hcju/proofwikiqa huggingface dataset:
$\BB \subseteq \powerset A$ is a synthetic basis for $T$ {{iff}}: :'''B1:''' $A$ is a union of sets from $\BB$ :'''B2:''' If $B_1, B_2 \in B$, then $B_1 \cap B_2$ is a union of sets from $\BB$. Let $A = \mathbb S$ is a union of sets from $\BB$. Then: :$\displaystyle A = \bigcup_{S \mathop \in \mathbb S} S$ for some $\mathbb S \subseteq \BB$. Hence: {{begin-eqn}} {{eqn | l = H       | r = A \cap H       | c = Intersection with Subset is Subset }} {{eqn | r = \paren {\bigcup_{S \mathop \in \mathbb S} S} \cap H       | c =  }} {{eqn | r = \bigcup_{S \mathop \in \mathbb S} \paren {S \cap H}       | c = Intersection Distributes over Union }} {{end-eqn}} But if $S \in \BB$, then $S \cap H \in \BB_H$ So $H$ is a union of sets from $\BB_H$. In the same way we investigate $U_1$ and $U_2$. Let $U_1, U_2 \in \BB_H$. Then $U_1 = B_1 \cap H, U_2 = B_2 \cap H$ for some $B_1, B_2 \in \BB$. Then $U_1 \cap U_2 = B_1 \cap H \cap B_2 \cap H = \paren {B_1 \cap B_2} \cap H$. As $\BB$ is a (synthetic) basis for $A$, we have that: :$\displaystyle B_1 \cap B_2 = \bigcup_{S \mathop \in \mathbb S} S$ for some $\mathbb S \subseteq \BB$. Hence: {{begin-eqn}} {{eqn | l = U_1 \cap U_2       | r = \paren {B_1 \cap B_2} \cap H       | c =  }} {{eqn | r = \paren {\bigcup_{S \mathop \in \mathbb S} S} \cap H       | c =  }} {{eqn | r = \bigcup_{S \mathop \in \mathbb S} \paren {S \cap H}       | c = Intersection Distributes over Union }} {{end-eqn}} But if $S \in \BB$, then $S \cap H \in \BB_H$. So $U_1 \cap U_2$ is a union of sets from $\BB_H$. So $\BB_H$ fulfils the conditions to be a synthetic basis for $H$. {{qed}}
$(1): \quad$ Suppose that $J$ is the set of non-units of $R$. Then by Ideal of Unit is Whole Ring, every ideal not equal to $R$ is contained in $J$. Therefore $J$ is the unique maximal ideal of $R$, so $\tuple {R, J}$ is local. $(2): \quad$ Let $x \in R \setminus J$. Then: :$J \subsetneq \map I {J \cup \set x} \subseteq R$ where $\map I S$ is the ideal generated by $S$. Since $J$ is maximal, by definition, $x$ and $J$ generate all of $R$. Therefore $t x + m = 1$ for some $m \in J$, $t \in R$. Thus $t x = 1 - m \in 1 + J$ is a unit by hypothesis. Therefore $x$ is a unit. Now use part $(1)$. {{qed}} Category:Local Rings rndfook9i5ozjv58ox7cg93kkm7xjke
Let $f$ be a bijection. Let $g := f^{-1}$. By Bijection iff Inverse is Bijection we have that $g$ is a bijection and that $g^{-1} = f$. Let $f$ be open. Then by definition of open mapping: :$\forall H \in \tau_1: f \sqbrk H \in \tau_2$ taking $H \in \tau_1$ by definition of open in $T_1$. But $f = g^{-1}$ and so: :$\forall H \in \tau_1: g^{-1} \sqbrk H \in \tau_2$ which is exactly the definition for $g$ to be continuous. The argument works the other way. Let $g$ be continuous. Then by definition of continuous mapping: :$\forall H \in \tau_1: g^{-1} \sqbrk H \in \tau_2$ But $g^{-1} = f$ and so: :$\forall H \in \tau_1: f \sqbrk H \in \tau_2$ which is exactly the definition for $f$ to be open. {{qed}}
Let $f$ be a bijection. Suppose $f$ is an open mapping. From the definition of open mapping: :$\forall H \in \tau_1: f \sqbrk H \in \tau_2$ As $f$ is a bijection: :$f \sqbrk {S_1 \setminus H} = f \sqbrk {S_1} \setminus f \sqbrk H = S_2 \setminus f \sqbrk H$ By definition of closed set: :$S_1 \setminus H$ is closed in $T_1$ and as $f$ is an open mapping: :$f \sqbrk {S_1 \setminus H} = S_2 \setminus f \sqbrk H$ is closed in $T_2$. Hence by definition $f$ is a closed mapping. A similar argument demonstrates that if $f$ is closed then it is open. {{qed}}
From the definition of $T_2$ (Hausdorff) space: :$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ As $U \cap V = \O$ it follows from the definition of disjoint sets that: :$x \in U \implies x \notin V$ :$y \in V \implies y \notin U$ So if $x \in U, y \in V$ then: :$\exists U \in \tau: x \in U, y \notin U$ :$\exists V \in \tau: y \in V, x \notin V$ which is precisely the definition of a $T_1$ (Fréchet) space. {{qed}}
Let $\struct {S, \tau}$ be a $T_1$ space. Let $x, y \in S: x \ne y$. From the definition of $T_1$ space: :'''Both''' ::$\exists U \in \tau: x \in U, y \notin U$ :'''and''': ::$\exists V \in \tau: y \in V, x \notin V$ From the Rule of Simplification: : $\exists U \in \tau: x \in U, y \notin U$ From the Rule of Addition: :'''Either''' ::$\exists U \in \tau: x \in U, y \notin U$ :'''or''': ::$\exists V \in \tau: y \in V, x \notin V$ which is precisely the definition of a Kolmogorov ($T_0$) space. {{qed}}
Let $\struct {S, \tau}$ be a $T_5$ space. From the definition of $T_5$ space: :$\forall A, B \subseteq S, A^- \cap B = A \cap B^- = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$  where $A^-$ is the closure of $A$ in $T$. Let $C, D \subseteq S$ be disjoint sets which are closed in $T$. Thus $C, D \in \map \complement \tau$ from the definition of closed set. From Topological Closure is Closed: :$C^- = C, D^- = D$ and so from $C \cap D = \O$: :$C^- \cap D = C \cap D^- = \O$ Thus from the definition of $T_5$ space: :$\forall C, D \in \map \complement \tau, C \cap D = \O: \exists U, V \in \tau: C \subseteq U, D \subseteq V$  which is precisely the definition of a $T_4$ space. {{qed}}
Let $\struct {S, \tau}$ be a completely normal space. From the definition, $\struct {S, \tau}$ is a completely normal space {{iff}}: :$\struct {S, \tau}$ is a $T_5$ space :$\struct {S, \tau}$ is a $T_1$ (Fréchet) space. We have that a $T_5$ space is a $T_4$ space. So: :$\struct {S, \tau}$ is a $T_4$ space :$\struct {S, \tau}$ is a $T_1$ (Fréchet) space. which is precisely the definition of a normal space. {{qed}}
Let $T = \struct {S, \tau}$ be a regular space. From the definition of regular space: :$\struct {S, \tau}$ is a $T_3$ space :$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space. Let $x, y \in S$. As $T$ is $T_0$, it follows that either: :$\exists V \in \tau: x \in V, y \notin V$ or: :$\exists V \in \tau: y \in V, x \notin V$ that is, there exists $V$, an open set, containing one but not the other. {{WLOG}}, suppose that $\exists V \in \tau: y \in V, x \notin V$. Then by definition of relative complement: :$x \in \relcomp S V$ Let $F := \relcomp S V$. As $V$ is open, by definition of closed set we have that $F = \relcomp S V$ is closed. That is: :$\relcomp S V \in \tau$ As $y \in V$ it follows that $y \notin F$, that is: :$y \in \relcomp S F$ Now $\struct {S, \tau}$ is a $T_3$ space, and so: :$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$ So we have that: :$x \in F \subseteq U \implies x \in U$ :$y \notin F, y \in V$ such that $U \cap V = \O$. So: :$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ which is precisely the definition of a $T_2$ (Hausdorff) space. {{qed}}
Let $T = \struct {S, \tau}$ be a normal space. From the definition of normal space: :$\struct {S, \tau}$ is a $T_4$ space :$\struct {S, \tau}$ is a Fréchet ($T_1$) space. Let $F$ be any closed set in $T$. Let $y \in \relcomp S F$, that is, $y \in S$ such that $y \notin F$. As $T$ is a Fréchet ($T_1$) space it follows from Equivalence of Definitions of $T_1$ Space that $\set y$ is closed. As $T = \struct {S, \tau}$ is a normal space, we have that: :$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$  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. But $F$ and $\set y$ are disjoint closed sets. So: :$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$  which is precisely the definition of a $T_3$ space. {{qed}}
Let $T = \struct {S, \tau}$ be a completely Hausdorff space. From the definition: :$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U^- \cap V^- = \O$  We have that Set is Subset of its Topological Closure and so $U \subseteq U^-$ and $V \subseteq V^-$. This leads to: :$\forall x, y \in S: x \ne y: \exists U, V \in \tau: x \in U, y \in V: U \cap V = \O$ which is precisely the definition of a Hausdorff ($T_2$) space. {{qed}}
Let $T = \struct {S, \tau}$ be a regular space. Let $x, y \in S$ with $x \ne y$. From the definition of a completely Hausdorff Space: :$T$ is completely Hausdorff if $x$ and $y$ are separated by closed neighborhoods. From the definition: :$\struct {S, \tau}$ is a $T_3$ space :$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space. From Regular Space is $T_2$ Space, there exist disjoint open sets containing $x$ and $y$ respectively: :$\exists U_x, U_y \in \tau: x \in U_x, y \in U_y, U_x \cap U_y = \O$ From definition of $T_3$ space, each open set contains a closed neighborhood around each of its points, specifically for $x$ and $y$: :$\exists N_x: \relcomp S {N_x} \in \tau: \exists V_x \in \tau: x \in V_x \subseteq N_x \subseteq U_x$ :$\exists N_y: \relcomp S {N_y} \in \tau: \exists V_y \in \tau: y \in V_y \subseteq N_y \subseteq U_y$ From Subsets of Disjoint Sets are Disjoint: :$N_x \cap N_y = \O$ Hence $x$ and $y$ are separated by closed neighborhoods. The result follows. {{qed}}
=== Definition by Open Sets implies Definition by Limit Points === Let $T = \struct {S, \tau}$ be a topological space for which: :$\forall x, y \in S$, either: ::$(1): \quad \exists U \in \tau: x \in U, y \notin U$ ::$(2): \quad \exists U \in \tau: y \in U, x \notin U$ Let $x, y \in S$ such that $x \ne y$. By the definition of limit point of a point: :$(3): \quad x$ is a limit point of $y$ if every open set $U \in \tau$ such that $x \in U$ contains $y$. :$(4): \quad y$ is a limit point of $x$ if every open set $U \in \tau$ such that $y \in U$ contains $x$. Thus: :If $(3)$ holds then $(1)$ is false. :If $(4)$ holds then $(2)$ is false. So both $(3)$ and $(4)$ can not both hold at once. Hence $T = \struct {S, \tau}$ is a topological space for which no two points can be limit points of each other. {{qed|lemma}} === Definition by Limit Points implies Definition by Open Sets === Let $T = \struct {S, \tau}$ be a topological space for which no two points can be limit points of each other. Let $x, y \in S$ such that $x \ne y$. Suppose that: : $(1): \quad \nexists U \in \tau: x \in U, y \notin U$ and : $(2): \quad \nexists U \in \tau: y \in U, x \notin U$ Expressing this in different terms: :$(3): \quad$ Every open set $U \in \tau$ such that $x \in U$ contains $y$. :$(4): \quad$ Every open set $U \in \tau$ such that $y \in U$ contains $x$. That is, by definition: :$x$ is a limit point of $y$ and $y$ is a limit point of $x$. From this contradiction we deduce that: :$\forall x, y \in S$, either: ::$\exists U \in \tau: x \in U, y \notin U$ :or: ::$\exists U \in \tau: y \in U, x \notin U$ {{qed}}
=== Definition by Open Sets implies Definition by Closed Points === Let $T = \struct {S, \tau}$ be a topological space for which: :$\forall x, y \in S$, both: ::$(1): \quad \exists U \in \tau: x \in U, y \notin U$ and: ::$(2): \quad \exists U \in \tau: y \in U, x \notin U$ Let $x, y \in S$. By the definition of limit point of a point, the above condition means: :$(3): \quad y$ is a limit point of $x$ if every open set $U \in \tau$ such that $y \in U$ contains $x$. Thus $(2)$ and $(3)$ give us that $y$ is ''not'' a limit point of $x$. As $x$ and $y$ are any two points in $S$, it follows that $x$ has no limit points. Thus it holds vacuously that all the limit point of $x$ are in $\set x$. By Closed Set iff Contains all its Limit Points, we have that $\set x$ is a closed set in $T$. This holds for all $x \in S$. Hence $T = \struct {S, \tau}$ is a topological space for which all points are closed. {{qed|lemma}} === Definition by Closed Points implies Definition by Open Sets === $T = \struct {S, \tau}$ is a topological space for which all points are closed. Let $x, y \in S$. By Closed Set iff Contains all its Limit Points, we have that: :all the limit points of $x$ are in $\set x$ :all the limit points of $y$ are in $\set y$. It follows by definition of limit point (reversing the above argument) that: :$\exists U \in \tau: x \in U, y \notin U$ and :$\exists U \in \tau: y \in U, x \notin U$ {{qed}}
Let $T = \struct {S, \tau}$ be a $T_{3 \frac 1 2}$ space. From the definition of $T_{3 \frac 1 2}$ space: :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 $\set y$. Let $F \subseteq S$ be a closed set in $T$ and let $y \in \relcomp S F$. An Urysohn function for $F$ and $\set y$ is a continuous mapping $f: S \to \closedint 0 1$ where: :$\forall a \in F: \map f a = 0$ :$\map f y = 1$ Let: :$U = \map {f^{-1} } 0$ :$V = \map {f^{-1} } 1$ As $f$ is continuous, both $U$ and $V$ are open in $T$. Suppose $x \in U \cap V$. Then we would have: :$x \in U \implies \map f x = 0$ :$x \in V \implies \map f x = 1$ which contradicts the many-to-one nature of a mapping. So $U \cap V = \O$. Thus we have: :$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$  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 $F \subseteq U$, $y \in V$. which is precisely the definition of a $T_3$ space. {{qed}}
Let $T = \struct {S, \tau}$ be a Tychonoff space. 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. We have that a $T_{3 \frac 1 2}$ space is a $T_3$ space. From the definition, a regular space is: :a $T_3$ space :a $T_0$ (Kolmogorov) space. So a Tychonoff space is a regular space. Then we have a regular space is a $T_2$ (Hausdorff) space. Then we have a $T_2$ (Hausdorff) space is a $T_1$ (Fréchet) space. {{qed}}
Let $T = \struct {S, \tau}$ be a normal space. From the definition of normal space: :$\struct {S, \tau}$ is a $T_4$ space :$\struct {S, \tau}$ is a $T_1$ (Fréchet) space. Let $F$ be any closed set in $T$. Let $y \in \relcomp S F$, that is, $y \in S$ such that $y \notin F$. As $T$ is a $T_1$ (Fréchet) space it follows from Equivalence of Definitions of $T_1$ Space that $\set y$ is closed. As $T = \struct {S, \tau}$ is a $T_4$ space, we have that for any two disjoint closed sets $A, B \subseteq S$ there exists an Urysohn function for $A$ and $B$. But $F$ and $\set y$ are disjoint closed sets. So there exists an Urysohn function for $F$ and $\set y$. This is the definition of a $T_{3 \frac 1 2}$ space. Next we note that as $T$ is a $T_1$ (Fréchet) space, from $T_1$ Space is $T_0$ Space it follows that $T$ is a $T_0$ (Kolmogorov) space. So $T$ is both a $T_{3 \frac 1 2}$ space and $T_0$ (Kolmogorov) space. So, by definition, $T$ is a Tychonoff space. {{qed}}
Let $T = \struct {S, \tau}$ be a $T_4$ space which is also a $T_3$ space. From it being $T_3$: :$\forall F \subseteq S: \relcomp S F \in \tau, y \in \relcomp S F: \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \O$ Consider this $U \in \tau$, which is disjoint from $\set y$. Then $\relcomp S U$ is a closed set which is disjoint from $F$ but such that $\set y \subseteq \relcomp S U$. As $T$ is a $T_4$ space, we have that from Urysohn's Lemma there exists an Urysohn function $f$ for $F$ and $\relcomp S U$. As $\set y \subseteq \relcomp S U$, this function $f$ is a Urysohn function for $F$ and $\set y$ as well. So: :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 $\set y$. which is precisely the definition of a $T_{3 \frac 1 2}$ space. {{qed}}
By definition of homeomorphism, $\phi$ is a closed continuous bijection. The result follows from $T_0$ (Kolmogorov) Space is Preserved under Closed Bijection. {{qed}}
By definition of homeomorphism, $\phi$ is a closed continuous  bijection. The result follows from $T_1$ (Fréchet) Space is Preserved under Closed Bijection. {{qed}}
By definition of homeomorphism, $\phi$ is a closed continuous bijection. The result follows from $T_2$ (Hausdorff) Space is Preserved under Closed Bijection. {{qed}}
By definition of homeomorphism, $\phi$ is a closed continuous bijection. The result follows from $T_{2 \frac 1 2}$ (Completely Hausdorff) Space is Preserved under Closed Bijection. {{qed}}
We have that $\struct {S, \tau}$ is a regular space {{iff}}: :$\struct {S, \tau}$ is a $T_3$ space :$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space. From $T_3$ Space is Preserved under Homeomorphism: :If $T_A$ is a $T_3$ space, then so is $T_B$. From $T_0$ (Kolmogorov) Space is Preserved under Homeomorphism: :If $T_A$ is a $T_0$ (Kolmogorov) space, then so is $T_B$. Hence the result. {{qed}}
Suppose that $T_A$ is a $T_3$ space. Let $F$ be closed in $T_B$. Let $y \in S_B$ such that $y \notin F$. From Preimage of Intersection under Mapping it follows that $\phi^{-1} \sqbrk F$ and $\map {\phi^{-1}} y$ are disjoint. Also, as $\phi$ is a homeomorphism, it is a fortiori continuous. Thus by Continuity Defined from Closed Sets, $\phi^{-1} \sqbrk F$ is closed. Now as $T_A$ is a $T_3$ space, we find disjoint open sets $U_1$ containing $\phi^{-1} \sqbrk F$ and $U_2$ containing $\map {\phi^{-1}} y$. From Image of Subset is Subset of Image, we have $F = \phi \sqbrk {\phi^{-1} \sqbrk F} \subseteq \phi \sqbrk {U_1}$. Here, the first equality follows from Subset equals Image of Preimage iff Mapping is Surjection, as $\phi$ is a fortiori surjective, being a homeomorphism. Mutatis mutandis, we deduce also $\set y \subseteq \phi \sqbrk {U_2}$. From Image of Intersection under Injection it follows that $\phi \sqbrk {U_1}$ and $\phi \sqbrk {U_2}$ are disjoint. Since $\phi$ is a homeomorphism, they are also both open in $T_B$. Therewith, we have construed two disjoint open sets in $T_B$, one containing $F$, and the other containing $y$. Hence $T_B$ is shown to be a $T_3$ space as well. {{qed}}
Let $F \subseteq S_B$ be closed, and let $y \in S_B$ such that $y \notin F$. Let $G = \phi^{-1} \sqbrk F$ and let $z = \map {\phi^{-1} } y$. Since $T_A$ is a $T_{3 \frac 1 2}$ space, there exists an Urysohn function $f: X_A \to \closedint 0 1$ for $G$ and $\set z$. Define $g: S_B \to \closedint 0 1$ by: :$\map g x = \map f {\map {\phi^{-1} } x}$ By Composite of Continuous Mappings is Continuous, $g$ is continuous. Also, for all $x \in F$: :$\map g x = 0$ as $\map {\phi^{-1} } x \in G$. Similarly, because $\map {\phi^{-1} } y = z$: :$\map g y = 1$ Hence $g$ is an Urysohn function for $F$ and $\set y$. Since $F$ and $y$ were arbitrary, it follows that $T_B$ is a $T_{3 \frac 1 2}$ space. {{qed}}
We have that $\struct {S, \tau}$ is a Tychonoff space {{iff}}: :$\struct {S, \tau}$ is a $T_{3 \frac 1 2}$ space :$\struct {S, \tau}$ is a $T_0$ (Kolmogorov) space. From $T_{3 \frac 1 2}$ Space is Preserved under Homeomorphism: :If $T_A$ is a $T_{3 \frac 1 2}$ space, then so is $T_B$. From $T_0$ (Kolmogorov) Space is Preserved under Homeomorphism: :If $T_A$ is a $T_0$ (Kolmogorov) space, then so is $T_B$. Hence the result. {{qed}}
We have that $\struct {S, \tau}$ is a normal space {{iff}}: :$\struct {S, \tau}$ is a $T_4$ space :$\struct {S, \tau}$ is a $T_1$ (Fréchet) space. From $T_4$ Space is Preserved under Homeomorphism: :If $T_A$ is a $T_4$ space, then so is $T_B$. From $T_1$ Space is Preserved under Homeomorphism: :If $T_A$ is a $T_1$ (Fréchet) space, then so is $T_B$. Hence the result. {{qed}}
Suppose that $T_A$ is a $T_4$ space. Let $B_1$ and $B_2$ be closed in $T_B$ that are disjoint. Denote by $A_1 := \phi^{-1} \sqbrk {B_1}$ and $A_2 := \phi^{-1} \sqbrk {B_2}$ the preimages of $B_1$ and $B_2$ under $\phi$, respectively. From Preimage of Intersection under Mapping it follows that $A_1$ and $A_2$ are disjoint. Also, as $\phi$ is a homeomorphism, it is a fortiori continuous. Thus Continuity Defined from Closed Sets applies to yield that both $A_1$ and $A_2$ are closed. Now as $T_A$ is a $T_4$ space, we find disjoint open sets $U_1$ containing $A_1$ and $U_2$ containing $A_2$. From Image of Subset is Subset of Image, we have $B_1 = \phi \sqbrk {\phi^{-1} \sqbrk {B_1} } = \phi \sqbrk {A_1} \subseteq \phi \sqbrk {U_1}$. Here, the first equality follows from Subset equals Image of Preimage iff Mapping is Surjection, as $\phi$ is a fortiori surjective, being a homeomorphism. Mutatis mutandis, we deduce also $B_2 \subseteq \phi \sqbrk {U_2}$. From Image of Intersection under Injection it follows that $\phi \sqbrk {U_1}$ and $\phi \sqbrk {U_2}$ are disjoint. Since $\phi$ is a homeomorphism, they are also both open in $T_B$. Therewith, we have construed two disjoint open sets in $T_B$, one containing $B_1$, and the other containing $B_2$. Hence $T_B$ is shown to be a $T_4$ space as well. {{qed}}
We have that $\struct {S, \tau}$ is a completely normal space {{iff}}: :$\struct {S, \tau}$ is a $T_5$ space :$\struct {S, \tau}$ is a $T_1$ (Fréchet) space. From $T_5$ Space is Preserved under Homeomorphism: :If $T_A$ is a $T_5$ space, then so is $T_B$. From $T_1$ Space is Preserved under Homeomorphism: :If $T_A$ is a $T_1$ (Fréchet) space, then so is $T_B$. Hence the result. {{qed}}
By definition, a homeomorphism is a bijection such that both $f$ and $f^{-1}$ are continuous. As $f$ is a bijection then by Bijection iff Inverse is Bijection, so is $f^{-1}$. So by definition $f^{-1}$ is a bijection such that both $f^{-1}$ and $\left({f^{-1}}\right)^{-1}$ are continuous. The result follows from Inverse of Inverse of Bijection. {{qed}} Category:Homeomorphisms 47c5e8g6whe2mnurj633c2axmzllsxc
Let $T_A$ be a $T_0$ (Kolmogorov) space. 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. By definition:  :$\forall x, y \in S_A$, either: ::$\exists U \in \tau_A: x \in U, y \notin U$ :or: ::$\exists U \in \tau_A: y \in U, x \notin U$ Suppose that: :$\exists x, y \in S_B: \forall V \in \tau_B: x, y \in V \lor x, y \notin V$ That is: :$\exists x, y \in S_B: \forall V \in \tau_B: \set {x, y} \subseteq V \lor \set {x, y} \cap V = \O$ From Image of Subset under Relation is Subset of Image: Corollary 3 it follows that: :$\forall V \in \tau_B: \phi^{-1} \sqbrk {\set {x, y} } \subseteq \phi^{-1} \sqbrk V \lor \phi^{-1} \sqbrk {\set {x, y} } \cap \phi^{-1} \sqbrk V = \O$ that is: :$\forall V \in \tau_B: \map {\phi^{-1} } x, \map {\phi^{-1} } y \in \phi^{-1} \sqbrk V \lor \map {\phi^{-1} } x, \map {\phi^{-1} } y \notin \phi^{-1} \sqbrk V$ But by definition of continuous mapping, $U = \phi^{-1} \sqbrk V$ is open in $T_A$ {{iff}} $V$ is open in $T_B$. Thus: :$\forall U \in \tau_A: \map {\phi^{-1} } x, \map {\phi^{-1} } y \in U \lor \map {\phi^{-1} } x, \map {\phi^{-1} } y \notin U$ This contradicts the condition that $T_A$ is a $T_0$ (Kolmogorov) space. It follows that $T_B$ must also be a $T_0$ (Kolmogorov) space. {{qed}}