hcju/proofwikiqa huggingface dataset:
It is sufficient to find $a, q$ not necessarily coprime. Once such an $a, q$ have been found, then a coprime pair can be found by dividing numerator and denominator of $\dfrac a q$ by the GCD of $a$ and $q$. Let $X = \left\lfloor{x}\right\rfloor$ be the integer part of $x$. Let $\alpha_q = \alpha q - \left\lfloor{\alpha q}\right\rfloor \in \left[{0 \,.\,.\, 1}\right)$ for $q = 1, \ldots, X$. Consider the disjoint real intervals: :$I_k = \left[{\dfrac{k-1} {X + 1} \,.\,.\, \dfrac k {X+1}} \right), \quad k=1,\ldots, X + 1$ Suppose there exists $q$ such that: :$\alpha_q \in \left[{0 \,.\,.\, \dfrac 1{X+1}} \right)$ Then: :$0 \le \alpha q - \left\lfloor{\alpha q}\right\rfloor < \dfrac 1 {X+1}$ Hence: :$\left|{\alpha - \dfrac{\left\lfloor{\alpha q}\right\rfloor} q}\right| < \dfrac 1 {q \left({X + 1}\right)} < \dfrac 1 {q x}$ Taking $a = \left\lfloor{\alpha q}\right\rfloor$, the construction is complete. {{qed|lemma}} Suppose that not to be the case, but suppose that there exists $q$ such that: : $\alpha_q \in \left[{\dfrac X {X + 1} \,.\,.\, 1}\right)$ Then similarly: :$\dfrac X {X+1} < \alpha q - \left\lfloor{\alpha q}\right\rfloor < 1$ so: :$-\dfrac 1 {X+1} < \alpha q - \left\lfloor{\alpha q}\right\rfloor - 1 < 0$ and: :$\left|{\alpha  - \dfrac {\left\lfloor{\alpha q}\right\rfloor + 1} q}\right| < \dfrac 1 {q \left({X+1}\right)} < \dfrac 1 {q x}$ Taking $a = \left\lfloor{\alpha q}\right\rfloor + 1$, again the construction is complete. {{qed|lemma}} If neither of the above cases holds, then the $X - 1$ remaining intervals $I_k$ contain the $X$ values of $\alpha_q$. Therefore, by the Pigeonhole Principle, there are $k_0 \in \left\{{2, \ldots, X}\right\}$ and $q_1 < q_2$ such that: :$\alpha_{q_1}, \alpha_{q_2} \in I_{k_0}$ Then for $i = 1, 2$: :$\dfrac{k_0 - 1} {X + 1} \le \alpha q_i - \left\lfloor{\alpha q_i}\right\rfloor < \dfrac {k_0} {X+1}$ Therefore: :$\left|{\alpha q_2 - \alpha q_1 - \left\lfloor{\alpha q_2}\right\rfloor + \left\lfloor{\alpha q_1}\right\rfloor}\right| < \dfrac 1 {X + 1}$ and therefore: :$\left|{\alpha - \dfrac{\left\lfloor{\alpha q_2}\right\rfloor - \left\lfloor{\alpha q_1}\right\rfloor} {q_2 - q_1}}\right| < \dfrac 1 {\left({X + 1}\right) \left({q_2 - q_1}\right)} < \dfrac 1 {x \left({q_2 - q_1}\right)}$ Taking $a = \left\lfloor{\alpha q_2}\right\rfloor - \left\lfloor{\alpha q_1}\right\rfloor$ and $q = q_2 - q_1$ the construction is complete. {{qed}} {{Namedfor|Johann Peter Gustav Lejeune Dirichlet|cat = Dirichlet}} Category:Diophantine Equations 1tm1fhvu4mf49qcj6b5cwklvhc05tit
By Exponential is Entire, the partial sums: :$\displaystyle f_N \left({s}\right) = \sum_{n \mathop = 1}^N a_n n^{-s}$ are analytic. So by Uniform Limit of Analytic Functions is Analytic it is sufficient to show locally uniform convergence. For $0 < A < \pi / 2$, $\delta > 0$ we let: :$D_{A, \delta} = \left\{{s \in \C : -A < \arg \left({s}\right) < A,\ \Re \left({s}\right) > \delta}\right\}$ Then for any $s \in \C$ such that $\Re \left({s}\right) > 0$, we can choose $A$, $\delta$ such that $s \in D_{A,\delta}$. So it is sufficient to prove locally uniform convergence in this region. Also note that if $\lambda_n = \log n$, then: :$\displaystyle f \left({s}\right) = \sum_{n \mathop \ge 1} a_n e^{-\lambda_n s}$ By Abel's Lemma: Formulation 2, for $N, M \in \N$ we have: :$\displaystyle \sum_{n \mathop = M}^N a_n e^{-\lambda_n s} = \sum_{n \mathop = M}^{N - 1} \left({\sum_{k \mathop = M}^n a_n}\right) \left[{e^{-\lambda_n s} - e^{-\lambda_{n + 1} s}}\right] + e^{-\lambda_N s} \left({\sum_{n \mathop = M}^N a_n}\right)$ Let $\epsilon > 0$ be arbitrary, and choose $N_0 \in \N$ such that: :$\displaystyle \forall n \ge N_0: \left|{e^{-\lambda_n s}}\right| < \epsilon$ Now for $N, M \ge N_0$, by the above we obtain: :$(1): \quad \displaystyle \left\vert \sum_{n \mathop = M}^N a_n e^{-\lambda_n s} \right\vert \le B \sum_{n \mathop = M}^{N - 1} \left\vert e^{-\lambda_n s} - e^{-\lambda_{n+1} s} \right\vert + B \epsilon$ For any $\alpha, \beta \in \R$ we have: {{begin-eqn}} {{eqn | l = \left\vert{e^{-\alpha s} - e^{-\beta s} }\right\vert       | r = \left\vert{s \int_\alpha^\beta e^{-x s} \ \mathrm d x}\right\vert       | c =  }} {{eqn | o = \le       | r = \left\vert{s}\right\vert \int_\alpha^\beta e^{-x \sigma} \ \mathrm d x       | c = where $\sigma = \Re \left({s}\right)$ }} {{eqn | r = \frac{\left\vert{s}\right\vert} \sigma \left({e^{-\alpha \sigma} - e^{-\beta \sigma} }\right)       | c =  }} {{end-eqn}} Therefore: {{begin-eqn}} {{eqn | l = \left\vert{\sum_{n \mathop = M}^N a_n e^{-\lambda_n s} }\right\vert       | o = \le       | r = \frac {B \left\vert{s}\right\vert} \sigma \sum_{n \mathop = M}^{N - 1} \left({e^{-\lambda_n \sigma} - e^{-\lambda_{n + 1} \sigma} }\right) + B \epsilon       | c = by $(1)$ }} {{eqn | r = \frac {B \left\vert{s}\right\vert} \sigma \left({e^{-\lambda_N \sigma} - e^{-\lambda_M \sigma} }\right) + B \epsilon       | c =  }} {{eqn | o = \le       | r = \frac {B \epsilon \left\vert{s}\right\vert} \sigma + B \epsilon       | c =  }} {{end-eqn}} Finally letting $\Im \left({s}\right) = t$: :$\dfrac{\left\vert{s}\right\vert} \sigma = \dfrac{\sqrt{\sigma^2 + t^2} } \sigma = \sqrt{1 + \dfrac {t^2} {\sigma^2} }$ and: :$\dfrac {\pi} 2 > A \ge \arg \left({s}\right) = \arctan \left({\dfrac t \sigma}\right)$ so $\dfrac t \sigma < \tan A$ is bounded uniformly in $D_{A, \delta}$. Thus letting $N \to \infty$, we have shown that: :$\displaystyle \left\vert{\sum_{n \mathop = M}^\infty a_n e^{-\lambda_n s} }\right\vert \to 0$ as $M \to \infty$ uniformly in $D_{A, \delta}$. {{qed}} Category:Analytic Number Theory lqwt0v3yzx14ccjck17qzdu4uz9jwvo
By Exponential is Entire, the partial sums: :$\displaystyle f_N \left({s}\right) = \sum_{n \mathop = 1}^N a_n n^{-s}$ are analytic. So by Uniform Limit of Analytic Functions is Analytic it is sufficient to show locally uniform convergence. Let $B$ be a bound for the $a_n$: :$\forall n \in \N: \left\vert{a_n}\right\vert \le B$ Let $D$ be any open subset of $\Re \left({s}\right) > 1$. So for some $\kappa > 0$: :$\forall s \in D: \Re \left({s}\right) \ge 1 + \kappa$ Now: {{begin-eqn}} {{eqn | l = \left\vert{f_N \left({s}\right)}\right\vert       | o = \le       | r = \sum_{n \mathop = 1}^N \left\vert{a_n}\right\vert \left\vert{n^{-s} }\right\vert       | c =  }} {{eqn | o = \le       | r = B \sum_{n \mathop = 1}^N \frac 1 {n^{1 + \kappa} }       | c =  }} {{end-eqn}} which we know to be finite (!). {{MissingLinks|I would imagine the statement (!) is proved somewhere, but can't think what it'd be called}} Category:Analytic Number Theory jeq3skah0g8mnus3u7r39tc3pmc1d7q
By definition: :$\map {\chi_0} a = \begin{cases} 1 & : \gcd \set {a, q} = 1 \\ 0 & : \text{otherwise} \end{cases}$ {{explain|Worth specifying the domain of $a$ for added clarity here.}} Therefore: {{begin-eqn}} {{eqn | l = \map L {s, \chi_0}       | r = \prod_p \paren {1 - \map \chi p p^{-s} }^{-1}       | c = Euler Product: $p$ ranges over the primes }} {{eqn | r = \prod_{p \mathop \nmid q} \paren {1 - p^{-s} }^{-1}       | c = Fundamental Theorem of Arithmetic: $\gcd \set {p, q} > 1$ {{iff}} $p \divides q$ }} {{eqn | r = \prod_{p \mathop \divides q} \paren {1 - p^{-s} } \prod_p \paren {1 - p^{-s} }^{-1}       | c = }} {{eqn | r = \map \zeta s \prod_{p \mathop \divides q} \paren {1 - p^{-s} }       | c = Euler Product }} {{end-eqn}} {{explain|Not sure whether Fundamental Theorem of Arithmetic is where you need to go for $\gcd \set {p, q} > 1$ {{iff}} $p \divides q$ -- I'm fairly sure there's a result on {{ProofWiki}} demonstrating it more directly}} Hence the result. {{Qed}} Category:Analytic Number Theory 1o3ubcvfr7e7dkorv40cltzcfylb59q
$T$ is separable {{iff}} there exists a countable subset of $S$ which is everywhere dense. $T$ is metacompact {{iff}} every open cover of $S$ has an open refinement which is point finite. $T$ is a Lindelöf space if every open cover of $S$ has a countable subcover. Having established the definitions, we proceed. Let $T = \left({S, \tau}\right)$ be separable and metacompact. {{AimForCont}} there exists an open cover $\UU$ of $S$ which has no countable subcover. As $T$ is metacompact, $\UU$ has an open refinement $\VV$ which is point finite. By nature of $\UU$, which has no countable subcover, $\VV$ is uncountable. {{explain|How would $\VV$ being countable lead to a countable subcover of $\UU$? ACC is helpful for this.}} By hypothesis, $T$ is separable. Let $\SS$ be a countable subset of $S$ which is everywhere dense. Then each $V \in \VV$ contains some $s \in \SS$. {{explain|why? Also, this is not true if $\VV$ contains the empty set.}} So some $s \in \SS$ is contained in an uncountable number of elements of $\VV$. Thus, by definition, $\VV$ is not point finite. Thus no uncountable open refinement $\VV$ of $\UU$ exists which is point finite. It follows that $\VV$ must be countable. Thus $\UU$ has a countable subcover. That is, by definition, $T$ is a Lindelöf space. {{qed}}
$T$ is metacompact {{iff}} every open cover of $S$ has an open refinement which is point finite. $T$ is a Lindelöf space if every open cover of $S$ has a countable subcover. Having established the definitions, we proceed. Let $\mathcal U$ be an open cover of $S$. Let $\mathcal V$ be a point finite open refinement of $\mathcal U$. By Point Finite Set of Open Sets in Separable Space is Countable, $\mathcal V$ is countable. Define a mapping $H$ on $\mathcal V$ thus: :$\forall V \in \mathcal V: H \left({V}\right) = \left\{{U \in \mathcal U: V \subseteq U}\right\}$ By Image of Countable Set under Mapping is Countable, the image of $H$ is countable. Call this image $I$. Since $\mathcal V$ is a refinement of $\mathcal U$, $\varnothing \notin I$. By the Axiom of Countable Choice, $I$ has a choice function $c$. Then $G = c \circ H: \mathcal V \to \mathcal U$ is a mapping such that: :$\forall V \in \mathcal V: V \subseteq G \left({V}\right)$ Then $\mathcal Q = G \left({\mathcal V}\right)$ is countable by Image of Countable Set under Mapping is Countable. Each element of $\mathcal Q$ is an element of $\mathcal U$ by the definition of $G$. Let $x \in S$. Then since $\mathcal V$ is a cover for $S$: : $\exists V \in \mathcal V: x \in V$ Then $x \in V \subseteq G \left({V}\right) \in \mathcal Q$. Thus $\mathcal Q$ is a countable subcover of $\mathcal U$. Thus each open cover of $S$ has a countable subcover, so $T$ is a Lindelöf space. {{qed}}
If $\psi = \chi_0$, then it is straightforward that: {{begin-eqn}} {{eqn | l = \sum_{x \mathop \in G} \psi \left({x}\right)       | r = \sum_{x \mathop \in G} \chi_0 \left({x}\right)       | c = Assumption }} {{eqn | r = \sum_{x \mathop \in G} 1       | c = {{Defof|Trivial Character}} }} {{eqn | r = \vert G \vert       | c =  }} {{end-eqn}} If $\psi \neq \chi_0$, then $\exists y \in G$ such that $\psi \left({y}\right) \neq 1$. As $x$ runs through $G$ in the summation, $yx$ also runs through $G$. So: {{begin-eqn}} {{eqn | l = \sum_{x \mathop \in G} \psi \left({x}\right)       | r = \sum_{x \mathop \in G} \psi \left({y x}\right)       | c = by the above claim }} {{eqn | r = \sum_{x \mathop \in G} \psi \left({y}\right) \psi \left({x}\right)       | c = $\psi$ as a character is by definition a homomorphism }} {{eqn | r = \psi \left({y}\right) \sum_{x \mathop \in G} \psi \left({x}\right)       | c =  }} {{end-eqn}} Since by assumption $\psi \left({y}\right) \ne 1$, it must be true that: :$\displaystyle \sum_{x \mathop \in G} \psi \left({x}\right) = 0$ {{qed}} {{refactor|Suggestion: Move the dual space version into a corollary, as it is a direct application of the main theorem on the dual space, with identifying $x \in G$ as a character: $x: \chi \mapsto \chi \left({x}\right)$}} Category:Analytic Number Theory abb5loovdpeici0lkr1ik5j3bcfxspw
Let $\chi$ be the trivial character. Then by Dirichlet L-Function from Trivial Character: :$\displaystyle L \left({s, \chi}\right) = \zeta \left({s}\right) \cdot \prod_{p \mathop \backslash q} \left(1 - p^{-s}\right)$ where $\backslash$ denotes divisibility. Also, by Poles of Riemann Zeta Function, $\zeta$ is analytic on $\C$ except for a simple pole at $s = 1$. Since $L \left({s, \chi}\right)$ is just $\zeta$ times some constant, the same holds for this function. If $\chi$ is nontrivial, then by the Orthogonality Relations for Characters: :$\displaystyle \sum_{x \mathop \in G} \chi \left({x}\right) = 0$ By definition, $\chi$ is $q$-periodic, and zero on integers not coprime to $q$. So for any $M \in \N$: :$\displaystyle \sum_{n \mathop = M + 1}^{M + Q} \chi \left({n}\right) = 0$ Let $M, N \in \N$ be arbitrary. Let $d$ be such that $M + q d \le N \le M + q \left({d + 1}\right)$. Then: {{begin-eqn}} {{eqn | l = \sum_{n \mathop = M}^N \chi \left({n}\right)       | r = \sum_{k \mathop = 0}^{d-1} \sum_{n \mathop = 0}^{q-1} \chi \left({M + kq + n}\right) + \sum_{n \mathop = M + q d}^N \chi \left({n}\right)       | c = }} {{eqn | r = \sum_{n \mathop = M \mathop + q d}^N \chi \left({n}\right)       | c = because $\chi$ is $q$-periodic, and zero on integers not coprime to $q$ }} {{eqn | o = \le       | r = q       | c = because $\left\vert{N - M + q d}\right\vert \le q$ }} {{end-eqn}} So the coefficients $\chi \left({n}\right)$ of $L \left({s, \chi}\right)$ have bounded partial sums. Therefore, by Convergence of Dirichlet Series with Bounded Partial Sums, $L \left({s, \chi}\right)$ converges locally uniformly to an analytic function on $\Re \left({s}\right) > 0$. {{qed}} {{expand|Should be extended to all of $\C$, put the above down to get to Dirichlet's theorem}} Category:Analytic Number Theory 75jlpx5ar0zsewjbqdfhqqn3sri6k4q
Let $G^*$ be the group of characters modulo $q$. {{explain|"group of characters modulo $q$" - we have the definition for character, and we have (somewhere) the definition of the ring of integers modulo $q$, but not this entity.}} Let $\displaystyle \map {\zeta_q} s = \prod_{\chi \mathop \in G^*} \map L {s, \chi} R$. Among the factors of $\zeta_q$ is the the $L$-function associated to the trivial character, which by Analytic Continuation of Dirichlet L-Function we know to have a simple pole at $s = 1$. {{AimForCont}} $\map L {1, \psi} = 0$.  Then the zero of this factor kills the pole of the principal $L$-function. So by Analytic Continuation of Dirichlet L-Function $\zeta_q$ is analytic on $\map \Re s > 0$. For $\map \Re s > 1$ we have: {{begin-eqn}} {{eqn | l = \map {\zeta_q} s       | r = \prod_{\chi \mathop \in G^*} \prod_p \frac 1 {1 - \map \chi p p^{-s} }       | c = Euler Product, where $p$ ranges over the primes }} {{eqn | r = \prod_{p \mathop \nmid q} \prod_{\chi \mathop \in G^*}  \frac 1 {1 - \map \chi p p^{-s} }       | c = because the Euler Product is absolutely convergent }} {{end-eqn}} For any prime $p$, let $f_p$ be the order of $p$ mod $q$. {{explain|order}} Then: :$\map \chi p^{f_p} = \map \chi {p^{f_p} } = \map \chi 1 = 1$ So $\map \chi p$ is an $f_p$th root of unity. Moreover by the Orthogonality Relations for Characters each distinct such root occurs $\map \phi q / f_p$ times among the numbers $\map \chi p$ where $\chi \in G^*$. Also, letting $\xi$ be a primitive $f_p$th root of unity we find that for any $u \in \C$: :$\displaystyle \prod_{i \mathop = 0}^{f_p} \paren {1 - \xi^i u} = 1 - u^\xi$ Putting these facts together: :$\displaystyle \prod_{\chi \mathop \in G^*} \frac 1 {1 - \map \chi p p^{-s} } = \paren {\frac 1 {1 - p^{-f_p s} } }^{\map \phi q / f_p}$ Therefore: {{begin-eqn}} {{eqn | l = \map {\zeta_q} s       | r = \prod_{p \mathop \divides q} \paren {\frac 1 {1 - p^{-f_p s} } }^{\map \phi q / f_p}       | c = }} {{eqn | r = \prod_{p \mathop \divides q} \paren {1 + p^{-f_p s} + p^{-2 f_p s} + \cdots}^{\map \phi q / f_p}       | c = }} {{end-eqn}} Also, if $\chi_0$ is the trivial character modulo $q$, by Euler Product we have: :$\displaystyle \map L {\map \phi q s, \chi_0} = \prod_{p \mathop \nmid q} \paren {1 + p^{-\map \phi q s} + p^{-2 \map \phi q s} + \cdots}$ from which we see that for $s \in \R_{\ge 0}$: :$\map {\zeta_q} s \ge \map L {\map \phi q s, \chi_0}$ However, by Analytic Continuation of Dirichlet L-Function, $\map L {\map \phi q s, \chi_0}$ diverges for $s = \map \phi q^{-1}$, and therefore so does $\map {\zeta_q} s$. But we showed above that $\map {\zeta_q} s$ converges for $\map \Re s > 0$, a contradiction. {{qed}} Category:Analytic Number Theory 9xcwrmj6yiktscec0jn9tgxixhzz0vh
By Convergence of Dirichlet Series with Bounded Coefficients, $\map f s$ converges absolutely on $\map \Re s > 1$ to an analytic function. For fixed $s \in \set {\map \Re s > 1}$: {{begin-eqn}} {{eqn | l = \sum_{n \mathop \ge 1} \sum_p \frac {\map \chi p^n} {n p^{n s} }       | r = \sum_p \paren {\frac {\map \chi p} {p^s} + \frac {\map \chi p^2} {2 p^{2 s} } + \cdots}       | c = Manipulation of Absolutely Convergent Series }} {{eqn | r = \sum_p \map \ln {\frac 1 {1 - \map \chi p p^{-s} } }       | c = Taylor Series of Logarithm; this is the branch of $\ln$ with $\map \ln {1 + x} = x - \dfrac {x^2} 2 + \cdots$ for small $x$ }} {{eqn | r = \ln \prod_p \paren {\frac 1 {1 - \map \chi p p^{-s} } }       | c = Logarithm is Continuous; this not necessarily the same branch of $\ln$ }} {{eqn | r = \ln \map L {s, \chi}       | c = Euler Product }} {{end-eqn}} Hence $\map f s$ is an analytic branch of $\ln$ on $\map \Re s > 1$. {{qed}} Category:Analytic Number Theory pfys7o7jkgvmengnjn3eetzutl6ls03
=== Lemma 1 === {{:Dirichlet's Theorem on Arithmetic Sequences/Lemma 1}}{{qed|lemma}} Define: :$\eta_{a, q} : n \mapsto \begin{cases} 1 & : n \equiv a \pmod q\\ 0 & : \text{otherwise} \end{cases}$ === Lemma 2 === {{:Dirichlet's Theorem on Arithmetic Sequences/Lemma 2}}{{qed|lemma}} We have: {{begin-eqn}} {{eqn | l = \sum_{p \mathop \in \PP_{a, q} } p^{-s}       | r = \sum_p \map {\eta_{a, q} } p p^{-s}       | c =  }} {{eqn | r = \sum_p \sum_{\chi \mathop \in G^*} \frac {\map {\overline \chi} a} {\map \phi q} \, \map \chi p p^{-s}       | c = Lemma 2 }} {{eqn | r = \frac 1 {\map \phi q} \sum_p \frac {\map {\chi_0} p} {p^s} + \sum_{\substack {\chi \mathop \in G^* \\ \chi \mathop \ne \chi_0} } \frac {\map {\overline \chi} a} {\map \phi q} \sum_p \map \chi p p^{-s}       | c = where $\chi_0$ is the trivial character on $G$ }} {{end-eqn}} By Lemma 1, the first term grows like $\dfrac 1 {\map \phi q} \ln \dfrac 1 {s - 1}$ as $s \to 1$, while all other terms are bounded. That is: :$\displaystyle \sum_{p \mathop \in \PP_{a, q} } \frac 1 {p^s} \sim \frac 1 {\map \phi q} \, \map \ln {\dfrac 1 {s - 1} }$ as $s \to 1$.  {{qed}} {{Namedfor|Johann Peter Gustav Lejeune Dirichlet|cat = Dirichlet}}
{{begin-eqn}} {{eqn|l = \frac 1 {\phi \left({q}\right)} \sum_{\chi \mathop \in G^*} \langle \eta, \chi \rangle_G \chi \left({y}\right)      |r = \frac 1 {\phi \left({q}\right)} \sum_{\chi \mathop \in G^*} \sum_{x \mathop \in G} \eta \left({x}\right) \overline \chi \left({x}\right) \chi \left({y}\right)      |c = }} {{eqn|r = \frac 1 {\phi \left({q}\right)} \sum_{x \mathop \in G} \eta(x) \sum_{\chi \mathop \in G^*} \overline \chi \left({x}\right) \chi \left({y}\right)      |c = }} {{eqn|r = \frac 1 {\phi \left({q}\right)} \eta \left({y}\right) \phi \left({y}\right)      |c = Orthogonality Relations for Characters }} {{eqn|r = \eta \left({y}\right)      |c = }} {{end-eqn}} {{qed}} Category:Fourier Analysis Category:Representation Theory dlnm1gr781v18q9tma21hb0389hk2s3
Let $\FF$ be the set of all ordered pairs $\tuple {Z, W}$ such that: :$Z, W \in \tau$ :$V_1 \subseteq Z$ :$Z \cap W = \O$ By the lemma, $\Img \FF$ covers $V_2$. {{explain|It is not clear what mapping or relation $\Img \FF$ refers to.}} By the definition of compact space, there exists a finite subset $K$ of $\Img \FF$ which also covers $V_2$. By the definition of topology, $\displaystyle \bigcup K$ is open. By the Principle of Finite Choice, there exists a bijection $\GG \subseteq \FF$ such that $\Img \GG = K$. {{explain}} Then $\GG$, and hence its preimage, will be finite. Let $\displaystyle J = \bigcap \Preimg \GG$. By Intersection is Largest Subset, $V_1 \subseteq J$. By the definition of a topology, $J$ is open. Then $\displaystyle \bigcup K$ and $J$ are disjoint open sets such that $\displaystyle V_2 \subseteq \bigcup K$ and $V_1 \subseteq J$. {{qed}}
Let $A \subseteq S$ be compact in $T$. Let $U \in \tau$ such that $A \subseteq U$. Since $T$ is $T_3$: :Each open set contains a closed neighborhood around each of its points: ::$\forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V_x \in \tau: x \in V_x \subseteq N_x \subseteq U$ Note that $\set {V_x: x \in A}$ forms an open cover for $A$. By compactness of $A$, $\set {V_x: x \in A}$ has a finite subcover. Therefore there is a finite subset $I \subseteq A$ such that:  :$\set {V_x: x \in I}$ is an open cover for $A$. Let $V = \displaystyle \bigcup_{i \mathop \in I} V_x$, $N = \displaystyle \bigcup_{i \mathop \in I} N_x$. By definition of a topology, the set $\displaystyle \bigcup_{i \mathop \in I} V_x$ is open since it is a  union of open sets. By Finite Union of Closed Sets is Closed in Topological Space, the set $\displaystyle \bigcup_{i \mathop \in I} N_x$ is closed since it is a finite union of closed sets. By Set Union Preserves Subsets, since $V_x \subseteq N_x \; \forall x \in I \subseteq U$: :$V \subseteq N$ By Set Closure is Smallest Closed Set in Topological Space: :$V \subseteq V^- \subseteq N$ Thus: {{begin-eqn}} {{eqn | l = A       | o = \subseteq       | r = V       | c = {{Defof|Open Cover}} }} {{eqn | o = \subseteq       | r = V^- }} {{eqn | o = \subseteq       | r = N }} {{eqn | o = \subseteq       | r = U       | c = Union of Subsets is Subset: $\forall x \in I \subseteq U: N_x \subseteq U$ }} {{end-eqn}} satisfies our claim. {{qed}}
We have that a Compact Subspace of Hausdorff Space is Closed. We also have that a Closed Subspace of Compact Space is Compact. We also have that Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods. $T$ is a $T_4$ space when any two disjoint closed subsets of $S$ are separated by neighborhoods. Hence the result. {{qed}}
{{AimForCont}} there exists a topology $\tau'$ on $S$ such that: :$\tau' \subseteq \tau$ but $\tau' \ne \tau$ :$\tau'$ is a Hausdorff space. From Equivalence of Definitions of Finer Topology: :the identity mapping $I_S: \struct {S, \tau} \to \struct {S, \tau'}$ is continuous. Let $A \in \tau$. Then $S \setminus A \subseteq S$ is closed in $\struct {S, \tau}$. By Closed Subspace of Compact Space is Compact, $S \setminus A$ is compact in $\struct {S, \tau}$. From Continuous Image of Compact Space is Compact: :$I_S \sqbrk {S \setminus A}$ is also compact. By hypothesis, $\struct {S, \tau'}$ is a Hausdorff space. From Compact Subspace of Hausdorff Space is Closed: :$I_S \sqbrk {S \setminus A}$ is closed in $\struct {S, \tau'}$. Hence: :$A = S \setminus I_S \sqbrk {S \setminus A} \in \tau'$ By definition of subset: :$\tau \subseteq \tau'$ Thus we have that $\tau \subseteq \tau'$ and $\tau' \subseteq \tau$. Hence by definition of set equality: :$\tau' = \tau$ But this contradicts our hypothesis that $\tau' \ne \tau$. By Proof by Contradiction, it follows that no topology which is strictly coarser than $\tau$ can be Hausdorff. {{qed}}
Let $\tau'$ be a topology on $S$ such that $\tau \subseteq \tau'$ but that $\tau \ne \tau'$. Consider the identity mapping $I_S: \struct {S, \tau'} \to \struct {S, \tau}$. From Separation Properties Preserved in Subspace, $I_S$ is a continuous bijection from a Hausdorff space to a compact Hausdorff space. {{AimForCont}} $\struct {S, \tau'}$ is compact. Then $I_S$ is also open, and so $\tau' \subseteq \tau$. {{explain|Why does the fact that $\struct {S, \tau'}$ is compact mean $I_S$ is open?}} Thus we have that $\tau \subseteq \tau'$ and $\tau' \subseteq \tau$. Hence by definition of set equality: :$\tau' = \tau$ But this contradicts our hypothesis that $\tau' \ne \tau$. By Proof by Contradiction, it follows that no topology which is strictly finer than $\tau$ can be compact. {{qed}}