hcju/proofwikiqa huggingface dataset:
=== Lemma === {{:Tietze Extension Theorem/Lemma}}{{qed|lemma}} First suppose that for any continuous mapping on a closed set there is a continuous extension. Let $C$ and $D$ be disjoint sets which are closed in $S$. Define $f: C \cup D \to \R$ by: :$\map f x = \begin {cases} 0 & : x \in C \\ 1 & : x \in D \end {cases}$ Now $f$ is continuous and we can extend it to a continuous mapping $g: X \to \R$. By Urysohn's Lemma, $S$ is normal because $g$ is a continuous mapping such that $\map g x = 0$ for $x \in C$ and $\map g x = 1$ for $x \in D$. Conversely, let $S$ be normal and $A$ be closed in $S$. By the lemma, there exists a continuous mapping $g_0: S \to \R$ such that: :$\forall x \in S: \size {\map {g_0} x} \le \dfrac 1 3$ :$\forall x \in A: \size {\map f x − \map {g_0} x} \le \dfrac 2 3$ Since $\paren {f − g_0}: A \to \R$ is continuous, the lemma tells us there is a continuous mapping $g_1: X \to \R$ such that: :$\forall x \in S: \size {\map {g_1} x} \le \dfrac 1 3 \paren {\dfrac 2 3}$ :$\forall x \in A: \size {\map f x − \map {g_0} x - \map {g_1} x} \le \dfrac 2 3 \paren {\dfrac 2 3}$ By repeated application of the lemma we can construct a sequence of continuous mappings $g_0, g_1, g_2, \ldots$ such that: :$\forall x \in S: \size {\map {g_n} x} \le \dfrac 1 3 \paren {\dfrac 2 3}^n$ :$\forall x \in A: \size {\map f x − \map {g_0} x − \map {g_1} x - \map {g_2} x - \cdots - \map {g_n} x} \le \dfrac 2 3 \paren {\dfrac 2 3}^n$ Define: :$\displaystyle \map g x = \sum_{n \mathop = 0}^\infty \map {g_n} x$ We have that: :$\size {\map {g_n} x} \le \dfrac 1 3 \paren {\dfrac 2 3}^n$ and: :$\displaystyle \sum_{n \mathop = 0}^\infty \dfrac 1 3 \paren {\dfrac 2 3}^n$ converges as a geometric series. Thus $\displaystyle \sum_{n \mathop = 0}^\infty \map {g_n} x$ converges absolutely and uniformly. So $g$ is a continuous mapping defined everywhere. Also, $\displaystyle \sum_{n \mathop = 0}^\infty \dfrac 1 3 \paren {\dfrac 2 3}^n = 1$ implies that $\size {\map g x} \le 1$. Now for $x \in A$, we have that: :$\displaystyle \size {\map f x − \sum_{n \mathop = 0}^k \map {g_n} x} \le \paren {\dfrac 2 3}^{k + 1}$ As $k \to \infty$, the {{RHS}} $\to 0$ and so the sum goes to $\map g x$. Thus: :$\size {\map f x − \map g x} = 0$ Therefore $g$ extends $f$.  {{qed}} {{Namedfor|Heinrich Franz Friedrich Tietze|cat = Tietze}}
Let $\hat H$ be a Hermitian operator on an inner product space $V$ over the field of complex numbers, $\C$. That is, $\hat H = \hat H^\dagger$. Then for an eigenvector $\left\vert{x}\right\rangle \in V, \left\vert{x}\right\rangle \ne \left\vert{0}\right\rangle$ and eigenvalue $\lambda \in \C$: :$\left.{\hat H}\middle\vert{x}\right\rangle = \left.{\lambda}\middle\vert{x}\right\rangle$  We know for a general operator $\hat A$ on $V$, the following holds: :$\forall \left\vert{x}\right\rangle, \left\vert{y}\right\rangle \in V: \left\langle{x}\middle\vert{\hat A}\middle\vert{y}\right\rangle = \left\langle{y}\middle\vert{\hat A^\dagger}\middle\vert{x }\right\rangle^*$ where $^*$ denotes the complex conjugate. Noting $\hat H = \hat H^\dagger$ gives: :$\left\langle{x}\middle\vert{\hat H}\middle\vert{y}\right\rangle = \left\langle{y}\middle\vert{\hat H}\middle\vert{x}\right\rangle^*$ Now we compute: {{begin-eqn}} {{eqn | l = \left\langle{x}\middle\vert{\hat H}\middle\vert{x}\right\rangle       | r = \left\langle{x}\middle\vert{ \left({\hat H}\middle\vert{x}\right\rangle}\right)       | c =  }} {{eqn | r = \left\langle{x}\middle\vert{\lambda}\middle\vert{x}\right\rangle       | c =  }} {{eqn | r = \lambda \left\langle{x}\middle\vert{x}\right\rangle       | c =  }} {{end-eqn}} Using our previous result: {{begin-eqn}} {{eqn | l = \left\langle{x}\middle\vert{\hat H}\middle\vert{x}\right\rangle       | r = \left\langle{x}\middle\vert{\hat H}\middle\vert{x}\right\rangle^*       | c =  }} {{eqn | r = \left({\lambda \left\langle{x}\middle\vert{x}\right\rangle}\right)^*       | c =  }} {{end-eqn}} Equating the two previous equations gives: :$\lambda \left\langle{x}\middle\vert{x}\right\rangle = \left({\lambda \left\langle{x}\middle\vert{x}\right\rangle}\right)^*$ Recalling the conjugate symmetry property of the inner product, we can see that: :$\left\langle{x}\middle\vert{x}\right\rangle = \left\langle{x}\middle\vert{x}\right\rangle^*$  which is true {{iff}} $\left\langle{x}\middle\vert{x}\right\rangle \in \R$. So :$\lambda \left\langle{x}\middle\vert{x}\right\rangle = \lambda^* \left\langle{x}\middle\vert{x}\right\rangle$ and so: :$\lambda = \lambda^*$ Therefore $\lambda \in \R$. {{qed}} Category:Linear Algebra fvvr0rwh5xnun62ea9bf3lt0ll3idue
Let $\hat H$ be a Hermitian operator on an inner product space $V$ over the complex numbers $\C$, with a simple spectrum: :$\hat H \left\vert{x_i}\right\rangle = \lambda_i \left\vert{x_i}\right\rangle$  :$\lambda_i \ne \lambda_j$   :$\forall i, j \in \N: i \ne j$ Now we compute the following: {{begin-eqn}} {{eqn | l = \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle       | r = \left\langle{x_j}\middle \vert{\left({\hat H}\middle \vert{x_i}\right\rangle}\right)       | c =  }} {{eqn | r = \left\langle{x_j}\middle \vert{\lambda_i}\middle \vert{x_i}\right\rangle       | c =  }} {{eqn | r = \lambda_i \left\langle{x_j}\middle \vert{x_i}\right\rangle       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^*       | r = \left({\left\langle{x_i}\middle \vert{\left({\hat H}\middle \vert{x_j}\right\rangle}\right)}\right)^*       | c =  }} {{eqn | r = \left\langle{x_i}\middle \vert{\lambda_j}\middle \vert{x_j}\right\rangle^*       | c =  }} {{eqn | r = \left({\lambda_j \left\langle{x_i}\middle \vert{x_j}\right\rangle}\right)^*       | c =  }} {{end-eqn}} {{explain|Justification for the above is needed.}} From the property $\lambda_j = \lambda_j^*$ and the conjugate symmetry of the inner product: :$\left\langle{x_i}\middle \vert{x_j}\right\rangle = \left\langle{x_j}\middle \vert{x_i}\right\rangle^*$ this becomes: :$\left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^* = \lambda_j \left\langle{x_j}\middle \vert{x_i}\right\rangle$ It can be shown that the following relation holds since $\hat H = \hat H^\dagger$: :$\left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \left\langle{x_i}\middle \vert{\hat H}\middle \vert{x_j}\right\rangle^*$ {{handwaving|"can be shown" -- link to a page which demonstrates this}} This now gives us the equations: :$(1): \quad \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \lambda_i \left\langle{x_j}\middle \vert{x_i}\right\rangle$ :$(2): \quad \left\langle{x_j}\middle \vert{\hat H}\middle \vert{x_i}\right\rangle = \lambda_j \left\langle{x_j}\middle \vert{x_i}\right\rangle$ Subtracting $(2)$ from $(1)$ gives: :$\left({\lambda_i - \lambda_j}\right) \left\langle{x_j}\middle \vert{x_i}\right\rangle = 0$ Note that $\left({\lambda_i - \lambda_j}\right) \ne 0$ since we were given $\lambda_i \ne \lambda_j$. Therefore: :$\left\langle{x_j}\middle \vert{x_i}\right\rangle = 0$ Two vectors have inner product $0$ if and only if they are orthogonal. Therefore the eigenvectors of $\hat H$ are orthogonal. {{qed}} Category:Linear Algebra tvwvf85maehs2tcdqnbwhyifkqvys85
Suppose $T$ is a $T_2$ (Hausdorff) space. Let $\Delta: S \to S \times S$ be the diagonal mapping on $S$. We have that $\map \Delta S$ consists of all the elements of $S \times S$ whose coordinates are equal. :$\map \Delta S = \set{\tuple{x, x} \in S \times S: x \in S} = \set{\tuple{x, y} \in S \times S: x = y}$ By definition, $\Delta_S$ is the same set as $\map \Delta S$. Now consider $H := \relcomp {S \times S} {\map \Delta S}$, its complement in $S \times S$. We have that $H$ consists of all the elements of $S \times S$ whose coordinates are different: :$H = \set{\tuple{x, y} \in S \times S: x \ne y}$ Let $\tuple{x, y} \in H$. Because $T$ is a $T_2$ space: :$\exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ Because $U \cap V = \O$ it follows that $U \times V \cap \map \Delta S = \O$. That is: :$U \times V \subseteq H$ Now from Natural Basis of Tychonoff Topology of Finite Product, we have that $U \times V$ is an element of the natural basis of $T^2 = \struct{S \times S, \TT}$. Hence $H$ is open in $T^2$. Therefore, by definition, $\Delta_S$ is closed in $T^2 = \struct{S \times S, \TT}$. {{qed|lemma}} Now suppose $\Delta_S$ is closed in $T^2 = \struct{S \times S, \TT}$. Then $H = \set{\tuple{x, y} \in S \times S: x \ne y}$ is open in $T^2$. Let $\tuple{x, y} \in H$. We have that $x \ne y$ by the nature of $H$. Then there exists an element $U \times V$ in the natural basis of $T^2$, and therefore open in $T^2$. As $U \times V \subseteq H$ it follows that $U \cap V = \O$. From Natural Basis of Tychonoff Topology of Finite Product, as $U \times V$ is open in $T^2$, then $U$ and $V$ are both open in $T$. So, for any $\tuple{x, y}$ we have $U, V \subseteq S$ such that $x \in U, y \in V, U \cap V = \O$. So $T$ is a $T_2$ (Hausdorff) space. {{qed}}
Let $T$ be a $T_0$ (Kolmogorov) space. That is: :$\forall x, y \in S$ such that $x \ne y$, either: ::$\exists U \in \tau: x \in U, y \notin U$ :or: ::$\exists U \in \tau: y \in U, x \notin U$ We have that the set $\tau_H$ is defined as: :$\tau_H := \set {U \cap H: U \in \tau}$ Let $x, y \in H$ such that $x \ne y$. Then as $x, y \in S$ we have that: :$\exists U \in \tau: x \in U, y \notin U$ or: :$\exists U \in \tau: y \in U, x \notin U$ Then either: :$U \cap H \in \tau_H: x \in U \cap H, y \notin U \cap H$ or: :$U \cap H \in \tau_H: y \in U \cap H, x \notin U \cap H$ and so the $T_0$ axiom is satisfied. {{qed}}
Let $T$ be a $T_1$ (Fréchet) space. That is: :$\forall x, y \in S$ such that $x \ne y$, both: ::$\exists U \in \tau: x \in U, y \notin U$ :and: ::$\exists U \in \tau: y \in U, x \notin U$ We have that the set $\tau_H$ is defined as: :$\tau_H := \set {U \cap H: U \in \tau}$ Let $x, y \in H$ such that $x \ne y$. Then as $x, y \in S$ we have that: :$\exists U \in \tau: x \in U, y \notin U$ and: :$\exists U \in \tau: y \in U, x \notin U$ Then both: :$U \cap H \in \tau_H: x \in U \cap H, y \notin U \cap H$ and: :$U \cap H \in \tau_H: y \in U \cap H, x \notin U \cap H$ and so the $T_1$ axiom is satisfied. {{qed}}
Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) space. Then: :$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$  That is, for any two distinct elements $x, y \in S$ there exist disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively. We have that the set $\tau_H$ is defined as: :$\tau_H := \set {U \cap H: U \in \tau}$ Let $x, y \in H$ such that $x \ne y$. Then as $x, y \in S$ we have that: :$\exists U, V \in \tau: x \in U, y \in V, U \cap V = \O$ As $x, y \in H$ we have that: :$x \in U \cap H, y \in V \cap H: \paren {U \cap H} \cap \paren {V \cap H} = \O$ and so the $T_2$ axiom is satisfied in $H$. {{qed}}
Let $T = \struct {S, \tau}$ be a $T_{2 \frac 1 2}$ (completely Hausdorff) space. Then: :$\forall x, y \in S, x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$  That is, for any two distinct elements $x, y \in S$ there exist open sets $U, V \in \tau$ containing $x$ and $y$ respectively whose closures are disjoint. We have that the set $\tau_H$ is defined as: :$\tau_H := \set {U \cap H: U \in \tau}$ Let $x, y \in H$ such that $x \ne y$. Then as $x, y \in S$ we have that: :$\exists U, V \in \tau: x \in U, y \in V, U^- \cap V^- = \O$ As $x, y \in H$ we have that: :$x \in U \cap H, y \in V \cap H: \paren {U^- \cap H} \cap \paren {V^- \cap H} = \O$ From Corollary to Closure in Subspace we have that: :$\paren {U \cap H}^- \subseteq U^- \cap H$ :$\paren {V \cap H}^- \subseteq V^- \cap H$ where $\paren{U \cap H}^-$ and $\paren{V \cap H}^-$ are the closures in $H$ of the respective sets. So: :$x \in U \cap H, y \in V \cap H: \paren {U \cap H}^- \cap \paren {V \cap H}^- = \O$ and so the $T_{2 \frac 1 2}$ (completely Hausdorff) axiom is satisfied in $H$. {{qed}}
Let $\mathbb K$ be defined as: :$\mathbb K := \set{K \subseteq S: U \subseteq K, K \text { closed in } T}$ Then from Set Closure as Intersection of Closed Sets: :$\displaystyle \map {\cl_T} U = \bigcap \mathbb K$ Then: {{begin-eqn}} {{eqn | l = \map {\cl_T} U \cap H       | r = \paren{\bigcap \mathbb K} \cap H       | c = }} {{eqn | r = \bigcap_{K \mathop \in \mathbb K} \paren{K \cap H}       | c = Intersection is Idempotent }} {{end-eqn}} From Closed Set in Topological Subspace it follows that $K \cap H$ is closed in $T_H$ for all $K \in \mathbb K$. Let $K'$ be a closed set in $T_H$ containing $U$. From Closed Set in Topological Subspace, there exists a closed set $K$ in $T$ such that $K' = K \cap H$. Since $U \subseteq K' \subseteq K$ then $K \in \mathbb K$. Thus $\set{K \cap H : K \in \mathbb K}$ is the set of all closed sets of $T_H$ which contain $U$. From Set Closure as Intersection of Closed Sets $\displaystyle \bigcap_{K \mathop \in \mathbb K} \paren {K \cap H}$ is the closure of $U$ in $T_H$. The result follows. {{qed}} Category:Set Closures Category:Topological Subspaces kq23ugifg8ryrgroq24jy8q5vi0cg49
Let $T = \struct {S, \tau}$ be a $T_3$ space. Then: :$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$  That is, for any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$ there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$. We have that the set $\tau_H$ is defined as: :$\tau_H := \set {U \cap H: U \in \tau}$ Let $F \subseteq H$ such that $F$ is closed in $H$. Let $y \in H$ such that $y \notin F$. From Closed Set in Topological Subspace $F$ is also closed in $T$. Because $T$ is a $T_3$ space, we have that: :$\exists U, V \in \tau: F \subseteq U, y \in V, U \cap V = \O$ As $F \subseteq H$ and $y \in H, y \notin F$ we have that: :$F \subseteq U \cap H, y \in V \cap H: \paren {U \cap H} \cap \paren {V \cap H} = \O$ and so the $T_3$ axiom is satisfied in $H$. {{qed}}
Let $T = \struct {S, \tau}$ be a $T_{3 \frac 1 2}$ space. Then: :For any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$, there exists an Urysohn function for $F$ and $\set y$. We have that the set $\tau_H$ is defined as: :$\tau_H := \set {U \cap H: U \in \tau}$ Let $F \subseteq H$ such that $F$ is closed in $H$. Let $y \in H$ such that $y \notin F$. From Closed Set in Topological Subspace $F$ is also closed in $T$. Because $T$ is a $T_{3 \frac 1 2}$ space, we have that there exists an Urysohn function for $F$ and $\set y$: That is, there exists a continuous mapping $f: S \to \closedint 0 1$, where $\closedint 0 1$ is the closed unit interval, such that: :$f {\restriction_F} = 0, f {\restriction_{\set y} } = 1$ where $f {\restriction_F}$ denotes the restriction of $f$ to $F$. That is: :$\forall a \in F: \map f a = 0$ :$\forall b \in \set y: \map f b = 1$ From Continuity of Composite with Inclusion, as $f$ is continuous on $S$, then $f {\restriction_H}$ is continuous on $H$. Thus $f {\restriction_H}$ is an Urysohn function for $F$ and $\set y$ in $H$. So the $T_{3 \frac 1 2}$ axiom is satisfied in $H$. {{qed}}
Let $T = \struct {S, \tau}$ be a $T_5$ space. Then: :$\forall A, B \subseteq S, \map {\cl_S} A \cap B = A \cap \map {\cl_S} B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$  where $\map {\cl_S} A$ denotes the closure of $A$ in $S$. That is: :For any two separated sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively. We have that the set $\tau_H$ is defined as: :$\tau_H := \set {U \cap H: U \in \tau}$ Let $A, B \subseteq H$ such that $\map {\cl_H} A \cap B = A \cap \map {\cl_H} B = \O$. That is, $A$ and $B$ are separated in $H$. Then: {{begin-eqn}} {{eqn | l = \map {\cl_S} A \cap B       | r = \map {\cl_S} A \cap \paren {H \cap B}       | c = as $B \subseteq H$ }} {{eqn | r = \paren {\map {\cl_S} A \cap H} \cap B       | c = Intersection is Associative }} {{eqn | r = \map {\cl_H} A \cap B       | c = Closure in Subspace }} {{eqn | r = \O       | c = Assumption }} {{end-eqn}} Similarly: :$A \cap \map {\cl_S} B = \O$ So $A$ and $B$ are separated in $S$. Because $T$ is a $T_5$ space, we have that: :$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ It follows that: :$\exists U \cap H, V \cap H \in \tau_H : A \subseteq U \cap H, B \subseteq V \cap H, \paren {U \cap H} \cap \paren {V \cap H} = \O$ and so the $T_5$ axiom is satisfied in $H$. {{qed}}
=== Necessary Condition === Suppose $\struct {S, \tau}$ is $T_5$.  Take an arbitrary subspace $\struct {Y, \tau_Y}$ of $S$.  Let $A, B \subset Y$ be closed sets with $A \cap B = \O$.  Let $R^-$ denote the closure of $R \subset Y$.  From Closure in Subspace we have that: :$\paren {A \cap Y}^- = A^- \cap Y$ But as $A \subseteq Y$ we have from Intersection with Subset is Subset that $A \cap Y = A$, and so: :$A^- = A^- \cap Y$ Since $A, B \subset S$, we look at $A^- \cap B$. Since $A, B \subset Y$, we have:  :$A^- \cap B = A^- \cap B \cap Y \subset A^- \cap Y = A^-$. But from Closed Set Equals its Closure we have $A^- = A$, since $A$ is closed in $Y$. Thus we obtain:  :$A^- \cap B \subset A$  and similarly:  :$A \cap B^- \subset B$  Since $A \cap B = \O$, we obtain: :$A^- \cap B = \O$ and similarly: :$A \cap B^- = \O$ Thus we now have: :$A, B \subset S$ and: :$A^- \cap B = B^- \cap A = \O$ Thus, since $S$ is $T_5$, there exist $U_0, U_1$ disjoint open sets in $S$, such that $A \subset U_0$ and $B \subset U_1$. Then $V_i = U_i \cap Y$ is an open set in $Y$ for $i = 0, 1$. Also: :$V_0 \cap V_1 = \O$ and trivially: :$A \subset V_0$ and: :$B \subset V_1$ Thus $Y$ is $T_4$. Since $Y \subset S$ was arbitrary, we have found that every subspace of $\struct {S, \tau}$ is $T_4$. {{qed|lemma}} === Sufficient Condition === Suppose every subspace of $\struct {S, \tau}$ is $T_4$. Let $A, B \subset S$ with $A^- \cap B = B^- \cap A = \O$. Then because every subspace of $\struct {S, \tau}$ is $T_4$, the subspace: :$\struct {S \setminus \paren {A^- \cap B^-}, \tau_{S \setminus \paren {A^- \cap B^-} } }$ is $T_4$. Because we have $A^- \cap B = B^- \cap A = \O$, from Set is Subset of its Topological Closure we see that: :$A \cap B = \O$ Also: :$A \subset S \setminus \paren {A^- \cap B^-}$ and: :$B \subset S \setminus \paren {A^- \cap B^-}$ Then, because: :$\struct {S \setminus \paren {A^- \cap B^-}, \tau_{S \setminus \paren {A^- \cap B^-} } }$ is $T_4$, there exist $U_0, U_1 \subset S \setminus \paren {A^- \cap B^-}$ disjoint open sets in $S \setminus \paren {A^- \cap B^-}$ such that $A \subset U_0$ and $B \subset U_1$.  Because: :$U_0 = V_0 \cap \paren {S \setminus \paren {A^- \cap B^-} }$ for some open set $V_0 \subset S$ in $S$ and similarly: :$U_1 = V_1 \cap \paren {S \setminus \paren {A^- \cap B^-} }$ for some open set $V_1 \subset S$ in $S$ we find that: :$A \subset V_0$ and: :$B \subset V_1$ and: :$V_0 \cap V_1 = \O$ Thus $\struct {S, \tau}$ is a $T_5$-space. {{qed}}
=== Necessary Condition ===  Let $T$ be a $T_0$ (Kolmogorov) space. From: :Subspace of Product Space Homeomorphic to Factor Space :Separation Properties Preserved in Subspace :Separation Axioms Preserved under Homeomorphism it follows that $T_A$ and $T_B$ are both $T_0$. {{qed|lemma}} === Sufficient Condition === Now suppose that $T$ is not $T_0$. Then $\exists p, q \in S_A \times S_B$ such that: :$p \ne q$ and: :$\forall U \in \tau: \set {p, q} \subseteq U$ or $\set {p, q} \cap U = \O$ If $p \ne q$, then they are different in at least one coordinate. {{WLOG}}, let $p_A \ne q_A$. Let $V \in \tau_A$ such that $p_A \in V$. Then from Projection from Product Topology is Continuous: :$\pr_A^{-1} \sqbrk V \in \tau$. Since $p \in \pr_A^{-1} \sqbrk V$, we have that: :$q \in \pr_A^{-1} \sqbrk V$ From Projection is Surjection: :$\map {\pr_i} q = q_A \in V$ So: :$\forall V \in \tau_A: \set {p_A, q_A} \subseteq V$ or: :$\set {p_A, q_A} \cap V = \O$ Thus, by definition, $T_A$ is not $T_0$. {{qed}}
=== Necessary Condition === Suppose that for some $\beta$, $\struct {S_\beta, \tau_\beta}$ is not a $T_1$ space. Then $\exists a, b \in S_\beta$ such that $\forall U_\beta \in \tau_\beta$, $a \in U_\beta \implies b \in U_\beta$. Consider the elements $y, z \in S$ defined as: :$y = \family {x_\alpha}: x_\alpha = \begin{cases} s_\alpha & : \alpha \ne \beta \\ a & : \alpha = \beta \end{cases}$ :$z = \family {x_\alpha}: x_\alpha = \begin{cases} s_\alpha & : \alpha \ne \beta \\ b & : \alpha = \beta \end{cases}$ That is, $y$ and $z$ match on all coordinates except that for $\beta$. Let $H \subseteq S: y \in H$ be open. Then $z \in H$ as $\forall U_\beta \in \map {\pr_\beta} H: a \in U_\beta \implies b \in U_\beta$ So $T$ is not a $T_1$ (Fréchet) space. {{qed|lemma}} === Sufficient Condition === Suppose $T$ is not a $T_1$ (Fréchet) space. Then $\exists a, b \in S, a \ne b$ such that for all $U \in \tau$, $a \in U \implies b \in U$. Then $a$ and $b$ are different in at least one coordinate. Suppose, $a_\alpha = p, b_\alpha = q$ for some coordinate $\alpha$. Then for each $U_\alpha \in \tau_\alpha$: :$\displaystyle a \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta \implies b \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta$ But: :$\displaystyle a \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta$ {{iff}} $p \in U_\alpha$ :$\displaystyle b \in U_\alpha \times \prod_{\beta \ne \alpha} S_\beta$ {{iff}} $q \in U_\alpha$ from which we infer, for each $U_\alpha \in \tau_\alpha$: :$p \in U_\alpha \implies q \in U_\alpha$ It follows that $\struct {S_\alpha, \tau_\alpha}$ is not a $T_1$ (Fréchet) space. {{qed}}
{{AimForCont}} that such a $T$ is consistent and complete. By the Undecidability Theorem, since $T$ is consistent and contains $Q$, it is not recursive. But, by Complete Recursively Axiomatized Theories are Recursive, since $T$ is complete and is the set of theorems of a recursive set, it is recursive. The result follows by Proof by Contradiction. {{qed}}
Let $\Theta$ be the set of Gödel numbers of the theorems in $T$. We have that $T$ is a consistent extension of $Q$. By Set of Gödel Numbers of Arithmetic Theorems Not Definable in Arithmetic, $\Theta$ is not a definable set in $T$. Since Recursive Sets are Definable in Arithmetic, this means that $\Theta$ is not recursive. {{AimForCont}} $T$ were recursive. Then by Gödel Numbering is Recursive, we could recursively go from $\Theta$ to $T$. Then using the recursivity of $T$, we could then recursively go back to $\Theta$. Thus $\Theta$ would be recursive. This would be a contradiction. Thus, by Proof by Contradiction, $T$ is not recursive. {{qed}}