
{{eqn | l = U \cap H
      | o = \ne
      | r = \O
}}
{{eqn | l = V \cap H
      | o = \ne
      | r = \O
}}
{{end-eqn}}
showing that $H$ is disconnected.
This is a contradiction.
Hence by Proof by Contradiction it must be the case that $H$ is dense-in-itself. 
{{qed}}
Theorem: Scattered Space is T0

Let $T = \struct {S, \tau}$ be a scattered topological space.
Then $T$ is also a $T_0$ space.

Proof:
{{Recall|T0 Space|$T_0$ space|index = 2}}
{{:Definition:T0 Space/Definition 2}}
Suppose $T$ is not a $T_0$ space.
Hence by definition there exist $x, y \in S$ such that $x$ and $y$ are both limit points of each other.
So by definition of isolated point, neither $x$ nor $y$ are isolated in $\set {x, y}$.
Thus we have found a subset $\set {x, y} \subseteq T$ such that $\set {x, y}$ is by definition dense-in-itself.
So $T$ is not scattered.
Hence the result by Rule of Transposition.
{{qed}}
Theorem: Set with Dispersion Point is Biconnected

Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a connected set in $T$.
Let $p \in H$ be a dispersion point of $H$.
Then $H$ is biconnected.

Proof:
{{Recall|Biconnected Set|biconnected set}}
{{:Definition:Biconnected Set}}
{{AimForCont}} $H$ is not biconnected.
Then by definition there exist disjoint non-degenerate connected sets $U, V$ such that $H = U \cup V$.
{{WLOG}}, let $p \in U$.
Then $V \subset H \setminus \set p$.
As $p$ is a dispersion point of $H$, $H \setminus \set p$ is totally disconnected.
Thus, by definition, $H \setminus \set p$ contains no non-degenerate connected sets.
But this contradicts our definition of $V$, as $V \subset H \setminus \set p$.
It follows by Proof by Contradiction that $H$ is biconnected.
{{qed}}
Theorem: Totally Disconnected Space is Punctiform

Let $T = \struct {S, \tau}$ be a topological space which is totally disconnected.
Then $T$ is punctiform.

Proof:
Let $T = \struct {S, \tau}$ be totally disconnected.
Then by definition its components are singletons.
Thus by definition each of its connected sets are degenerate.
{{qed}}
Theorem: Distance in Pseudometric is Non-Negative

Let $X$ be a set on which a pseudometric $d: X \times X \to \R$ has been imposed.
Then:
:$\forall x, y \in X: \map d {x, y} \ge 0$

Proof:
By definition of pseudometric, we have that:
{{begin-axiom}}
{{axiom | n = \text M 1
        | q = \forall x \in A
        | m = \map d {x, x} = 0
}}
{{axiom | n = \text M 2
        | q = \forall x, y, z \in A
        | m = \map d {x, y} + \map d {y, z} \ge \map d {x, z}
}}
{{axiom | n = \text M 3
        | q = \forall x, y \in A
        | m = \map d {x, y} = \map d {y, x}
}}
{{end-axiom}}
Hence:
{{begin-eqn}}
{{eqn | q = \forall x, y \in X
      | l = \map d {x, y} + \map d {y, x}
      | o = \ge
      | r = \map d {x, x}
      | c = {{Metric-space-axiom|2}}
}}
{{eqn | ll= \leadsto
      | l = 2 \map d {x, y}
      | o = \ge
      | r = 0
      | c = {{Metric-space-axiom|1}} and {{Metric-space-axiom|3}}
}}
{{eqn | ll= \leadsto
      | l = \map d {x, y}
      | o = \ge
      | r = 0
}}
{{end-eqn}}
{{qed}}
Category:Pseudometrics
Category:Pseudometric Spaces
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