hcju/proofwikiqa huggingface dataset:
Let $T_A$ be a $T_{2 \frac 1 2}$ (completely Hausdorff) space. Then: :$\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A^- \cap V_A^- = \O$ where $U_A^-$ signifies the closure of $U_A$. Suppose that $T_B$ is not completely Hausdorff. Then: :$\exists a, b \in S_B: a \ne b: \forall U_B, V_B \in \tau_B: a \in U_B, b \in V_B \implies U_B^- \cap V_B^- \ne \O$ That is, there exists at least one pair of points $a$ and $b$ for which the closures of all the open sets containing $a$ and $b$ are not disjoint. From Topological Closure is Closed, both $U_B^-$ and $V_B^-$ are closed. From Topology Defined by Closed Sets, it follows that $W_B = U_B^- \cap V_B^-$ is also a closed set. From Preimage of Intersection under Mapping, we have: :$\phi^{-1} \sqbrk {U_B^- \cap V_B^-} = \phi^{-1} \sqbrk {U_B^-} \cap \phi^{-1} \sqbrk {V_B^-}$ Let $H_A = \phi^{-1} \sqbrk {U_B^-}, K_A = \phi^{-1} \sqbrk {V_B^-}, W_A = \phi^{-1} \sqbrk {W_B}$. Then we have: :$H_A \cap K_A = W_A \ne \O$ By Bijection is Open iff Closed, $\phi$ is an open bijection. By Bijection is Open iff Inverse is Continuous, it follows that $\phi^{-1}$ is continuous. From Continuity Defined from Closed Sets, we have that $H_A$, $K_A$ and $W_A$ are all closed. Also, let $U_A = \phi^{-1} \sqbrk {U_B}$ and $V_A = \phi^{-1} \sqbrk {V_B}$. From Continuity Defined by Closure, we have that: :$\phi^{-1} \sqbrk {U_B^-} \subseteq \paren {\phi^{-1} \sqbrk {U_B} }^-$ :$\phi^{-1} \sqbrk {V_B^-} \subseteq \paren {\phi^{-1} \sqbrk {V_B} }^-$ that is: :$H_A \subseteq U_A^-$ :$K_A \subseteq V_A^-$ As $H_A \cap K_A \ne \O$ it follows that $U_A^- \cap V_A^- \ne \O$. From the bijective nature of $\phi$, we have that: :$a \in U_B \implies \map {\phi^{-1} } a \in U_A$ :$b \in V_B \implies \map {\phi^{-1} } b \in V_A$ Let $x = \map {\phi^{-1} } a, y = \map {\phi^{-1} } b$. Thus we have that: :$\exists x, y \in X_A: x \ne y: \forall U_A, V_A \in \tau_A: x \in U_A, y \in V_A \implies U_A^- \cap V_A^- \ne \O$ contradicting the fact that $T_A$ is a $T_{2 \frac 1 2}$ (completely Hausdorff) space. Hence $T_B$ must after all be a $T_{2 \frac 1 2}$ (completely Hausdorff) space. {{qed}}
Let $T_A$ be a $T_1$ (Fréchet) space. By definition, all points in $T_A$ are closed. Let $a \in S_A$. Then $\set a$ is a closed set. As $\phi$ is a closed mapping it follows directly that $\phi \sqbrk {\set a}$ is closed. As $\phi$ is a bijection it follows that every point in $S_B$ is the image under $\phi$ of a single point in $S_A$. Hence every point in $S_B$ is closed. That is, $T_B$ is a $T_1$ (Fréchet) space. {{qed}}
Let $T_A$ be a $T_2$ (Hausdorff) space. Then: :$\forall x, y \in S_A, x \ne y: \exists U_A, V_A \in \tau_A: x \in U_A, y \in V_A: U_A \cap V_A = \O$ Suppose that $T_B$ is not Hausdorff. Then: :$\exists a, b \in S_B: a \ne b: \forall U_B, V_B \in \tau_B: a \in U_B, b \in V_B \implies U_B \cap V_B \ne \O$ That is, there exists at least one pair of points $a$ and $b$ for which all the open sets containing $a$ and $b$ are not disjoint. As $\tau_B$ is a topology, it follows that $W_B = U_B \cap V_B$ is also an open set. Let $U_A = \phi^{-1} \sqbrk {U_B}, V_A = \phi^{-1} \sqbrk {V_B}, W_A = \phi^{-1} \sqbrk {W_B}$. From Preimage of Intersection under Mapping, we have: :$\phi^{-1} \sqbrk {U_B \cap V_B} = \phi^{-1} \sqbrk {U_B} \cap \phi^{-1} \sqbrk {V_B}$ that is: :$U_A \cap V_A = W_A$ By Bijection is Open iff Closed, $\phi$ is an open bijection. By Bijection is Open iff Inverse is Continuous, it follows that $\phi^{-1}$ is continuous. As $\phi$ is continuous, all of $U_A, V_A, W_A$ are open in $T_A$. From the bijective nature of $\phi$, we have that: :$a \in U_B \implies \map {\phi^{-1} } a \in U_A$ :$b \in V_B \implies \map {\phi^{-1} } b \in V_A$ Let $x = \map {\phi^{-1} } a, y = \map {\phi^{-1} } b$. Thus we have that: :$\exists x, y \in S_A: x \ne y: \forall U_A, V_A \in \tau_A: x \in U_A, y \in V_A \implies U_A \cap V_A \ne \O$ contradicting the fact that $T_A$ is a $T_2$ (Hausdorff) space. Hence $T_B$ must after all be a $T_2$ (Hausdorff) space. {{qed}}
Let $S$ be a set. Let $\struct {S, \tau_1}$ and $\struct {S, \tau_2}$ be topological spaces based on $S$ such that $\tau_2$ is an expansion of $\tau_1$. That is, let $\tau_1$ and $\tau_2$ be topologies on $S$ such that $\tau_1 \subseteq \tau_2$. Let $I_S: \struct {S, \tau_1} \to \struct {S, \tau_2}$ be the identity mapping from $\struct {S, \tau_1}$ to $\struct {S, \tau_2}$. From Identity Mapping to Expansion is Closed, we have that $I_S$ is closed. We also have Identity Mapping is Bijection. So we can directly apply: :$T_0$ (Kolmogorov) Space is Preserved under Closed Bijection :$T_1$ (Fréchet) Space is Preserved under Closed Bijection :$T_2$ (Hausdorff) Space is Preserved under Closed Bijection :$T_{2 \frac 1 2}$ (Completely Hausdorff) Space is Preserved under Closed Bijection and hence the result. {{qed}}
$\tau_1 \subseteq \tau_2$ means that every open set of $\struct {S, \tau_1}$ is also an open set of $\struct {S, \tau_2}$. Let $A \subseteq S$ be closed in $\struct {S, \tau_1}$ Then by definition $S \setminus A$ is open in $\struct {S, \tau_1}$. Then $\map {I_S} {S \setminus A} = S \setminus A$ is open in $\struct {S, \tau_2}$. So, by definition, $\map {I_S} A = A$ is closed in $\struct {S, \tau_2}$. So for all $A \subseteq S$ closed in $\struct {S, \tau_1}$, it holds that $\map {I_S} A$ is closed in $\struct {S, \tau_2}$. So by definition of closed mapping, the result follows. {{qed}}
:Product Space is $T_0$ iff Factor Spaces are $T_0$ :Product Space is $T_1$ iff Factor Spaces are $T_1$ :Product Space is $T_2$ iff Factor Spaces are $T_2$ :Product Space is Completely Hausdorff iff Factor Spaces are Completely Hausdorff :Product Space is $T_3$ iff Factor Spaces are $T_3$ :Product Space is $T_{3 \frac 1 2}$ iff Factor Spaces are $T_{3 \frac 1 2}$ :Factor Spaces are $T_4$ if Product Space is $T_4$ :Factor Spaces are $T_5$ if Product Space is $T_5$
:$T_0$ Property is Hereditary :$T_1$ Property is Hereditary :$T_2$ Property is Hereditary :Completely Hausdorff Property is Hereditary :$T_3$ Property is Hereditary :$T_3 \frac 1 2$ Property is Hereditary :$T_5$ Property is Hereditary {{qed}}
Let $T = \struct {S, \tau}$ be a $T_4$ space. Then: :$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$  That is, for any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively. We have that the set $\tau_K$ is defined as: :$\tau_K := \set {U \cap K: U \in \tau}$ where $K$ is closed in $T$. Let $A, B \subseteq K$ be closed in $K$ such that $A \cap B = \O$. From Intersection with Subset is Subset we have that $A \subseteq K \iff A \cap K = A$. As $K$ is itself closed in $T$, it follows that so is $A \cap K = A$ from Topology Defined by Closed Sets. Similarly, so is $B \cap K = B$. Because $T$ is a $T_4$ space, we have that: :$\exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ As $A, B \subseteq K$ such that  we have that: :$A \subseteq U \cap K, B \subseteq V \cap K: \paren {U \cap K} \cap \paren {V \cap K} = \O$ From the definition of topological subspace, both $U \cap K$ and $V \cap K$ are open in $K$. Thus the $T_4$ axiom is satisfied in $K$. {{qed}}
From the definitions, we have that: :$T$ is a completely normal space {{iff}}: ::$\struct {S, \tau}$ is a $T_5$ space ::$\struct {S, \tau}$ is a $T_1$ (Fréchet) space. :$T$ is a normal space {{iff}}: ::$\struct {S, \tau}$ is a $T_4$ space ::$\struct {S, \tau}$ is a $T_1$ (Fréchet) space. From $T_1$ Property is Hereditary, any subspace of a $T_1$ space is also a $T_1$ space. Then we have that a space is $T_5$ iff every subspace is $T_4$. Hence the result. {{qed}}
Let $T = \struct {S, \tau}$ be a Tychonoff space. Let $x, y \in S: x \ne y$. From the definition of Tychonoff space: :$\struct {S, \tau}$ is a $T_{3 \frac 1 2}$ space :$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space. From Tychonoff Space is Regular, $T_2$ and $T_1$ we use the fact that $T$ is a $T_1$ (Fréchet) space. Clearly $\set x$ and $\set y$ are disjoint. From the definition of $T_1$ (Fréchet) space, we have that $\set x$ and $\set y$ are closed. So by definition of a $T_{3 \frac 1 2}$ space, there exists an Urysohn function for $\set x$ and $\set y$. This is exactly the definition of an an Urysohn space. {{qed}}
Let $T = \struct {S, \tau}$ be an Urysohn space. Then for any distinct points $x, y \in S$ (i.e. $x \ne y$), there exists an Urysohn function for $\set x$ and $\set y$. {{proof wanted|Then we do some stuff.}} Thus: :$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$  which is precisely the definition of a $T_{2 \frac 1 2}$ (completely Hausdorff) space. {{qed}}
Let $T = \struct {S, \tau}$ be a perfectly normal space. From the definition: :$T$ is a perfectly $T_4$ space :$T$ is a $T_1$ (Fréchet) space. {{ProofWanted|Don't know where to start at the moment. I'm tired, it's been a busy day.}}
Let $\BB = \set {B \subseteq S: B^{- \circ} = B}$. In other words, $\BB$ is the collection of regular open sets contained in $S$. It follows immediately from the definition of a basis that our theorem is proved if we can show that: :$\BB$ is a cover for $S$ :$\forall U, V \in \BB: \forall x \in U \cap V: \exists W \in \BB: x \in W \subseteq U \cap V$ First we show that $\BB$ is a cover for $S$.  Since $S$ is open: :$S^\circ = S$ Since $S$ is closed: :$S^- = S$ Therefore: :$S^{- \circ} = S$ and so: :$S \in \BB$ From Set is Subset of Union: :$S \subseteq \bigcup \BB$ Thus $\BB$ covers $S$. {{qed|lemma}} Next we demonstrate the second condition for $\BB$ to be a basis. Let $U, V \in \BB$, $x \in U \cap V$. We will show that for some $W \in \BB$: :$x \in W \subseteq U \cap V$ By General Intersection Property of Topological Space, $U \cap V \in \tau$.  Since $T$ is $T_3$, there is a closed neighborhood $N_x$ around $x$ that is contained in $U \cap V$: :$\exists N_x: \relcomp S {N_x} \in \tau: \exists Q \in \tau: x \in Q \subseteq N_x \subseteq \paren {U \cap V}$ Let $W := {N_x}^{- \circ}$.  Then by Interior of Closure is Regular Open, $W^{- \circ} = W$ and $W \in \BB$. On the other hand, since $N_x$ is closed: :${N_x}^- = N_x$, and thus $W = {N_x}^\circ$ By definition of interior, we have $Q \subseteq W \subseteq N_x$. Therefore we have $x \in W \subseteq N_x \subseteq U \cap V$, as desired. It follows that $\BB$ is a basis for $\tau$. Hence the result by definition of semiregular space. {{qed}}
Let $T = \struct {S, \tau}$ be a regular space. From the definition: :$\struct {S, \tau}$ is a $T_3$ space :$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space. We also have that a $T_3$ Space is Semiregular. Hence the result, by definition of semiregular space. {{qed}}
Let every open cover of $S$ have a finite subcover. Let $\AA$ be any set of closed subsets of $S$ satisfying $\displaystyle \bigcap \AA = \O$. We define the set: :$\VV := \set {S \setminus A : A \in \AA}$ which is trivially an open cover of $S$. From De Morgan's Laws: Difference with Union: :$\displaystyle S \setminus \bigcup \VV = \bigcap \set {S \setminus V : V \in \VV} = \bigcap \set {A : A \in \AA} = \O$ and therefore: :$S = \displaystyle \bigcup \VV$ By definition, there exists a finite subcover $\tilde \VV \subseteq \VV$. We define: :$\tilde \AA := \set {S \setminus V : V \in \tilde \VV}$ then $\tilde \AA \subseteq \AA$ by definition of $\VV$. Because $\tilde \VV$ covers $S$, it follows directly that: :$\displaystyle \bigcap \tilde \AA = \bigcap \set {S \setminus V : V \in \tilde \VV} = S \setminus \bigcup \tilde \VV = \O$ Thus, in every set $\AA$ of closed subsets of $S$ satisfying $\displaystyle \bigcap \AA = \O$ exists a finite subset $\tilde \AA$ such that $\displaystyle \bigcap \tilde \AA = \O$. That is, $S$ satisfies the Finite Intersection Axiom. {{qed|lemma}} The converse works exactly as the previous, but with the roles of the open cover and $\AA$ reversed. {{qed}}
Let every open cover of $S$ have a finite subcover. Let $\BB$ be a sub-basis of $\tau$.  By definition of a compact space, from every cover of $S$ by elements of $\BB$, a finite subcover can be selected. {{qed|lemma}} Let the space $T$ have a sub-basis $\BB$ such that every cover of $S$ by elements of $\BB$ has a finite subcover. {{AimForCont}} $T$ is not such that every open cover of $S$ has a finite subcover. Use Zorn's Lemma to find an open cover $\CC$ which has no finite subcover that is maximal among such open covers. So if: :$V$ is an open set and: :$V \notin \CC$ then $\CC \cup \set V$ has a finite subcover, necessarily of the form: :$\CC_0 \cup \set V$ for some finite subset $\CC_0$ of $\CC$. Consider $\CC \cap \BB$, that is, the sub-basic subset of $\CC$. Suppose $\CC \cap \BB$ covers $S$. Then, by hypothesis, $\CC \cap \BB$ would have a finite subcover. But $\CC$ does not have a finite subcover. So $\CC \cap \BB$ does not cover $S$. Let $x \in S$ that is not covered by $\CC \cap \BB$. We have that $\CC$ covers $S$, so: :$\exists U \in \CC: x \in U$ We have that $\BB$ is a sub-basis. So for some $B_1, \ldots, B_n \in \BB$, we have that: :$x \in B_1 \cap \cdots \cap B_n \subseteq U$ {{explain|Could that be rewritten as: "for some $B_i \in \BB$, we have that $x \in B_i \subseteq U$"}} Since $x$ is not covered, $B_i \notin \CC$. {{explain|"Since $x$ is not covered" -- by exactly which set? $\CC \cap \BB_i$? It is not completely clear what is going on here.}} As noted above, this means that for each $i$, $B_i$ along with a finite subset $\CC_i$ of $\CC$, covers $S$. But then $U$ and all the $\CC_i$ cover $S$. Hence $\CC$ has a finite subcover. This contradicts our supposition that we can construct $\CC$ so as to have no finite subcover. It follows that we cannot construct an open cover $\CC$ of $S$ which has no finite subcover. {{qed}}
By definition, a $\sigma$-compact space is the union of countably many compact sets. A compact space is the union of exactly one compact space. Hence the result. {{qed}}
Let $T = \struct {S, \tau}$ be a $\sigma$-compact space. By definition: :$T$ is a Lindelöf space {{iff}} every open cover of $X$ has a countable subcover. By definition of $\sigma$-compact space, $T = \bigcap \TT$ where $\TT$ is the union of countably many compact subspaces. Let $\CC$ be an open cover of $T$. Each element of $\TT$ is covered by a finite number of elements of $\CC$. Hence $T$ is covered by a countable union of a finite number of elements of $\CC$. Hence $\CC$ has a countable subcover. Hence the result. {{qed}}
Let every countable open cover of $S$ have a finite subcover. Let $\AA$ be any set of closed sets of $T$ satisfying $\bigcap \AA = \O$. We define the set: :$\VV := \set {S \setminus A : A \in \AA}$ which is clearly an open cover of $S$. From De Morgan's Laws: Difference with Union: :$\displaystyle S \setminus \bigcup \VV = \bigcap \set {S \setminus V : V \in \VV} = \bigcap \set{A : A \in \AA} = \O$ and therefore $S = \bigcup \VV$. By definition, there exists a finite subcover $\tilde \VV \subseteq \VV$. We define: :$\tilde \AA := \set {S \setminus V : V \in \tilde \VV}$ Then $\tilde \AA \subseteq \AA$ by definition of $\VV$. Because $\tilde \VV$ covers $S$, it follows directly that: :$\displaystyle \bigcap \tilde \AA = \bigcap \set {S \setminus V : V \in \tilde \VV} = S \setminus \bigcup \tilde \VV = \O$ Thus, in every countable set $\AA$ of closed sets of $T$ satisfying $\displaystyle \bigcap \AA = \O$ exists a finite subset $\tilde \AA$ such that $\displaystyle \bigcap \tilde \AA = \O$. That is, $S$ satisfies the Countable Finite Intersection Axiom. {{qed|lemma}} The converse works exactly as the previous, but with the roles of the open cover and $\AA$ reversed.
=== $(1) \iff (2)$: Countably Compact Space satisfies Countable Finite Intersection Axiom === {{:Countably Compact Space satisfies Countable Finite Intersection Axiom}} {{qed|lemma}} === $(1) \implies (4)$: Countably Infinite Set in Countably Compact Space has $\omega$-Accumulation Point === {{:Countably Infinite Set in Countably Compact Space has Omega-Accumulation Point}} This is proven in Countably Infinite Set in Countably Compact Space has $\omega$-Accumulation Point. {{qed|lemma}} === $(4) \implies (3)$ === Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $S$. Let $A \subseteq S$ be the range of $\sequence {x_n}$. If $A$ is finite, then consider the equality: :$\displaystyle \N = \bigcup_{y \mathop \in A} \set {n \in \N: x_n = y}$ Therefore, there exists a $y \in A$ such that $\set {n \in \N: x_n = y}$ is an infinite set. Hence, $y$ is an accumulation point of $\sequence {x_n}$. Otherwise, $A$ is countably infinite. By assumption, $A$ has an $\omega$-accumulation point $x \in S$. It follows that $x$ is an accumulation point of $\sequence {x_n}$. {{qed|lemma}} === $(5) \implies (4)$ === {{qed|lemma}} === $(3) \implies (1)$ === We use a Proof by Contradiction. {{AimForCont}} $S$ has a countably infinite open cover $\set {U_n: n \in \N}$ which does not have a finite subcover. Then, using the axiom of countable choice, we can obtain a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $S$ such that: :$\displaystyle \forall n \in \N: x_n \notin \bigcup_{k \mathop = 0}^n U_k$. Let $x \in S$. Then, by the definition of a cover, there exists an $n \in \N$ such that $x \in U_n$. By construction, $U_n \subseteq \set {x_0, x_1, \ldots, x_{n - 1} }$. Hence, $x$ is not an accumulation point of $\sequence {x_n}$. {{qed|lemma}} === $(4) \implies (5)$ === Follows directly from Infinite Set has Countably Infinite Subset and the definition of an $\omega$-accumulation point. {{qed}} {{ACC}}
Let $T = \struct {S, \tau}$ be a sequentially compact topological space. By definition of countably compact space, it suffices to show that every infinite sequence in $S$ has an accumulation point in $S$. By the definition of a sequentially compact space, every infinite sequence in $S$ has a subsequence which converges in $S$. The result follows from Limit Point of Sequence is Accumulation Point. {{qed}} {{ACC|Equivalence of Definitions of Countably Compact Space}}
Let $T = \struct {S, \tau}$ be a countably compact space. By definition: :$T$ is weakly countably compact {{iff}} every infinite subset of $S$ has a limit point in $S$. By definition of countably compact space: :every infinite subset of $S$ has an $\omega$-accumulation point in $S$. By definition, an $\omega$-accumulation point is a limit point. Hence the result. {{qed}}
=== Sufficient Condition === We have that a Countably Compact Space is Weakly Countably Compact whatever the separation properties. {{qed|lemma}} === Necessary Condition === We need to show that a $T_1$ space which is weakly countably compact is also necessarily countably compact. So, let $T = \struct {S, \tau}$ be a $T_1$ space which is weakly countably compact. Let $A \subseteq S$. Suppose $x$ to be a limit point but not an $\omega$-accumulation point of $A$. Then there exists some open set $U_x \in \tau$ such that $x \in U_x$ which contains only a finite number of points of $A$, say: :$U_x = \set {a_1, a_2, \ldots, a_n}$ But this, we have specified, is a $T_1$ space. So this would imply that $x$ has an open neighborhood which contains no points of $A$. That is, $x$ is not a limit point of $A$ after all. Hence if $x$ is a limit point of $A$, it is an $\omega$-accumulation point of $A$. Thus $S$ is countably compact. {{qed}}
Let $T = \struct {S, \tau}$ be a countably compact space. Then every countable open cover of $S$ has a finite subcover. By definition: :$T$ is pseudocompact {{iff}} every continuous real-valued function on $S$ is bounded. Let $f: S \to \R$ be such that $f$ is continuous. Consider the sets: :$S_n = \set {x \in S: \size {\map f x} < n}$ $S_n$ forms a countable cover of $S$ whose finite subcover yields a bound for the absolute value of $f$. Hence the result. {{qed}}
Let $T = \struct {S, \tau}$ be a normal space. By Countably Compact Space is Pseudocompact we already have that if $T$ is countably compact then $T$ is pseudocompact. It remains to prove that if $T$ is pseudocompact then $T$ is countably compact. {{AimForCont}} $T$ is not countably compact. By definition of countably compact space, $S$ contains an infinite subset $H = \set {x_n}$ with no $\omega$-accumulation point in $T$. Since, by definition of normal, $S$ is a $T_1$ space, $H$ is closed and discrete in the subspace topology. {{Explain|$H$ is closed and discrete in the subspace topology.}} By definition, $T$ is pseudocompact {{iff}} every continuous real-valued function on $S$ is bounded. By definition of normal, we have that $S$ is a $T_4$ space. Hence the Tietze Extension Theorem guarantees a continuous extension to $S$ of the unbounded continuous mapping $f: H \to \R$ defined by $\map f {x_n} = n$. This contradicts our hypothesis that $S$ is pseudocompact. It follows by Proof by Contradiction that $T$ is countably compact. {{qed}}
By the definitions: :If $T = \struct {S, \tau}$ is a Lindelöf space then every open cover of $S$ has a countable subcover. :If $T = \struct {S, \tau}$ is a countably compact space then every countable open cover of $S$ has a finite subcover. It follows trivially that every open cover of $S$ has a finite subcover. Hence the result by definition of compact. {{qed}}
Let $T$ be a compact space. By definition, $T$ is weakly locally compact {{iff}} each point of $T$ has a compact neighborhood in $T$. Let $x \in S$ be an arbitrary point of $T$. By Space is Neighborhood of all its Points, $T$ is a neighborhood of $x$. Thus $x$ has a compact neighborhood in $T$. Hence the result by definition of weakly locally compact. {{qed}}
Follows directly from: :Compact Space is Weakly $\sigma$-Locally Compact :a weakly $\sigma$-locally compact space is weakly locally compact by definition. {{qed}}
We are required to prove that: :$\xi \left({s}\right) = \dfrac 1 2 s \left({s - 1}\right) \pi^{-s/2} \Gamma \left({\dfrac s 2}\right) \zeta \left({s}\right) \ll \exp \left({|s|^\beta}\right)$ for all $\beta > 1$, where $\ll$ is the order notation. Note that by the Functional Equation for Riemann Zeta Function, it is sufficient to check this for $\Re \left({s}\right) \ge \dfrac 1 2$. We simply check this fact for each factor. Evidently: {{begin-eqn}} {{eqn | l = \frac 1 2 s \left({s - 1}\right) \pi^{-s/2}       | o = \ll       | r = \exp\left({2 \log s - \frac s 2 \log \pi}\right) }} {{eqn | o = \ll       | r = \exp\left({c_1 \left\vert{s}\right\vert}\right) }} {{end-eqn}} for some $c_1 > 0$. For the gamma factor, we have Stirling's Formula for Gamma Function: :$\displaystyle \log \Gamma \left({s}\right) = \left({s - \frac 1 2}\right)\log s - s + \frac {\log 2 \pi} 2 + \sum_{n \mathop = 1}^{d-1} \frac{B_{2 n}}{ 2 n \left({2 n - 1}\right)s^{2 n-1}} + \mathcal O \left({s^{1 - 2 d} }\right)$ This is valid only away from the poles of $\Gamma$ at $s = 0, -1, -2, \ldots$ However, it is assumed that $\Re \left({s}\right) \ge \dfrac 1 2$, so this is not a problem. The error term $\mathcal O \left({s^{1 - 2d} }\right)$ is small for large $s$. More generally, the largest contribution is the term $\left({s - \dfrac 1 2}\right)\log s$, so we have: :$\log \Gamma\left({\dfrac s 2}\right) \ll \left\vert{s}\right\vert \log \left\vert{s}\right\vert$ That is: :$\Gamma \left({\dfrac s 2}\right) \ll \exp \left({\left\vert{s}\right\vert \log \left\vert{s}\right\vert}\right)$ Finally, from Integral Representation of Riemann Zeta Function in terms of Fractional Part, for $\Re \left({s}\right) > \dfrac 1 2$: :$\displaystyle \zeta \left({s}\right) = \frac s {s-1} - s \int_1^\infty \left\{{x}\right\} x^{-s-1} \ \mathrm d x$ It is seen that for $\Re \left({s}\right) > \dfrac 1 2$, the integral is bounded, and therefore: :$\left({1 - s}\right) \zeta \left({s}\right) \ll \mathcal O (\left\vert{s}\right\vert^2) \ll \exp \left({\left\vert{s}\right\vert}\right)$ Combining these facts, and using that $\log s \ll s^\epsilon$ for all $\epsilon > 0$ (shown by Upper Bound of Natural Logarithm), we have: :$\left|{\xi \left({s}\right)}\right| \ll \exp\left({\left\vert{s}\right\vert^{1 + \epsilon} }\right)$  for all $\epsilon > 0$, and the proof is complete. {{qed}} Category:Riemann Zeta Function jk1z74v284tro7oyhtwcf4lspax9183