hcju/proofwikiqa huggingface dataset:
We have that: :$\forall x \in S: \exists \set x \in \CC: x \in \set x$ and so $\CC$ is a cover for $S$. Then from Set in Discrete Topology is Clopen, it follows that $\CC$ is an open cover of $T$. From Point in Discrete Space is Neighborhood, every point $x \in S$ has a neighborhood $\set x$. This neighborhood $\set x$ intersects exactly one element of $\CC$, that is: $\set x$ itself. As $1$ is a finite number, the result follows from definition of locally finite. Now consider any cover $\VV$ of $S$. By definition: :$\forall x \in S: \exists V \in \VV: x \in V$ That is: :$\forall x \in S: \exists V \in \VV: \set x \subseteq V$ That is, every element of $\CC$ is contained in some element of $\VV$. Thus by definition, $\CC$ is a refinement of $\VV$. {{qed}}
Let $\VV$ be any open cover of $S$. Consider the set $\CC$ of all singleton subsets of $S$: :$\CC := \set {\set x: x \in S}$ From Discrete Space has Open Locally Finite Cover, $\CC$ is an open cover which is locally finite. This result also shows that $\CC$ is the finest cover on $T$. So $\CC$ is an open refinement of $\VV$ which is locally finite. So $T$ is paracompact, by definition. {{qed}}
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\set x$. Then from Interior Equals Closure of Subset of Discrete Space we have that $\set x$ equals its closure. As $\set {\set x}$ is (trivially) an open cover of $\set x$, it follows by definition that $\set x$ is compact. {{qed}}
We have that Singleton Set in Discrete Space is Compact. We also have that $S$ is the union of all its singleton sets: :$\displaystyle S = \bigcup_{x \mathop \in S} \set x$ As $S$ is countable, it is the union of countably many compact sets. Hence the result, by definition of $\sigma$-compact. {{qed}}
We have: :Countable Discrete Space is $\sigma$-Compact :$\sigma$-Compact Space is Lindelöf So if $S$ is countable, $T$ is a Lindelöf space. {{qed}}
Let $T = \struct {S, \tau}$ be the discrete space on $S$.  Let $U \in \tau$. Then: :$\displaystyle U = \bigcup_{x \mathop \in U} \set x$ Hence: :$\forall x \in U: \exists \set x \in \BB: \set x \subseteq U$ Thus $U$ is the union of elements of $\BB$. Hence by definition $\BB$ is a basis for $T$. {{qed}} Category:Discrete Topology Category:Topological Bases q66i6e4itpusdv0jhwo5feqfpte6yfm
From Basis for Discrete Topology, the set: :$\BB := \set {\set x: x \in S}$ is a basis for $T$. There is a trivial one-to-one correspondence $\phi: S \leftrightarrow \BB$ between $S$ and $\BB$: :$\forall x \in S: \map \phi x = \set x$ Let $S$ be countable. Then $\BB$ is also countable by definition of countability. So we have that $T$ has a countable basis, and so is second-countable by definition. {{qed}}
Let $T = \left({S, \tau}\right)$ be a countable discrete topological space. From Countable Discrete Space is Second-Countable: :$T$ is second-countable. From Second-Countable Space is Separable: :$T$ is separable. {{qed}}
By definition, $T$ is separable {{iff}} there exists a countable subset of $S$ which is everywhere dense in $T$. Let $H \subseteq S$ be everywhere dense in $T$. Then by definition of everywhere dense, $H^- = S$ where $H^-$ denotes the closure of $H$. However, as $T$ is a discrete space, $H^- = H$ from Interior Equals Closure of Subset of Discrete Space. So $H^- = S \implies H = S$. But $S$ is uncountable. So there exists no $H \subseteq S$ such that $H$ is both countable and everywhere dense. Hence by definition of separable space, if $T$ is an uncountable discrete space it can not be separable. {{qed}}
We have that an Uncountable Discrete Space is not Separable. From Second-Countable Space is Separable, it follows that $T$ can not be second-countable. {{qed}}
Consider the set $\CC$ of all singleton subsets of $S$: :$\CC := \set {\set x: x \in S}$ From Discrete Space has Open Locally Finite Cover, $\CC$ is an open cover of $S$ which is finer than any other open cover of $S$. That is, $\CC$ is an open cover of $S$ which is uncountable and has no countable subcover. (Note that a subcover is a refinement of a cover.) So by definition $T$ can not be a Lindelöf space. {{qed}}
We have that an Uncountable Discrete Space is not Lindelöf. But a $\sigma$-compact space is Lindelöf. So an uncountable discrete space can not be $\sigma$-compact. {{qed}}
=== Necessary Condition === Let $T$ be a compact discrete space. {{AimForCont}} $T$ is infinite. Let $\mathcal C$ be the cover for $S$ defined as: :$\mathcal C = \left\{{\left\{{x}\right\}: x \in S}\right\}$ As $S$ is an infinite set then so is $\mathcal C$. Let $\mathcal C'$ be a proper subset of $\mathcal C$. Then: : $\exists y \in S: \left\{{y}\right\} \notin \mathcal C'$ and so $\mathcal C'$ is not a cover for $S$. So by definition $\mathcal C'$ is not a subcover of $\mathcal C$. So $\mathcal C$ can have no finite subcover. Hence by definition $T$ can not be compact. By Proof by Contradiction it follows that of $T$ is compact then $S$ must be finite. {{qed|lemma}} === Sufficient Condition === Let $T$ be a finite discrete space. Then from Finite Topological Space is Compact it follows that $T$ is a compact space. {{qed}}
We have: :Finite Topological Space is Compact :Compact Space is Countably Compact :Finite Space is Second-Countable :Second-Countable Space is First-Countable :First-Countable Space is Sequentially Compact iff Countably Compact. Hence the result. {{qed}} Category:Sequentially Compact Spaces Category:Finite Topological Spaces ax16keb5z9otsvz85ygtjga65ksbolz
We have that a Discrete Space is fully $T_4$. Then we note that from Discrete Space satisfies all Separation Properties, a discrete space is a $T_1$ (Fréchet) space. Therefore, by definition, $T$ is fully normal. {{qed}}
Consider the set $\CC$ of all singleton subsets of $S$: :$\CC := \set {\set x: x \in S}$ From Discrete Space has Open Locally Finite Cover: :$\CC$ is an open cover of $T$ which is locally finite. Let $x \in S$. The star of $x$ with respect to $\CC$ is defined as: :$\displaystyle x^* := \bigcup \set {U \in \CC: x \in U}$ That is, the union of all sets in $\CC$ which contain $x$. But here the only set of $\CC$ containing $x$ is $\set x$. Now let $\VV$ be an open cover of $T$. From the above, it follows that $\CC$ is a cover for $S$ such that: :$\forall x \in S: \exists U \in \VV: x^* \subseteq U$ where $x^* = \set x$ is the star of $x$ with respect to $\CC$. That is, $\CC$ is a star refinement of $\VV$. Therefore, by definition, $T$ is fully $T_4$. {{qed}}
Let $\mathcal V$ be an open cover of $T$. For each $x \in S$, define $\mathcal V_x$ to be $\left\{{V \in \mathcal V : x \in V}\right\}$ Since $S$ is finite, and since by definition of a cover, each $x\in S$ is contained in at least one $V$ in $\mathcal V$, we have that $\left\{{\mathcal V_x : x \in S}\right\}$ is a finite collection of nonempty sets. From Principle of Finite Choice, there is a choice function which selects one $V_x$ from $\mathcal V_x$ for each $x \in S$.  By definition of $\mathcal V_x$, such a $V_x$ contains $x$. Since there were only finitely many $\mathcal V_x$, this provides finitely many open sets $V_x \in \mathcal V$ such that $\displaystyle S \subseteq \bigcup_{x \in S} V_x$. Thus $\left\{{\mathcal V_x : x \in S}\right\}$ is a finite subcover of $\mathcal V$. The result follows by definition of compact. {{qed}}
From Power Set of Finite Set is Finite, the power set of $S$ is finite. Since the topology $\tau$ is by definition a set of subsets of $S$, it follows that $\tau$ is finite as well. Let $\VV$ be an open cover of $S$. By definition $\VV \subseteq \tau$ and so is also a finite set. From Set is Subset of Itself, $\VV \subseteq \VV$. Thus by definition $\VV$ is a finite subcover of $\VV$. The result follows by definition of compact. {{qed}}
Let $T = \struct {S, \tau}$ be a topological space where $S$ is finite. As $S$ is finite, then so is its power set $\powerset S$. So is $\tau \subseteq \powerset S$. Now $\tau$ is a basis for itself, as every set in $\tau$ is a union of sets from $\tau$ (just one). But $\tau$ is finite, and so countable. The result follows from definition of a second-countable space. {{qed}} Category:Second-Countable Spaces 2qgy61t10hcpmo8ryzugg9d0grhnhzb
A finite discrete space is by definition a topology on a finite set. A Discrete Space is Fully Normal. A fully normal space is fully $T_4$ by definition. The rest of the results follow directly from Finite Space Satisfies All Compactness Properties.
We have that: : A Finite Topological Space is Compact. : A Finite Space is Sequentially Compact. The remaining properties are demonstrated in: : Sequence of Implications of Global Compactness Properties : Sequence of Implications of Local Compactness Properties : Sequence of Implications of Paracompactness Properties {{qed}} Category:Finite Topological Spaces Category:Compact Spaces 41h380j70f2hy79vyo0sjcjwk1rof12
Let $U \subseteq S$ such that $U \ne \O$. From Interior Equals Closure of Subset of Discrete Space, we have: :$U^\circ = U = U^-$ where $U^\circ$ is the interior and $U^-$ the closure of $U$. So: :$\paren {U^-}^\circ = U \ne \O$ Thus, by definition, no non-empty subset of $S$ is nowhere dense. So $S$ can not be the union (countable or otherwise) of nowhere dense subsets. So by definition $S$ can not be meager. Hence the result. {{qed}}
Let $x \in S$ be an arbitrary point of $T$. From Set in Discrete Topology is Clopen it follows that $\left\{{x}\right\}$ is open in $T$. The result follows from Space with Open Point is Non-Meager. {{qed}}
Let $d: S \times S \to \R$ be the standard discrete metric. From Standard Discrete Metric induces Discrete Topology we have that $\struct {S, d}$ is a metric space whose induced topology is $\tau$. Consider now a Cauchy sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$. By the definition of Cauchy sequence: :$\forall \epsilon > 0: \exists N \in \N$ such that $\forall n, m > N: \map d {x_n, x_m} < \epsilon$ Take $\epsilon = \dfrac 1 2$. Then: :$\exists N \in \N: \forall n, m > N: \map d {x_n, x_m} < \epsilon = \dfrac 1 2$ From the definition of standard discrete metric: :$\exists N \in \N: \forall n, m > N: \map d {x_n, x_m} = 0 \implies x_n = x_m$ Thus the sequence $\sequence {x_n}_{n \mathop \in \N}$ is constant from some $N \in \N$. From Eventually Constant Sequence is Convergent: :$x_n \to x \in S$ Thus every Cauchy sequence in $S$ converges. The result follows by definition of complete metric space. {{qed}}
By definition, $T$ is dense-in-itself {{iff}} it contains no isolated points. The result follows from Topological Space is Discrete iff All Points are Isolated. {{qed}}
Let $T = \struct {S, \tau}$ be a non-trivial discrete space. Let $a \in S$. Let $A = \set a$ and $B = \relcomp S {\set a}$, where $\relcomp S {\set a}$ is the complement of $A$ in $S$. As $T$ is not trivial, $B \ne \O$. Then from Set in Discrete Topology is Clopen, $A$ and $B$ are both open. So $A \mid B$ is a separation of $S$. It follows, by definition, that $T$ is not connected. {{qed}}
From Set in Discrete Topology is Clopen, $\set a$ is open in $T$. From Basis for Discrete Topology, the set: :$\BB := \set {\set x: x \in S}$ is a basis for $T$. Let $\set x \in \BB$. From Point is Path-Connected to Itself, it follows that $\set x$ is path-connected. Hence $T$ has a basis consisting entirely of path-connected sets. So by definition $T$ is locally path-connected. {{qed}}
Consider the constant mapping on the closed unit interval $\mathbb I = \left[{0 \,.\,.\, 1}\right]$: :$\forall x \in \mathbb I: f_a \left({x}\right) = a$ Thus, in particular: :$f_a \left({0}\right) = a$ :$f_a \left({1}\right) = a$ As a Constant Mapping is Continuous, it follows that $f_a$ is a path in $X$. Thus $a$ is path-connected to itself. {{qed}} Category:Path-Connected Spaces 9qyrlejgzgz62ovjgvu2scn2ca7ud1j
We examine the uniformity axioms in turn: ;$\text U 1$: $\forall u \in \UU: \Delta_S \subseteq u$ This follows by definition of the discrete uniformity: :$\UU := \set {u \subseteq S \times S: \Delta_S \subseteq u}$ ;$\text U 2$: $\forall u, v \in \UU: u \cap v \in \UU$ We have that $\forall u, v \subseteq S \times S: \Delta_S \subseteq u, \Delta_S \subseteq v$ So $\Delta_S \subseteq u \cap v$ from Intersection is Largest Subset. So $u \cap v \in \UU$, and $\text U 2$ holds. ;$\text U 3$: $u \in \UU, u \subseteq v \subseteq S \times S \implies v \in \UU$ Let $u \subseteq v \subseteq S \times S$. We have that $\Delta_S \subseteq u$ and so $\Delta_S \subseteq v$ from Subset Relation is Transitive. So $v \in \UU$ by definition of $\UU$, and $\text U 3$ holds. ;$\text U 4$: $\forall u \in \UU: \exists v \in \UU: v \circ v \subseteq u$ where $\circ$ is defined as: ::$u \circ v := \set {\tuple {x, z}: \exists y \in S: \tuple {x, y} \in v, \tuple {y, z} \in u}$ We have by definition of $\UU$ that $\Delta_S \in \UU$. We note from Diagonal Relation is Equivalence that $\Delta_S$ is an equivalence relation and so by definition transitive. Therefore from Relation contains Composite with Self iff Transitive we have that $\Delta_S \circ \Delta_S \subseteq \Delta_S$. So: :$\forall u \in \UU: \exists \Delta_S \in \UU: \Delta_S \circ \Delta_S \subseteq \Delta_S \subseteq u$ and so $\text U 4$ holds. ;$\text U 5$: $\forall u \in \UU: \exists u^{-1} \in \UU$ where $u^{-1}$ is defined as: ::$u^{-1} := \set {\tuple {y, x}: \tuple {x, y} \in u}$ Let $u \in \UU$, so $\Delta_S \subseteq u \subseteq S \times S$. Thus: :$\forall x \in u: \tuple {x, x} \in u \implies \tuple {x, x} \in u^{-1}$ and so $\Delta_S \subseteq u^{-1}$ Thus $u^{-1} \in \UU$. {{qed|lemma}} All of the uniformity axioms are satisfied, and $\UU$ is demonstrated to be a uniformity. {{qed}}
From the construction, let $\tau \subseteq \powerset S$ be a subset of the power set of $S$, created from $\UU$ by: :$\tau := \set {\map u x: u \in \UU, x \in X}$ where: :$\forall x \in X: \map u x = \set {y: \tuple {x, y} \in u}$ We need to show that $\tau$ is the discrete topology. Consider $\Delta_S \in \UU$. Let $x \in S$. Then the set: :$U_x := \set {y: \tuple {x, y} \in \Delta_S}$ We have that: $\forall x \in S: U_x = \set x \in \UU$ Thus we have: :$\BB := \set {U_x: x \in S} = \set {\set x: x \in S}$ From Basis for Discrete Topology, we have that $\BB$ is a basis for the discrete topology on $S$. {{qed}}
Let $T = \struct {S, \set {\O, S} }$ be the indiscrete space on $S$. Confirming the criteria for $T$ to be a topology: :$(1): \quad$ Trivially, by definition, $\O \in \tau$ and $S \in \tau$. :$(2): \quad \O \cup \O = \O \in \tau$, $\O \cup S = S \in \tau$ and $S \cup S = S \in \tau$ from Union with Empty Set and Union is Idempotent. :$(3): \quad \O \cap \O = \O \in \tau$, $\O \cap S = \O \in \tau$ and $S \cap S = S \in \tau$ from Intersection with Empty Set and Intersection is Idempotent. {{qed}}
Let $\phi$ be any topology on $S$. Then by definition of topology, $\O \in \phi$ and $S \in \phi$ Hence by definition of subset, $\tau \subseteq \phi$.  Hence by definition of coarser topology, $\tau$ is coarser than $\phi$. {{qed}}
By definition, $x \in \struct {S, \tau}$ is a limit point of $H$ if every open set $U \in \tau$ such that $x \in U$ contains some point of $H$ other than $x$. Here, of course, there is only one open set that contains any points at all, and that is $S$. So as $S$ contains more than one point, it follows that every point of $T$ is a limit point of $H$. {{qed}}
Let $\alpha \in S$. By definition, $\sequence {s_n}$ converges to $\alpha$ if every open set in $T$ containing $\alpha$ contains all but a finite number of terms of $\sequence {s_n}$. But as $T$ has only one open set containing any points at all, '''every''' point of $\sequence {s_n}$ is contained in every open set in $T$ containing $\alpha$. Hence the result. {{qed}}
Let $x \in H$. Then as $H$ is not singleton, $\exists y \in H: y \ne x$. Then every neighborhood of $x$ contains $y$, as the only open set of $T$ is $S$, which also contains both $x$ and $y$. Hence $x$ is not isolated by definition. $x$ is general, so all points in $H$ are similarly not isolated. Hence the subset $H$ is dense-in-itself by definition. {{qed}}
Let $H \subseteq S$ such that $H \ne \O$. From Limit Points of Indiscrete Space, every point of $S$ is a limit point of $H$. So, by definition: :$H^- = S$ where $H^-$ is the closure of $H$ in $T$. Now the interior of $S$ is $S$ itself (trivially, by definition). So: :$\paren {H^-}^\circ = S \ne \O$ where $\paren {H^-}^\circ$ is the interior of the closure of $H$. So, by definition, $H$ is not nowhere dense. (Note that if $H = \O$, then Empty Set is Nowhere Dense applies, which is why the stipulation above that $H \ne \O$.) It follows that $H$ can not be the countable union of nowhere dense subsets of $S$. So $H$ is, by definition, not meager, so must be non-meager. {{qed}}
From Empty Set is Closed in Topological Space, $\O$ is closed in $T$. From Closed Set Equals its Closure: :$\O^- = \O$ where $\O^-$ is the the closure of $\O$. From Empty Set is Element of Topology, $\O$ is open in $T$. From the definition (trivially) we also have that: :$\O^\circ = \O$ where $\O^\circ$ is the interior of $\O$. So: :$\struct {\O^-}^\circ = \O$ and so by definition $\O$ is nowhere dense in $T$. {{qed}}
=== Necessary Condition === Let $U \subseteq T$ be open in $T$. From Set Interior is Largest Open Set: : $U \subseteq U^\circ$ But by definition of interior: : $U^\circ \subseteq U$ Hence $U^\circ = U$. {{qed|lemma}} === Sufficient Condition === Suppose $U^\circ = U$. By definition $U^\circ$ is the union of all subsets of $U$ which are open in $U$. By the definition of a topology, that means $U$ itself must also be open. {{qed}} Category:Set Interiors Category:Open Sets f6uidcbbematexftq7jgr2afde1q33s
As $H \subset S$, it follows that $H \ne T$. So the only open subset of $H$ is $\O$. So by definition: :$H^\circ = \O$ From Empty Set is Closed in Topological Space, $\O$ is closed in $T$. From Closed Set Equals its Closure: :$\O^- = \O$ From Empty Set is Element of Topology, $\O$ is open in $T$. From Interior of Open Set: :$\O^\circ = \O$ The result follows. {{qed}}
From Limit Points of Indiscrete Space, every point in $S$ is a limit point of $H$. So from the definition of closure, $H \ne \O \implies H^-= S$. From the open set axioms, $S$ is open in $T$. From Interior of Open Set, $S^\circ = S$. From Underlying Set of Topological Space is Clopen, $S$ is closed in $T$. From Closed Set Equals its Closure, $S^- = S$. The result follows. {{qed}}
From Closure of Subset of Indiscrete Space, $H^- = S$, where $H^-$ denotes set closure. From Interior of Subset of Indiscrete Space, $H^\circ = \O$, where $H^\circ$ denotes set interior. By definition: :$\partial H = H^- \setminus H^\circ = S \setminus \O = S$ From Open and Closed Sets in Topological Space, $\O$ and $S$ are both closed and open in $T$. So from Set Clopen iff Boundary is Empty: :$\partial H = \O$ {{qed}}
From Boundary of Subset of Indiscrete Space, either $\partial H = S$ or $\partial H = \O$, depending on whether $H = \O$ or $H = S$ or not. From Open and Closed Sets in Topological Space, $\O$ and $S$ are both closed and open in $T$. So from Set Clopen iff Boundary is Empty: :$\map \partial {\partial H} = \O$ {{qed}}
From Limit Points of Indiscrete Space, every point of $T$ is a limit point of $H$. Hence $H$ is everywhere dense by definition. {{qed}}
The only basis for $T$ is $\set S$ which is trivially countable. Hence the result. {{qed}}
From the definition of continuous: :$U \in \tau_2 \implies \phi^{-1} \sqbrk U \in \tau_1$ The only elements of $\tau_2$ are $S_2$ and $\O$, from which: :$\phi^{-1} \sqbrk {S_2} = S_1 \in \tau_1$ :$\phi^{-1} \sqbrk \O = \O \in \tau_1$ {{qed}}
Let $a, b \in S$. Consider any mapping $f: \closedint 0 1 \to S$ such that $\map f 0 = a$ and $\map f 1 = b$. From Mapping to Indiscrete Space is Continuous, we have that $f$ is continuous. The result follows by definition of path-connectedness. {{qed}}
We have that an Indiscrete Space is Path-Connected. Then we have that a Path-Connected Space is Connected. {{qed}}
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space such that $S$ is uncountable. Let $a, b \in S$. Consider an injection $f: \closedint 0 1 \to S$ such that $\map f 0 = a$ and $\map f 1 = b$. This can always be found because $S$ is itself uncountable. From Mapping to Indiscrete Space is Continuous, we have that $f$ is continuous. Thus $T$ is arc-connected. Now suppose $S$ is an indiscrete topological space which is arc-connected. Then there exists an injection $f: \closedint 0 1 \to S$ such that $\map f 0 = a$ and $\map f 1 = b$. This can only exist if $S$ is uncountable. {{qed}}
There is only one non-empty open set in $T$. So there can be no two open sets in $T$ which are disjoint. Hence (trivially) $T$ is irreducible. {{qed}}
There is only one non-empty closed set in $T$. So there can be no two closed sets in $T$ which are disjoint. Hence (trivially) $T$ is ultraconnected. {{qed}}
Let $a, b \in S$. By definition of indiscrete space, $S$ is the only open set in $T$ which is non-empty. So (trivially) there is no open set in $T$ containing $a$ and not $b$, or $b$ and not $a$. Hence the result, by definition of $T_0$ (Kolmogorov) space. {{qed}}
By definition, the only two separated sets in $T$ are $S$ and $\O$. Then there exist two disjoint open sets $S$ and $\O$ containing $S$ and $\O$ respectively. Hence (trivially) $T$ is a $T_5$ space. {{qed}}
We have that an indiscrete space is a $T_5$ space. Then we have that a $T_5$ Space is a $T_4$ Space. {{qed}}
Let $F \subseteq S$ be a closed set in $S$. Let $y \in S$ such that $y \notin F$. The only way this can happen is if $F = \O$. So there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$. That is, $U = \O$ and $V = S$. Hence (trivially) the result. {{qed}}
Let $d: S \times S \to \R$ be the mapping defined as: :$\forall x, y \in S: \map d {x, y} = 0$ Then clearly $d$ is a pseudometric. Let $\struct {S, \tau_{\struct {S, d} } }$ be the topological space induced by $d$. Since $\struct {S, \tau_{\struct {S, d} } }$ is a topological space, by Empty Set is Element of Topology we have that $\O$ is open. For any $\epsilon \in \R_{>0}$ we have: :$\map {B_\epsilon} a := \set {x \in S: \map d {x, a} < \epsilon} = S$ where $\map {B_\epsilon} a$ is the open $\epsilon$-ball of $a$. Let $U\subset S$ be a non-empty open set. By Open Sets in Pseudometric Space we see that for every $x \in U$, there must exist an $\epsilon > 0$ such that $\map {B_\epsilon} x \subset U$. However, $S \subset \map {B_\epsilon} x$. Thus $U = S$. Hence it is seen that $\O$ and $S$ are the only open sets of this resulting pseudometric space. {{qed}}
From Basis for Partition Topology, we have that $\PP$ is a basis for the partition topology. The result follows. {{qed}}
Checking the criteria for $\PP$ to be a synthetic basis for $\tau$: We have that $\displaystyle S = \bigcup \PP$ from the definition of a partition. Therefore, $\displaystyle S \subseteq \bigcup \PP$ and $\PP$ is a cover for $S$. Next, let $B_1, B_2 \in \PP$. Then as $\PP$ is a partition of $S$, we have that $B_1 \cap B_2 = \O$. But from Union of Empty Set we have that $\O$ is the (vacuous) union of sets of $\PP$. Hence $\PP$ is a synthetic basis for $\tau$. {{qed}}
Let $T = \struct {S, \tau}$ be a topological space. === Necessary Condition === Let $T$ be a partition space. Then there exists a partition $\PP$ which forms the basis for $T$. Let $U \in \tau$. Then $U$ is the union of elements of $\PP$. Then by definition $\relcomp S U$ is closed in $T$. Then $\relcomp S U$ is the union of all the other elements of $\PP$. That is, $\relcomp S U$ is also open in $T$. By definition, then, $\relcomp S {\relcomp S U} = U$ is closed in $T$. {{qed|lemma}} === Sufficient Condition === Now suppose that every open set in $T$ is closed, and every closed set in $T$ is open. {{AimForCont}} $T$ is not a partition space. Thus, by definition, there exists an open set of $T$ which is not the union of disjoint sets. {{finish|Anyone?}}
Let the partition $\PP$ be a basis of $T$. From Open Set in Partition Topology is also Closed, open sets are in fact clopen. So the elements of $\PP$ are clopen. The result follows from the definition of components. {{qed}}