hcju/proofwikiqa huggingface dataset:
Let: : $\left({R, +, \circ}\right)$ be a commutative ring : $\left({D, +, \circ}\right)$ be an integral subdomain of $R$ whose zero is $0_D$ and whose unity is $1_D$ : $X \in R$ be transcendental over $D$. It has been demonstrated that the Boubaker Polynomials sub-sequence $B_{4 n} \left({x}\right)$, defined in $D \left[{X}\right]$ as: :$\displaystyle B_{4 n} \left({x}\right) = 4 \sum_{p \mathop = 0}^{2 n} \frac {n - p} {4 n - p} \binom {4 n - p} p \left({-1}\right)^p x^{2 \left({2n - p}\right)}$ satisfies the properties: {{begin-eqn}} {{eqn | n = 1       | l = \sum_{k \mathop = 1}^N {p_n \left({0}\right)}       | r = -2N }} {{eqn | n = 2       | l = \sum_{k \mathop = 1}^N {p_n \left({\alpha_k}\right)}       | r = 0 }} {{eqn | n = 3       | l = \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d p_x \left({x}\right)} {\mathrm d x} }\right\vert_{x \mathop = 0}       | r = 0 }} {{eqn | n = 4       | l = \left.{\sum_{k \mathop = 1}^N \frac {\mathrm d {p_n}^2 \left({x}\right)} {\mathrm d x^2} }\right\vert_{x \mathop = 0}       | r = \frac 8 3 N \left({N^2 - 1}\right) }} {{end-eqn}} with $\left. {\alpha_k}\right\vert_{k \mathop = 1 \,.\,.\, N}$ roots of $B_{4 n}$. Suppose there exists another $4 n$-indexed polynomial $q_{4 n} \left({x}\right)$, with $N$ roots $\left.{\beta_k}\right\vert_{k \mathop = 1 \,.\,.\, N}$ in $F$ and which also satisfies simultaneously properties $(1)$ to $(4)$. Let: :$\displaystyle B_{4 n} \left({x}\right) = \sum_{p \mathop = 0}^{2 n} a_{4 n, p} x^{2 \left({2 n - p}\right)}$ and: :$\displaystyle q_{4 n} \left({x}\right) = \sum_{p \mathop = 0}^{2 n} b_{4 n, p} x^{2 \left({2 n - p}\right)}$ and: :$\displaystyle \mathrm d_{4 n, p} = a_{4 n, p} - b_{4 n, p}$ for $p = 0 \,.\,.\, 2 n$ then, simultaneous expressions of conditions $(1)$ and $(3)$ give: :$ \quad \displaystyle \sum_{k \mathop = 1}^N \mathrm d_{4 n, 2 n} = 0$ :$ \quad \displaystyle \sum_{k \mathop = 1}^N \mathrm d_{4 n, 2 n - 2} = 0$ It has also been demonstrated that $ B_{4 n}$ has exactly $4 n - 2$ real roots inside the domain $\left[{-2 \,.\,.\, 2}\right]$. So application of conditions $(3)$ and $(4)$ give $4n-2$ linear equation with variables $\left.{ d_{4n,p}}\right|_{p \mathop = 0 \,.\,.\, 2n-3}$. Finally, since $B_{4 n}$ contains $2 n$ monomial terms (see definition), we obtain a Cramer system in variables $\left.{ d_{4 n, p}}\right\vert_{p \mathop = 0 \,.\,.\, 2 n}$, with evident solution: :$\left.{\mathrm d_{4 n, p}}\right\vert_{p \mathop = 0 \,.\,.\, 2 n} = 0 $ and consequently: :$\left.{a_{4 n, p} }\right\vert_{p \mathop = 0 \,.\,.\, 2 n} = \left.{b_{4 n, p} }\right\vert_{p \mathop = 0 \,.\,.\, 2 n}$ which means: :$q_{4 n} \left({x}\right) = B_{4 n} \left({x}\right)$ {{qed}}
It will suffice to show that every open cover of $S_n^{\mathcal M}(A)$ by the basic open sets $[\phi]$ of the topology has a finite subcover. Let $\mathcal U = \{ [\phi_i] : i\in I \}$ be a cover of $S_n^{\mathcal M}(A)$ by basic open sets. This means that every complete $n$-type over $A$ contains some $\phi_i$. We will find a finite subcover of $\mathcal U$. Let $\Gamma = \{ \neg\phi_i : i\in I \}$. Then $\operatorname{Th}_\mathcal A(M) \cup \Gamma$ cannot be satisfied, since if $\mathcal N \models \operatorname{Th}_\mathcal A(M) \cup \Gamma (\bar{b})$, then the type $\operatorname{tp}_\mathcal N (\bar{b}/A)$ is a complete $n$-type in $S_n^{\mathcal M}(A)$ which does not contain any $\phi_i$. By the Compactness Theorem, $\operatorname{Th}_\mathcal A(M) \cup \Gamma$ must have a finite subset $\Delta$ which is not satisfiable. Since $\Delta$ is not satisfiable but $\operatorname{Th}_\mathcal A(M)$ is, $\Delta$ must contain some of the $\neg\phi_i$ from $\Gamma$. Furthermore, we must have that every model of $\operatorname{Th}_\mathcal A(M)$ fails to satisfy at least one of these finitely many $\neg\phi_i$ in $\Delta$. We claim that the finite set $\mathcal F = \{ [\phi_i] : \neg\phi_i \in \Delta \}$ is a finite subcover of $\mathcal U$. Since $\mathcal F$ is clearly a subset of $\mathcal U$, we need only show that every $p\in S_n^{\mathcal M}(A)$ is contained in some $[\phi_i]\in\mathcal F$. That is, we must show that each $p\in S_n^{\mathcal M}(A)$ contains one of these $\phi_i$. Let $p\in S_n^{\mathcal M}(A)$. By definition, this means that $p\cup \operatorname{Th}_\mathcal A(M)$ is satisfiable by some $\mathcal N$. But, as mentioned above, since $\mathcal N\models \operatorname{Th}_\mathcal A(M)$, we have that $\mathcal N\not\models \neg\phi_i$ for some $\neg\phi_i \in \Delta$. Since $p$ is complete, it must contain either $\phi_i$ or $\neg\phi_i$. However, if $p$ contained $\neg\phi_i$, then $\mathcal N\models \neg\phi_i$, which contradicts $\mathcal N\not\models \neg\phi_i$. Thus, $p$ contains $\phi_i$. This demonstrates that $\mathcal F$ is a finite subcover for $\mathcal U$. Thus, $S_n^{\mathcal M}(A)$ is compact. {{qed}} Category:Model Theory nn84c5wfi332meprrm3tew7ig41k53n
Let $x, y \in S$ such that $x \ne y$. {{WLOG}}, let $x \ne 0$. Then $U = \set x$ is open in $T$ from the definition of the either-or topology. We have that $x$ and $y$ are arbitrary. So: :$\forall x, y \in S: \exists U \in \tau: x \in U, y \notin U$ and the result follows by definition of $T_0$ (Kolmogorov) space. {{qed}}
Let $x = \dfrac 1 2$. We have that $V = \set x$ such that $x \in V, 0 \notin V$. However, by definition of the either-or topology, the only open sets of $T$ containing $0$ also contain $\openint {-1} 1$, and so must also contain $x$. So we have that $\nexists U, V \in \tau: 0 \in U, x \notin U, x \in V, 0 \notin V$. Hence the result by definition of $T_1$ (Fréchet) space. {{qed}}
=== Lemma 1 === {{:Vinogradov's Theorem/Minor Arcs/Lemma 1}} {{qed|lemma}} {{finish|The below need proofs}} === Lemma 2 === Let $a, q \in \Z$ such that: :$(1): \quad \displaystyle \size {\alpha - \frac a q} \le \frac 1 {q^2} \quad 1 \le q \le N, \quad \gcd \set {a, q} = 1$ Let $m, n \in \Z$ be integers. Then: :$\displaystyle\sum_{k mathop = 1}^m \min \set {\frac {m n} k, \frac 1 {\norm {\alpha k} } } \ll \paren {m + \frac {m n} q + q} \map \ln {2 q m}$ {{qed|lemma}} === Lemma 3 === Let $\alpha$ satisfy the condition $(1)$ of Lemma 2. Let $x, y \in \N$. Let $\beta_k, \gamma_k$ be any complex numbers such that $\size {\beta_k}, \size {\gamma_k} \le 1$. Then: :$\displaystyle \sum_{k \mathop = y}^{x/y} \sum_{\ell \mathop = y}^{x/k} \alpha_k \beta_\ell \map e {\alpha k \ell} \ll x^{1/2} \paren {\ln x}^2 \paren {\frac x y + y + \frac x q + q}^{1/2}$ {{qed|lemma}} {{qed}} Category:Vinogradov's Theorem dofrtafwri6ze1f5k2ftmsxii6qjyh7
=== Lemma 1 === {{:Vinogradov's Theorem/Major Arcs/Lemma 1}} {{qed|lemma}} {{finish|These need proof.}} === Lemma 2 === Let $N \in \N_{\ge 1}$. Let $\beta \in \R$. Let: :$\displaystyle \map u \beta = \sum_{n \mathop \le N} \map e {n \beta}$ For $P \ge 1$, define: :$\displaystyle \map {J_P} N = \int_{-P/N}^{P/N} \map u \beta^3 \map e {-N \beta} \rd \beta$ :$\map J N = \map {J_{N/2} } N$ {{explain|What is $P$? Real, integer ...?}} Then with $Q = \paren {\ln N}^B$ as above: {{explain|Above where? This page needs to be made standalone.}} :$\map {J_Q} N = \map J N + \map {\mathcal O} {\dfrac {N^2} {Q^2} }$ and: :$\map J N = \dfrac {N^2} 2 + \map {\mathcal O} N$ {{qed|lemma}} === Lemma 3 === Let $\alpha \in \map {\mathcal M} {q, a}$ for some $q, a$ such that: :$\map {\mathcal M} {q, a} \subseteq \mathcal M$ Let $\beta = \alpha - \dfrac a q$. Then: :$\map F \alpha^3 = \dfrac {\map \mu q} {\map \phi q^3} \map u \beta^3 + \map {\mathcal O} {N^3 \map \exp {-C \sqrt {\ln N} } }$ where $C$ is a constant that depends only on $B$. {{qed|lemma}} {{qed}} Category:Vinogradov's Theorem ejkfvj0y2jdoat0l4ygfprf6eag1rt1
Choose any $N > \dfrac{\left({k + 1}\right)!} {\alpha^{k+1}}$, where $!$ denotes the factorial. By Taylor Series Expansion for Exponential Function we have for any $x \in \R_{\ge 0}$: :$\displaystyle \exp \left({\alpha x}\right) = \sum_{m \mathop \ge 0} \frac{\left({\alpha x}\right)^m}{m!} > \frac{\left({\alpha x}\right)^{k+1}}{\left({k + 1}\right)!}$ Therefore, for any $x > N$ we have: {{begin-eqn}} {{eqn | l = \exp \left({\alpha x}\right)       | o = >       | r = \frac{\left({\alpha x}\right)^{k+1} } {\left({k + 1}\right)!} }} {{eqn | o = >       | r = \frac{N \alpha^{k+1} } {\left({k + 1}\right)!} x^k       | c = because $x > N$ }} {{eqn | o = >       | r = x^k       | c = by definition of $N$ }} {{end-eqn}} This establishes the result. {{qed}} Category:Exponential Function Category:Limits of Functions Category:Polynomial Theory toxnjiz0eq6qyr8zc8m7xcmvpyfys0g
Let $\epsilon \in \R_{>0}$ be a (strictly) positive real number. First we show that there exists $N$ such that $\ln n < n^\epsilon$ for all $n > N$,. By Exponential is Strictly Increasing, the real exponential function is strictly increasing. Therefore: :$\ln n < n^\epsilon \iff n < \exp \left({n^\epsilon}\right)$ Choose $k \in \N$ such that $k \epsilon > 2$ and $N = k!$. By Taylor Series Expansion for Exponential Function: :$\displaystyle \exp \left({n^\epsilon}\right) = \sum_{m \mathop \ge 0} \frac{n^{m \epsilon}}{m!} > \frac{n^{k \epsilon}}{k!}$ Then for all $n > N$: {{begin-eqn}} {{eqn | l = \exp \left({n^\epsilon}\right)       | r = \frac {n^{k \epsilon} } {k!}       | o = >       | c =  }} {{eqn | r = \frac N {k!} n^{k \epsilon - 1}       | o = >       | c = }} {{eqn | r = n       | o = >       | c = because $k \epsilon > 2$ and $N = k!$ }} {{end-eqn}} This completes the proof when $B = 1$. Now let $B \in \N$ be arbitrary. By the above we can find $N \in \N$ such that: :$\forall n > N: \ln n < n^{\epsilon / B}$ Then: :$\forall n > N: \left({\ln n}\right)^B < \left({n^{\epsilon / B} }\right)^B = n^\epsilon$ Hence the result. {{qed}} Category:Limits of Functions Category:Logarithms Category:Powers j8f0ldxcjxzqhv7zu4zcj5058qmpp24
{{AimForCont}} that the intervals of $\JJ$ are pairwise disjoint. Let $I = \closedint p q$ be the unique interval of $\JJ$ such that $a \in I$. Let $J = \closedint r s$ be the unique interval of $\JJ$ containing the least real number not in $I$. These choices are possible since $\JJ$ has at least two elements, and they are supposed disjoint. If $q = r$ then $I \cap J \ne \O$, a contradiction. If $r < q$ then $I \cap J \ne \O$, a contradiction. If $r > q$ then there is some real number with $q < \alpha < r$. Therefore $\alpha \notin I$. Since $J$ contains the least real number not in $I$ it follows that there is $\beta \in J$ with $\beta < \alpha$. But we also have that $\alpha < r \le \beta$ for all $\beta \in J$, a contradiction. This exhausts all the possibilities, and we conclude that the intervals of $\JJ$ are not pairwise disjoint. {{qed}} Category:Measure Theory 9wda0ujr5f54g8ssj2kpco4umgx8h92
We have that an Open Extension Space is Ultraconnected. That means none of its closed sets are disjoint. Hence, vacuously, any two of its disjoint closed subsets are separated by neighborhoods. The result follows by definition of $T_4$ space. {{qed}}
We have that an Excluded Point Space is Ultraconnected. That means none of its closed sets are disjount. Hence, vacuously, any two of its disjoint closed subsets of $S$ are separated by neighborhoods. The result follows by definition of $T_4$ space. {{qed}}
We have: : Excluded Point Topology is Open Extension Topology of Discrete Topology : Open Extension Topology is $T_4$ Hence the result. {{qed}}
We have: : Excluded Point Topology is $T_5$ : $T_5$ Space is $T_4$ {{qed}}
As $S$ is (trivially) open in $T$, it is also open in $T^*_{\bar p}$. As $S^*_p = S \cup \set p$, it follows that $\set p$ is closed in $T^*_{\bar p}$. But by definition, the only open set of $T^*_{\bar p}$ that contains $\set p$ is $S^*_p$ itself. So there can be no $x \in S^*_p: x \notin \set p$ contained in an open set of $T^*_{\bar p}$ which is disjoint from $S^*_p$. Hence the result. {{qed}}
:Excluded Point Topology is Open Extension Topology of Discrete Topology :Open Extension Topology is not T3 {{qed}}
Let $T = \struct {S, \tau}$ be a separable space. Then there exists a countable subset $H \subseteq S$ which is everywhere dense in $T$. That is, $H^- = S$ where $H^-$ is the closure of $H$ in $S$. So in $T_{\bar p}^*$, $S\subseteq H^-=S^-\subseteq S^*_p$. Hence $H^- = S$ or $H^- = S \cup \set p = S^*_p$. From Topological Closure is Closed we deduce that $H^-$ is closed in $T_{\bar p}^*$. From the definition of open extension topology, the closed sets of $T_{\bar p}^*$ are $S^*_p \setminus U$ where $U \in \tau$. Because $\set p \notin \tau$, $S$ is not closed. Thus it must be that $H^- = S^*_p$. Hence $H$ is a countable subset $H \subseteq S^*_p$ which is everywhere dense in $T^*_{\bar p}$. {{qed|lemma}} {{AimForCont}} $T^*_{\bar p} = \struct {S^*_p, \tau}$ is separable, but that $T = \struct {S, \tau}$ is not separable. Let $H \subseteq S^*_p$ be a countable subset which is dense in $T^*_{\bar p}$. If $H \subseteq S$, then $H^- = S$ in $T$, but this cannot be since $T$ is not separable. Then $p \in H$. Consider the subset $V = H \setminus \set p$. From Topological Closure is Closed we have that $V^-$ is closed in $T_{\bar p}^*$. From the definition of open extension space, all non-empty closed sets contain the point $p$. So $V \cup \set p = H \subseteq V^-$.  From Set Closure is Smallest Closed Set in Topological Space, $S^*_p = H^- \subseteq V^- \implies V$ is dense in $T_{\bar p}^*$. But because $V \subseteq S$, $V$ is dense in $S$ also. From Subset of Countably Infinite Set is Countable, $V$ is countable as it is a subset of a countable set. Thus $T$ is separable, which is a contradiction. Finally, our assumption is false: $T$ is indeed separable. {{qed}}
Let $\mathcal A$ be the set of finitely satisfiable extensions of $T$. By the lemma, for each element $S$ of $\mathcal A$ and each $\mathcal L$-sentence $\phi$, either $S \cup \left\{ {\phi}\right\} \in \mathcal A$ or $S \cup \left\{ {\neg \phi}\right\} \in \mathcal A$. $\mathcal A$ has finite character, by the following argument: Let $S \in \mathcal A$. Let $F$ be a finite subset of $S$. Then $S$ is satisfiable and hence finitely satisfiable. Thus in $\mathcal A$. Let $S$ be a theory on $\mathcal L$. Let every finite subset of $S$ be finitely satisfiable. Then every finite subset of $S$ is satisfiable. Therefore $S$ is finitely satisfiable. Thus $\mathcal A$ has finite character. By the Restricted Tukey-Teichmüller Theorem, $\mathcal A$ has an element $T'$ such that: :for each $\mathcal L$-sentence $\phi$, either $\phi \in T'$ or $\neg \phi \in T'$. {{qed}}
The set of all finitely satisfiable $\mathcal L$-theories containing $T$ forms an ordered set using subset inclusion as the ordering. {{explain|Link to "subset is ordering"}} Let $C$ be a nonempty chain in this ordered set. Let $\displaystyle T_C = \bigcup_{\Sigma \mathop \in C} \Sigma$. Let $\Delta$ be a finite subset of $T_C$. Then there exists a single $\Sigma$ in $C$ which contains $\Delta$. Since this $\Sigma$ is finitely satisfiable by definition, this means that $\Delta$ is satisfiable. Hence $T_C$ is finitely satisfiable. Since each $\Sigma \in C$ is contained in $T_C$, this means that $T_C$ is an upper bound for $C$ in the ordered set. Thus, by Zorn's Lemma, there is a finitely satisfiable $\mathcal L$-theory $T'$ containing $T$ such that $T'$ contains all other such theories. Let $\phi$ be an $\mathcal L$-sentence. Let $\phi \nsubseteq T'$. Aiming for a contradiction, suppose $T' \cup \left\{ {\phi}\right\}$ were finitely satisfiable. Then by definition of $T'$, $T'$ would contain $T' \cup \left\{ {\phi}\right\}$ as a subset. This would mean that $T'$ contains $\phi$, which contradicts the assumption. Thus $T' \cup \left\{ {\phi}\right\}$ is not finitely satisfiable. By the lemma, $T' \cup \left\{ {\neg \phi}\right\}$ is finitely satisfiable. Thus, by definition of $T'$, $T'$ contains $T' \cup \left\{{\neg \phi}\right\}$ as a subset. Hence $T'$ contains $\neg \phi$. {{qed}}
Let $T = \struct {S, \tau}$ be a first-countable space. Then every point in $S$ has a countable local basis. Every open set of $T$ is an open set of $T^*_{\bar p}$ by definition. So if $x$ has a countable local basis in $T$, then it has one in $T^*_{\bar p}$ as well. Finally we note that $p$ is by definition in exactly one open set of $T^*_{\bar p}$, that is, $S^*_p$, and thus (trivially) has a countable local basis in $T^*_{\bar p}$. Then every point in $S^*_p$ has a countable local basis in $T^*_{\bar p}$. So if $T$ is a first-countable space, then $T^*_{\bar p}$ is also a first-countable space. Now suppose $T = \struct {S, \tau}$ is not a first-countable space. Then $\exists x \in S$ such that $x$ has no countable local basis in $T$. As every open set of $T$ is an open set of $T^*_{\bar p}$ by definition, it follows that $x$ has no countable local basis in $T^*_{\bar p}$ either. So if $T $ is not a first-countable space, then neither is $T^*_{\bar p}$ a first-countable space. {{qed}}
Let $T = \struct {S, \tau}$ be a second-countable space. Then $\tau$ has a countable basis. Every open set of $T$ is an open set of $T^*_{\bar p}$ by definition. So if $\tau$ has a countable basis in $T$, then $\tau^*_{\bar p} = \tau \cup \set {S^*_p}$ has one as well. So if $T$ is a second-countable space, then $T^*_{\bar p}$ is also a second-countable space. Now suppose $T = \struct {S, \tau}$ is not a second-countable space. Then $\tau$ has no countable basis in $T$. As every open set of $T$ is an open set of $T^*_{\bar p}$ by definition, it follows that $\tau^*_{\bar p} = \tau \cup \set {S^*_p}$ has no countable basis either. So if $T$ is not a second-countable space, then neither is $T^*_{\bar p}$ a second-countable space. {{qed}}
Consider the set $\set 0$ which is closed from Closed Sets of Either-Or Topology. Let $x = \dfrac 1 2$. Let $U \in \tau$ such that $\set 0 \subseteq U$. Then by definition $\openint {-1} 1 \subseteq U$ and so $x \in U$. So we have found a closed set of $T$ and a point in $S$ which do not satisfy the conditions for $T$ to be a $T_3$ space. Hence the result. {{qed}}
We have that a Particular Point Space is $T_0$. Then from Condition for Open Extension Space to be $T_0$ Space, it follows that $T^*_{\bar q}$ is a $T_0$ (Kolmogorov) space. We have directly that: :An Open Extension Topology is not $T_1$. :An Open Extension Topology is $T_4$. Finally, we have that a Particular Point Topology with three points or more is not $T_4$. From $T_5$ Space is $T_4$ Space, it follows that $T$ is not a $T_5$ space. It follows from Condition for Open Extension Space to be $T_5$ Space that $T^*_{\bar q}$ is not a $T_5$ space. {{qed}}
Let $x \in S$ such that $x \ne 0$. Then, as $0 \notin \set x$, we have by definition of the either-or topology that $x$ is open in $T$. So whatever $H$ is, $\set x$ never contains any points of $H$ which are different from $x$. So $x$ can not be a limit point of $H$. However, every open set of $T$ which contains $0$ also contains the interval $\openint {-1} 1$ and so can contain points of $H$. It follows that the only non-empty sets of $T$ for which $0$ is ''not'' a limit point are $\set {-1}$, $\set 0$ and $\set 1$ and their unions. {{qed}}
Let $A, B \subseteq S$ such that $A \cap B \ne \O$. Suppose neither $A$ nor $B$ contain $0$. Then $A$ and $B$ are both open in $T$. From Limit Points of Either-Or Topology, the only limit point that either $A$ or $B$ may have is $0$. So either $A^- = A \cup \set 0$ or $A^- = A$, where $A^-$ is the closure of $A$ Similarly for $B$. So $A^- \cap B = \O$, and so are separated, and such that: :$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ where $U = A$ and $V = B$, fulfilling the condition for a $T_5$ space. Suppose one of $A$ and $B$ contains $0$. {{WLOG}}, suppose $0 \in A$. Then for $A$ and $B$ to be separated, $0 \notin B^-$ and so $B^-$ has to be $\set {-1}$ or $\set 1$ or their union. In any of these cases we have that $B$ and $S \setminus B$ are disjoint open sets containing $B$ and $A$ respectively. So in all cases we see that if $A$ and $B$ are separated, then: :$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ where $U = A$ and $V = B$. and so by definition $T$ is a $T_5$ space. {{qed}}
We have: :Either-Or Topology is Compact :Compact Space is Lindelöf Hence the result. {{qed}}
By definition of topological subspace, $U \subseteq H$ is open in $T_H$ {{iff}}: :$(1): \quad \set 0 \nsubseteq U$ or: :$(2): \quad \openint {-1} 1 \subseteq U$ But for all $U \subseteq H$, condition $(1)$ holds as $0 \notin H$. So $T_H$ is by definition a discrete space. As $T_H$ is uncountable, we have that Uncountable Discrete Space is not Lindelöf holds. Hence the result. {{qed}}
Let $x \in S$ such that $x \ne 0$. Then $\set x$ is open in $T$ and so on its own forms a local basis of $x$ which is (trivially) countable. Let $x = 0$. Let $U \in \tau$ be open in $T$ such that $x \in U$. Then by definition of the either-or space, $U$ contains the open set $\openint {-1} 1$. So $\openint {-1} 1$ forms a local basis of $0$ which is (trivially) countable. Hence the result by definition of first-countable space. {{qed}}
From Limit Points of Either-Or Topology, the only limit point of any set of $S$ is $0$. So the only set whose closure is $S$ are $S \setminus \set 0$ and $S$ itself. So these two are the only subsets of $S$ which are everywhere dense in $S$. Both of these are uncountable. Hence the result, by definition of separable space. {{qed}}
From the definition of the either-or space, we have that every point $x$ in $T$ (apart from $0$) forms an open set of $T$. So every non-empty subset of $T$ (apart from $\left\{{0}\right\})$ contains at least one open set of $T$. So no subset of $T$ is nowhere dense in $T$. So $T$ is not a countable union of subsets of $S$ which are nowhere dense in $T$. Hence the result by definition of non-meager. {{qed}}
From the definition of the either-or space, we have that every point $x$ in $T$ (apart from $0$) forms an open set of $T$. The result follows directly from Space with Open Point is Non-Meager. {{qed}}
Let $U \in \tau$ such that $0 \notin U$. Then: :$\displaystyle U = \bigcup_{x \mathop \in U} \left\{{x}\right\}$ where $x \ne 0$. Hence for all $x \in U$, we have $\left\{{x}\right\} \in \mathcal B$. Thus $U$ is the union of elements of $\mathcal B$. Now suppose $U \in \tau$ such that $0 \in U$. Then $\left({-1 \,.\,.\, 1}\right) \subseteq U$ by definition. So one of four cases holds: :$U = \left({-1 \,.\,.\, 1}\right)$ :$U = \left[{-1 \,.\,.\, 1}\right) = \left({-1 \,.\,.\, 1}\right) \cup \left\{{-1}\right\}$ :$U = \left({-1 \,.\,.\, 1}\right] = \left({-1 \,.\,.\, 1}\right) \cup \left\{{1}\right\}$ :$U = \left[{-1 \,.\,.\, 1}\right] = \left({-1 \,.\,.\, 1}\right) \cup \left\{{-1}\right\} \cup \left\{{1}\right\}$ All of these sets are in $\mathcal B$, so that $U$ is the union of elements of $\mathcal B$. Hence, by definition, $\mathcal B$ is a basis for $T$. {{qed}} Category:Either-Or Topology Category:Topological Bases hu2ftr2gslktr0yrxrxkre01xmye643
Consider the set: :$\BB := \set {\set x: x \in S, x \ne 0} \cup \openint {-1} 1$ Then by Basis for Either-Or Topology, $\BB$ is a basis for $T$. From Point is Path-Connected to Itself, all $x \in S, x \ne 0$ are path-connected elements of $\BB$. Finally we consider $\openint {-1} 1 \in \BB$. Let $p \in \openint {-1} 1$ and consider the mapping $f: \closedint 0 1 \to \openint {-1} 1$: :$\forall x \in \closedint 0 1: \map f x = \begin{cases} 0 & : x = 0 \\ p & : x \in \hointl 0 1 \end{cases}$ Then $f$ is a continuous mapping and so a path from $0$ to $p$. So $\openint {-1} 1$ is a path-connected element of $\BB$. Hence the result by definition of locally path-connected space. {{qed}}