hcju/proofwikiqa huggingface dataset:
By definition: :$\tau^*_p = \set {U \cup \set p: U \in \tau} \cup \set \varnothing$ Let $T = \struct {S, \tau}$ be a $T_0$ space. Then: :$\forall x, y \in S$ such that $x \ne y$, either: ::$\exists U \in \tau: x \in U, y \notin U$ :or: ::$\exists U \in \tau: y \in U, x \notin U$ Let $x, y \in S^*_p$ such that $x \ne y, x \ne p, y \ne p$. Then: :$\exists U \in \tau: x \in U, y \notin U$ or: :$\exists U \in \tau: y \in U, x \notin U$ Now let $U^* = U \cup \set p$. It follows that: :$x \in U^*, y \notin U^*$ or: :$y \in U^*, x \notin U^*$ Now consider $x, p \in S^*_p$ such that $x \ne p$. We have that $\set p \in \tau^*_p: p \in \set p, x \notin \set p$ So we have shown that: :$\forall x, y \in S^*_p$ such that $x \ne y$, either: ::$\exists U \in \tau^*_p: x \in U, y \notin U$ :or: ::$\exists U \in \tau^*_p: y \in U, x \notin U$ and so $T^*_p$ is a $T_0$ space. Now suppose that $T = \struct {S, \tau}$ is '''not''' a $T_0$ space. Let $x, y \in S$ such that: :$\forall U \in \tau: x \in U \implies y \in U$ :$\forall U \in \tau: y \in U \implies x \in U$ Now consider $U^* = U \cup \set p$ for any $U \in \tau$. Then: :As $x \in U \implies y \in U$ it follows that $x \in U^* \implies y \in U^*$. :As $y \in U \implies x \in U$ it follows that $y \in U^* \implies x \in U^*$. Hence $T^*_p$ is not a $T_0$ space. {{qed}}
Let $T$ be a trivial space. That is, let $S = \set p$. Then the result holds vacuously, as there are no two distinct points in $T$. Now suppose $T$ is not trivial. Then $\exists x \in S: x \ne p$. Now we have that $\set x \subseteq T$ is open in $T$ such that $p \notin \set x$ but $x \in \set x$. Finally, suppose that $x, y \in S: x \ne y, x \ne p, y \ne p$. Then we have that (for example) $\set x \subseteq T$ is open in $T$ such that $x \in \set x$ but $y \notin \set x$ Hence the result. {{qed}}
We have: : Excluded Point Topology is Open Extension Topology of Discrete Topology : Discrete Space satisfies all Separation Properties (including being a $T_0$ space) Then by Condition for Open Extension Space to be $T_0$ Space, as a discrete space is $T_0$ then so is its open extension. {{qed}}
By definition: :$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$ Let $x \in S^*_p, x \ne p$. Let $U = \set x$. Then $U \in \tau^*_p$ such that $x \in U, p \notin U$. But the only $v \in \tau^*_p$ such that $p \in V$ is the set $S^*_p$, and we have that $x \in S^*_p$. So there is no $V \in \tau^*_p$ such that $x \notin V, p \in V$, by definition of the open extension topology. Hence $T^*_{\bar p}$ can not be a $T_1$ (Fréchet) space. {{qed}}
{{AxiomReview|ZF version of this proof (use Continuous Image of Compact Space is Compact/Corollary 3)}} By Continuous Image of Compact Space is Compact, $f \sqbrk X \subseteq Y$ is compact. Therefore, by Compact Subspace of Metric Space is Bounded, $f$ is bounded. Let $\displaystyle A = \inf_{x \mathop \in X} \norm {\map f x}$. It follows from the definition of infimum that there exists a sequence $\sequence {y_n}$ in $X$ such that: :$\displaystyle \lim_{n \mathop \to \infty} \norm {\map f {y_n} } = A$ By Sequence of Implications of Metric Space Compactness Properties, $X$ is sequentially compact. So there exists a convergent subsequence $\sequence {x_n}$ of $\sequence {y_n}$. Let $\displaystyle x = \lim_{n \mathop \to \infty} x_n$. Since $f$ is continuous and a norm is continuous, it follows by Composite of Continuous Mappings is Continuous that: :$\displaystyle \norm {\map f x} = \norm {\map f {\lim_{n \mathop \to \infty} x_n} } = \norm {\lim_{n \mathop \to \infty} \map f {x_n} } = \lim_{n \mathop \to \infty} \norm {\map f {x_n} } = A$ So $\norm f$ attains its minimum at $x$. By replacing the infimum with the supremum in the definition of $A$, we also see that $\norm f$ attains its maximum by the same reasoning. {{qed}}
We have that the set of continuous mappings $X \to Y$ is a subset of the set $Y^X$ of ''all'' mappings $X \to Y$. Therefore by Vector Space of All Mappings is Vector Space, we need only show that $\CC$ is a subspace of $Y^X$. By the One-Step Vector Subspace Test we need only show that $\CC$ is closed under linear combinations (clearly $\CC$ contains $0$). But this is shown by the Combined Sum Rule for Continuous Functions. {{questionable|The above theorem has been proven only for $\Q, \R, \C$, not the general objects it is being asked to be applied to here.}} We have Supremum Norm is Norm. It remains to be shown that $\CC$ is complete. But this is precisely the statement of the Uniform Limit Theorem. {{questionable|The above statement says nothing about completeness -- just continuity.}} {{qed}} Category:Functional Analysis 2caq0ms1yyf6jpkbjdy6ztf89r7sayw
First: {{begin-eqn}} {{eqn | l = \norm f_\infty       | r = 0 }} {{eqn | ll= \leadstoandfrom       | l = \sup_{x \mathop \in S} \norm {\map f x}       | r = 0 }} {{eqn | ll= \leadstoandfrom       | lo= \forall x \in S:       | l = \norm {\map f x}       | r = 0       | c = since $\norm {\, \cdot \,}$ is a norm, and hence non-negative }} {{eqn | ll= \leadstoandfrom       | lo= \forall x \in S:       | l = \map f x       | r = 0       | c = since $\norm x = 0 \iff x = 0$ }} {{eqn | ll= \leadstoandfrom       | l = f       | r = 0       | c =  }} {{end-eqn}} Now let $\lambda \in K, f \in \BB$ We have: {{begin-eqn}} {{eqn | l = \norm {\lambda f}_\infty       | r = \sup_{x \mathop \in S} \norm {\lambda \map f x}       | c =  }} {{eqn | r = \size \lambda_K \sup_{x \mathop \in S} \norm {\map f x}       | c = Multiple of Supremum, and because $\norm {\, \cdot \,}$ is a norm }} {{eqn | r = \size \lambda_K \, \norm f_\infty       | c =  }} {{end-eqn}} Finally let $f, g \in \BB$. We have: {{begin-eqn}} {{eqn | l = \norm {f + g}_\infty       | r = \sup_{x \mathop \in S} \norm {\map f x + \map g x}       | c =  }} {{eqn | o = \le       | r = \sup_{x \mathop \in S} \paren {\norm {\lambda \map f x} + \norm {\lambda \map g x} }       | c = because $\norm {\, \cdot \,}$ is a norm }} {{eqn | o = \le       | r = \sup_{x \mathop \in S} \norm {\lambda \map f x} + \sup_{x \mathop \in S} \norm {\lambda \map g x}       | c = Supremum of Sum }} {{eqn | r = \norm f_\infty + \norm g_\infty       | c =  }} {{end-eqn}} Thus $\norm {\, \cdot \,}_\infty$ has the defining properties of a norm. {{qed}}
We need to verify the vector space axioms for $U$. We start with observing that the properties for a unitary module are true for all elements of $V$. Hence, since $U \subseteq V$, they hold for all elements of $U$ as well. The same holds for the axioms $(G1)$ and $(C)$. From Vector Inverse is Negative Vector, we have for all $u \in U$: :$u + \paren {- 1_K} u = 0_V$ which by assumption is an element of $U$. Since $U$ is non-empty, this means $0_V \in U$. Hence it is seen that axioms $(G2)$ and $(G3)$ are satisfied. The last axiom that remains is $(G0)$. To this end we employ the knowledge that for all $v \in U$, we have: :$1_K v = v$ and hence: :$u + 1_K v = u + v \in U$ Having verified all the vector space axioms, we conclude that $U$ is a subspace of $V$. {{qed}}
By definition: :$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$ Let $T = \struct {S, \tau}$ be a $T_0$ space. Let $\forall x, y \in S$ such that $x \ne y$. First, suppose $x \ne p \ne y$. Then by definition of $T$ as a $T_0$ space: :$\exists U \in \tau: x \in U, y \notin U$ or: :$\exists U \in \tau: y \in U, x \notin U$ Then as $U \in \tau^*_{\bar p}$, it follows that: :$\exists U \in \tau^*_{\bar p}: x \in U, y \notin U$ or: :$\exists U \in \tau^*_{\bar p}: y \in U, x \notin U$ Now suppose that $y = p$. Then by definition: :$\forall U \in \tau^*_{\bar p}: x \in U, p \notin U$ unless $U = S^*_p$. So we have shown that: :$\forall x, y \in S^*_p$ such that $x \ne y$, either: ::$\exists U \in \tau^*_{\bar p}: x \in U, y \notin U$ :or: ::$\exists U \in \tau^*_{\bar p}: y \in U, x \notin U$ and so $T^*_{\bar p}$ is a $T_0$ space. Now suppose that $T = \struct {S, \tau}$ is '''not''' a $T_0$ space. Let $x, y \in S$ such that: :$\forall U \in \tau: x \in U \implies y \in U$ :$\forall U \in \tau: y \in U \implies x \in U$ As $U \in \tau^*_{\bar p}$ it follows that: :$\forall U \in \tau^*_{\bar p}: x \in U \implies y \in U$ :$\forall U \in \tau^*_{\bar p}: y \in U \implies x \in U$ and if $U = S^*_p$ the same applies. Hence $T^*_{\bar p}$ is not a $T_0$ space. {{qed}}
Follows directly from: :Excluded Point Topology is Open Extension Topology of Discrete Topology :Open Extension Topology is not $T_1$ {{qed}}
Every open set of $T^*_p = \struct {S^*_p, \tau^*_{\bar p} }$ except $S^*_p$ does not contain the point $p$ by definition. So every open set $U \in \tau^*_{\bar p}$ such that $p \in U$ (there is only the one such open set) contains $x$. So by definition of the limit point of a point, $p$ is a limit point of $x$. {{qed}}
Let $U \subseteq S$. Let $x \in S$ such that $x \ne p$. From: :Excluded Point Topology is Open Extension Topology of Discrete Topology :Limit Points in Open Extension Space it follows that: :$p$ is a limit point of $U$ :$p$ is a limit point of $x$. Now suppose $y \in S$ such that $y \ne p$. We have by definition of excluded point space that $\set y$ is open in $T$. So there is no $z \in \set y: z \ne y, z \in U$.  Hence $y$ can not be a limit point of $U$. Similarly $x \notin \set y$. So $y$ can not be a limit point of $x$. Hence the result. {{qed}}
Let $A, B \subseteq S$ such that: :$A^- \cap B = A \cap B^- = \O$ where $A^-$ denotes the closure of $A$. Every closed set of $T$ contains $\set p$. From Limit Points in Excluded Point Space and the definition of closure: :$A^- = A \cup \set p, B^- = B \cup \set p$ So if $A^- \cap B = A \cap B^- = \O$ then $p \in A^-$ and $p \in B^-$. Thus, $p \notin A$ and $p \notin B$ and so by definition $A$ and $B$ are open sets in $T$. It follows that $A \cap B = \O$. Hence the result, as we have shown that: :$\forall A, B \subseteq X, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau_{\bar p}: A \subseteq U, B \subseteq V, U \cap V = \O$ where $U = A$ and $V = B$. {{qed}}
We have: : Excluded Point Topology is Open Extension Topology of Discrete Topology : Open Extension Space is Compact {{qed}}
By definition of excluded point space, the only open set of $T$ which contains $p$ is $S$. So any open cover $\CC$ of $T$ must have $S$ in it. So $\set S$ will be a subcover of $\CC$, whatever $\CC$ may be. And $\set S$ (having only one set in it) is trivially a finite cover of $T$. Hence the result, by definition of compact topological space. {{qed}}
By definition of open extension space, the only open set of $T^*_{\bar p}$ containing $p$ is $S^*_p$. So any open cover $\CC$ of $T^*_{\bar p}$ must have $S^*_p$ in it. So $\set {S^*_p}$ will be a subcover of $\CC$, whatever $\CC$ may be. And $\set {S^*_p}$ (having only one set in it) is trivially a finite cover of $T^*_{\bar p}$. Hence the result, by definition of compact space. {{qed}}
The only open set of $T$ which contains $p$ is $S^*_p$. Therefore it is impossible to set up a separation of $T$, as $S^*_p$ will always need to be an element of such a separation. {{qed}}
We have: : Excluded Point Topology is Open Extension Topology of Discrete Topology : Open Extension Space is Connected {{qed}}
The only open set of $T$ which contains $p$ is $S$. Therefore it is impossible to set up a separation of $T$, as $S$ will always need to be an element of such a separation. {{qed}}
: Excluded Point Space is Ultraconnected : Ultraconnected Space is Path-Connected : Path-Connected Space is Connected {{qed}}
Apart from $S^*_p$, every open set of $T^*_{\bar p}$ does not contain $p$, by definition of open extension space. So, apart from $\O$, every closed set of $T$ does contain $p$, by definition of closed set. So every pair of closed sets of $T$ has an intersection which contains at least $p$. So there are no non-empty disjoint closed sets of $T$. Hence the result, by definition of ultraconnected. {{qed}}
: Excluded Point Topology is Open Extension Topology of Discrete Topology : Open Extension Space is Ultraconnected {{qed}}
Apart from $S$, every open set of $T$ does not contain $p$, by definition of excluded point space. So, apart from $\O$, every closed set of $T$ does contain $p$, by definition of closed set. So every pair of closed sets of $T$ has an intersection which contains at least $p$. So there are no non-empty disjoint closed sets of $T$. Hence the result, by definition of ultraconnected. {{qed}}
By definition, open sets of $S$ are precisely the open sets of $S \setminus \set p$ under the discrete topology. Let $x, y \in S \setminus \set p: x \ne y$. Then $\set x$ and $\set y$ are both open sets of $T$ such that $\set x \cap \set y = \O$. Hence the result, by definition of irreducible. {{qed}}
: Excluded Point Topology is Open Extension Topology of Discrete Topology : Open Extension Space is Path-Connected {{qed}}
:Excluded Point Space is Ultraconnected :Ultraconnected Space is Path-Connected {{qed}}
:Open Extension Space is Ultraconnected :Ultraconnected Space is Path-Connected {{qed}}
Let $q \in S$ be such that $q \ne p$. Let $f: \closedint 0 1 \to T$ be an injection such that: :$\map f 0 = p$ :$\map f 1 = q$ Because $f$ is an injection, it must be that: :$f^{-1} \sqbrk {\set q} = \set 1$ where $f^{-1}$ denotes the preimage under $f$. Now since $p \notin \set q$, $\set q$ is open in $\tau_{\bar p}$, by definition of the excluded point topology. By Closed Real Interval is not Open Set, $\set 1 = \closedint 1 1$ is not open. Thus we have exhibited an open set whose preimage under $f$ is not open. Hence $f$ is not continuous. Since $f$ was an arbitrary injection, no arc between $p$ and $q$ can exist. Hence $T$ is not arc-connected. {{qed}}
Let $H \in \tau_{\bar p}$ be open in $T$. If $H = S$ then trivially $H$ is the union of elements of $\mathcal B$. So suppose $H \ne S$. Then by definition $p \notin H$ and so: : $\forall y \in H: \exists \left\{{y}\right\} \in \mathcal B$ Thus: :$\displaystyle H = \bigcup_{y \in H} \left\{{y}\right\}$ So $\mathcal B$ is an analytic basis for $T$. {{qed}} It could equally well be shown that $\mathcal B$ is also a synthetic basis for $T$. Category:Excluded Point Topology Category:Topological Bases mz0v244fyb4gjyq4r16iac42au531re
Consider the set $\BB$ defined as: :$\BB = \set {\set x: x \in S \setminus \set p} \cup \set S$ Then $\BB$ is a basis for $T$. Let $H \in \BB$. Then $\exists x \in S: H = \set x$. From Point is Path-Connected to Itself we have that $H$ is path-connected. Now consider the open set $S \in \BB$. Let $a, b \in S$. Let $\mathbb I$ be the closed unit interval in $\R$. Let $f: \mathbb I \to S$ be the mapping defined as: :$\forall x \in \mathbb I: \map f x = \begin{cases} a & : x \in \hointr 0 1 \\ b & : x = 1 \end{cases}$ Then $f^{-1} \sqbrk S = \closedint 0 1$ which is open in $\mathbb I$ because $\closedint 0 1 = \mathbb I$. As $a$ and $b$ are any two points of $S$ it follows that $S$ is path-connected. So we have shown that all elements of a basis of $T$ are path-connected. Hence by definition $T$ is locally path-connected. {{qed}}
Let $\BB \subseteq \tau_{\bar p}$ be a basis for $\tau_{\bar p}$. Since $\BB$ covers $S$, there must be an open set $B \in \BB$ such that $p \in B$. By definition of the excluded point topology, the only open set containing $p$ is $S$ itself. Hence necessarily $S \in \BB$. But by Excluded Point Space is not Arc-Connected, $S$ is not arc-connected. Hence $\BB$ does not consist only of arc-connected sets. Because $\BB$ was arbitrary, there cannot exist a basis for $\tau_{\bar p}$ comprising only arc-connected sets. Hence, by definition, $T$ is not locally arc-connected. {{qed}}
Let $\PP$ be the partition which is the basis for $T$. From Open Set in Partition Topology is also Closed, all the elements of $\PP$ are both closed and open. Hence the result, by definition of zero dimensional space {{qed}} Category:Zero Dimensional Spaces Category:Partition Topology cipbm5n8qxfxks85cia91feioap6cyt
We have from Partition of Singletons yields Discrete Topology that a discrete space is a partition space. The result follows from Partition Topology is Zero Dimensional. {{qed}}
Let $\BB$ be the set: :$\BB := \set {\set x: x \in S}$ From Basis for Discrete Topology, $\BB$ is a basis for $T$. From Set in Discrete Topology is Clopen, all the elements of $\mathcal B$ are both closed and open. Hence the result, by definition of zero dimensional space {{qed}}
We have that Topological Space is Discrete iff All Points are Isolated. So, by definition, no subset $H \subseteq S$ of $T$ such that $H \ne \varnothing$ is dense-in-itself. So, again, by definition, $T$ is scattered. {{qed}}
First we note that as Discrete Space satisfies all Separation Properties, $T$ is a $T_2$ (Hausdorff) space. Then from Interior Equals Closure of Subset of Discrete Space, it follows directly that the closure of every open set of $T$ is open. Hence by definition $T$ is extremally disconnected. {{qed}}
Let $T = \struct {S, \tau}$ be a totally separated space. Then for every $x, y \in S: x \ne y$ there exists a partition $U \mid V$ of $T$ such that $x \in U, y \in V$. Suppose $T$ were not totally disconnected. Then $\exists H \subseteq S: x, y \in U$ such that $H$ is connected. But by definition of connected, there can be no partition $U \mid V$ of $T$ such that $x \in U, y \in V$. Hence the result. {{qed}}
By definition: :$\tau^*_{\bar p} = \set {U: U \in \tau} \cup \set {S^*_p}$ We have that $S$ is an open set in $T$ and so open set in $T^*_{\bar p}$. So $\set p = S^*_p \setminus S$ is closed in $T^*_{\bar p}$. The only open set in $T^*_{\bar p}$ which contains $p$ is $S^*_p$. So $\set p$ can not be the intersection of open sets of $T^*_{\bar p}$, whether that intersection be countable or not. So by definition $T^*_{\bar p}$ is not a perfectly $T_4$ space. {{qed}}
:Excluded Point Topology is Open Extension Topology of Discrete Topology :Open Extension Topology is not Perfectly T4 {{qed}}
From Limit Points in Excluded Point Space, the only limit point of $H$ is $p$. So by definition, all points of $H$ are isolated in $H$ except $p$. So if  $H \ne \set p$, $H$ contains at least one point which is isolated in $H$. As for $p$ itself, from Singleton Point is Isolated we have that $p$ is itself isolated in $\set p$. So if  $H = \set p$, $H$ also contains one point which is isolated in $H$. Hence the result, by definition of dense-in-itself. {{qed}}
We have that Subset of Excluded Point Space is not Dense-in-itself. So, by definition, $T$ is scattered. {{qed}}
We have that the Excluded Point Topology is Open Extension Topology of Discrete Topology. So $S \setminus \set p$ is a discrete space. Then a discrete space is totally disconnected. The result follows from definition of dispersion point. {{qed}}
Let $x \in S$ such that $x \ne p$. By definition, the set $\set x$ is open in $T$. Let $U \in \tau_{\bar p}: x \in U$. Then $\set x \subseteq U$ and so $\set {\set x}$ is a local basis at $x$ which is trivially countable. Now if $x = p$ there is only one $U \in \tau_{\bar p}: p \in U$, and that is $S$. So $\set S$ is a local basis at $x$ which is trivially countable. Hence the result, by definition of first-countable. {{qed}}
:Excluded Point Space is Compact :Compact Space is Countably Compact :Excluded Point Space is First-Countable :First-Countable Space is Sequentially Compact iff Countably Compact {{qed}}
Consider the set $\mathcal B$ defined as: :$\mathcal B = \left\{{\left\{{x}\right\}: x \in S \setminus \left\{{p}\right\}}\right\} \cup \left\{{S}\right\}$ From Basis for Excluded Point Space, $\mathcal B$ is a basis for $T$, and trivially has the same cardinality as $S$. So by definition, if $S$ is countable, then $T$ is second-countable. {{qed}}
We have: : Countable Discrete Space is Second-Countable : Excluded Point Topology is Open Extension Topology of Discrete Topology The result follows from Condition for Open Extension Space to be Second-Countable. {{qed}}
Let $H = S \setminus \left\{{p}\right\}$ where $\setminus$ denotes set difference. By definition, $H$ is an uncountable discrete space. The result follows from Uncountable Discrete Space is not Second-Countable. {{qed}}
We have: : Uncountable Discrete Space is not Second-Countable : Excluded Point Topology is Open Extension Topology of Discrete Topology The result follows from Condition for Open Extension Space to be Second-Countable {{qed}}
The closure of the set $S$ (trivially) equals $S$. That is, $S$ is everywhere dense in $T$. But as $S$ is countable it follows by definition that $T$ is separable. {{qed}}
Let $H \subseteq S$ such that $H$ is countable. Then $H \ne S$ as $S$ is uncountable by hypothesis. From Limit Points in Excluded Point Space, the only limit point of $H$ is $p$. So, by definition, the closure of $H$ is $H \cup \set p$. From Countable Union of Countable Sets is Countable we have that $H \cup \set p$ is countable. So $H \cup \set p \ne S$. So $H$ is not everywhere dense in $T$. Thus $T$ can have no countable subset of $S$ which is everywhere dense in $T$. Hence the result by definition of separable space. {{qed}}
We have by definition that $S \in \tau_{\bar H}$. Also, as $H \cap \O = \O$, we have that $\O \in \tau_{\bar H}$. Now let $U_1, U_2 \in \tau_{\bar H}$. By definition: :$H \cap U_1 = \O$ and: :$H \cap U_2 = \O$ and so by definition of set intersection: :$H \cap \paren {U_1 \cap U_2} = \O$ So: :$U_1 \cap U_2 \in \tau_{\bar H}$ Now let $\UU \subseteq \tau_{\bar H}$. We have that: :$\forall U \in \UU: H \cap U = \O$ Hence from Subset of Union: :$H \cap \bigcup \UU = \O$ So all the conditions are fulfilled for $\tau_{\bar H}$ to be a topology on $S$. {{qed}}
Let $x, y \in H$ such that $x \ne y$. Then $x, y \in S$, but by definition $x$ and $y$ are in no other open sets of $T$. Hence there is no $U \in \tau_{\bar H}$ such that $x \in U, y \notin U$ or $y \in U, x \notin U$. Hence the result, by definition of $T_0$ (Kolmogorov) space. {{qed}}
From the definition: :$H \in \tau \iff \paren {\set 0 \nsubseteq H \lor \openint {-1} 1 \subseteq H}$ for any $H \subseteq S$. First note that: :$\set 0 \nsubseteq \O$ and so $\O \in \tau$ :$\openint {-1} 1 \subseteq S$ and so $S \in \tau$ Now suppose $H_1, H_2 \in \tau$. If either $\set 0 \nsubseteq H_1$ or $\set 0 \nsubseteq H_2$ then from De Morgan's Laws: Difference with Intersection (indirectly): :$\set 0 \nsubseteq H_1 \cap H_2$ Otherwise: :$\openint {-1} 1 \subseteq H_1$ and $\openint {-1} 1 \subseteq H_2$ and so from Intersection is Largest Subset: :$\openint {-1} 1 \subseteq H_1 \cap H_2$ So: :$H_1, H_2 \in \tau \implies H_1 \cap H_2 \in \tau$ Now let $\HH \subseteq \tau$ be a set of elements of $\tau$. Then either: :$\forall H \in \HH: \set 0 \nsubseteq H$ in which case $\set 0 \nsubseteq \bigcup \HH$ or: :$\exists H \in \HH: \openint {-1} 1 \subseteq H$ in which case $\openint {-1} 1 \subseteq \bigcup \HH$ In both cases $\displaystyle \bigcup \HH \in \tau$. Note that $\set 0 \nsubseteq H \implies \openint {-1} 1 \nsubseteq H$ so there does not exist the confusion of what happens if the conditions are contradictory. Hence the result, from the definition of topology. {{qed}}
From Open and Closed Sets in Topological Space we have that $\O$ and $S$ are closed sets trivially. From the definition of closed set, we have: :$U$ is open in $T$ {{iff}} $S \setminus U$ is closed in $T$ :$U$ is closed in $T$ {{iff}} $S \setminus U$ is open in $T$ where $S \setminus U$ denotes the relative complement of $U$ in $S$. Now we have that: :$\openint {-1} 1$ is open in $T$ so $S \setminus \openint {-1} 1 = \set {-1, 1}$ is closed in $T$. :Both $\hointr {-1} 1 \supseteq \openint {-1} 1$ and $\hointr {-1} 1 \supseteq \openint {-1} 1$ therefore are open in $T$, so $\set {-1}$ and $\set 1$ are closed in $T$. Apart from $S$ itself, there are no other subsets of $S = \closedint {-1} 1$ which have $\openint {-1} 1$ as a subset. Finally, let $U \in \tau: \set 0 \nsubseteq U$. Then by the definition of relative complement $\set 0 \subseteq S \setminus U$. Now suppose $\set 0 \subseteq V$. Then again by the definition of relative complement $\set 0 \nsubseteq S \setminus V$. So $S \setminus V$ is open in $T$ and so $V$ is closed in $T$. {{qed}}