hcju/proofwikiqa huggingface dataset:
First we show that $C_S$ is an integral extension of $A_S$. Let $x \in C_S$, and $\iota$ be the canonical inclusion from a ring to its localization. There exists $s \in S$ such that $sx \in \iota(C)$, say $sx = \iota(c)$. Since $c \in C$ is integral, there is an equation: :$c^n + a_{n-1}c^{n-1} + \cdots + a_1 c + a_0 = 0$ with the $a_i \in A$. Since $\iota$ is a homomorphism, we have: :$\left({s x}\right)^n + \iota \left({a_{n-1} }\right) \left({s x}\right)^{n-1} + \cdots + \iota \left({a_1}\right) \left({s x}\right) + \iota \left({a_0}\right) = 0$ {{explain|Just a link, probably, to indicate the nature of $\iota$ - so as to confirm that it is a (ring) homomorphism.}} Multiplying by $s^{-n}$ this gives: :$x^n + s^{-1}\iota \left({a_{n-1} }\right)x^{n-1} + \cdots + x^{1-n} \iota \left({a_1}\right) x + s^{-n}\iota \left({a_0}\right) = 0$ This is a monic polynomial with coefficients in $A_S$, so $x$ is integral over $A_S$. Now let $x \in B_S$ be integral over $A_S$. So we have an equation :$x^n + r_{n-1}x^{n-1} + \cdots + r_1 x + r_0 = 0$ Let: :$t \in S$ be such that $t x \in \iota \left({B}\right)$ :$u_i$ be such that $u_i r_i \in \iota \left({A}\right)$ for $i = 0, \ldots, n-1$ :$s = t u_0 \cdots u_{n-1}$. Then we have: :$\left({s x}\right)^n + r_{n-1} s \left({s x}\right)^{n-1} + \cdots + r_1 s^{n-1} \left({s x}\right) + r_0 s^n = 0$ Now let: : $s x = \iota \left({y}\right)$ for $y \in B$ : $r_{n-i} s^i = \iota \left({q_i}\right)$ for $i = 1, \ldots, n$ Then: :$\iota \left({y^n + q_1 y^{n-1} + \cdots + q_{n-1} y + q_n}\right) = 0$ Now: : $\ker \left({\iota: B \to B_S}\right) = \left\{{x \in B : s x = 0 \text{ for some } s \in S}\right\}$ So there exists $v \in S$ such that: :$\left({v y}\right)^n + q_1 v (\left({v y}\right)^{n-1} + \cdots + q_{n-1} v^{n-1} \left({v y}\right) + q_n v^n = 0$ Therefore $v y \in B$ is integral over $A$, so $vy \in C$. Therefore: :$\iota \left({v y}\right) \in C_S$ and: :$\iota \left({y}\right) \in C_S$ Thus: :$x = s^{-1} \iota \left({y}\right) \in C_S$ {{qed}} {{MissingLinks}} Category:Algebraic Number Theory fn2hcx9gs0fpovp7mrjceq789a5r8vh
=== $(1)$ implies $(2)$ === Let $\map Q R$ denote the field of quotients of an integral domain $R$. We have by Localization Preserves Integral Closure that: :$\map Q {A_{\mathfrak p}} = \map Q A$ Hence $A_{\mathfrak p}$ is integrally closed for all prime ideals $\mathfrak p$. {{qed|lemma}} === $(2)$ implies $(3)$ === This is true because a Maximal Ideal of Commutative and Unitary Ring is Prime Ideal. {{qed|lemma}} === $(3)$ implies $(1)$ === {{ProofWanted|Need some more results on localization first}} Category:Algebraic Number Theory qtf8nxkecuh3zjh3gb4g70co6oyzuoh
So, let $T = \struct {S, \tau}$ be that topological space which is both totally disconnected and locally connected. As $T$ is totally disconnected, every point is a component and therefore closed. As $T$ is locally connected, there exists a basis $\BB$ of $T$ such that every element of $\BB$ is a component of $T$. In order for $T$ to be covered by $\BB$, every singleton subset of $T$ must be in $\BB$. From the definition of topology, the union of any number of these singleton sets is an open set of $T$. That is, every subset of $S$ is open in $T$. That is, every element of the power set of $S$ is open in $T$. This is precisely the definition of the discrete space on $S$. {{qed}}
If $H$ is totally disconnected then its individual points are separated. If $H$ is connected it can not be represented as the union of two (or more) separated sets. So $H$ can have only one point in it. {{qed}}
Let $T = \struct {S, \tau}$ be '''totally separated''' by definition 1: :For every $x, y \in S: x \ne y$ there exists a separation $U \mid V$ of $T$ such that $x \in U, y \in V$. {{AimForCont}} there exist $x, y \in S$ such that $x \ne y$ such that $x$ and $y$ are in the same quasicomponent of $T$. Then, by definition of quasicomponent, every separation of $T$ includes a single open set $U \in \tau$ which contains both $x$ and $y$. But this contradicts our stipulation that there exists a separation $U \mid V$ of $T$ such that $x \in U, y \in V$. So there exists no quasicomponent of $T$ with more than one element. That is, $T = \struct {S, \tau}$ is '''totally separated''' by definition 2. {{qed|lemma}} Let $T = \struct {S, \tau}$ be '''totally separated''' by definition 2: :Each of its quasicomponents is a singleton set. {{AimForCont}} there exist $x, y$ such that there exists no separation $U \mid V$ of $T$ such that $x \in U, y \in V$. That is, each separation of $T$ contains an open set $U \in \tau$ which contains both $x$ and $y$. That is, there exists a quasicomponent which contains both $x$ and $y$. But this contradicts our stipulation that every quasicomponent of $T$ is a singleton. So for every $x, y \in S: x \ne y$ there exists a separation $U \mid V$ of $T$ such that $x \in U, y \in V$. That is, $T = \struct {S, \tau}$ is '''totally separated''' by definition 1. {{qed}}
Let $T = \struct {S, \tau}$ be a totally separated space. Then for every $x, y \in S: x \ne y$ there exists a separation $U \mid V$ of $T$ such that $x \in U, y \in V$. Consider a mapping $f: S \to \closedint 0 1$ such that: :$\Img {f {\restriction_U} } = \set 0$ :$\Img {f {\restriction_V} } = \set 1$ where: :$f {\restriction_U}$ and $f {\restriction_V}$ denote the restrictions of $f$ to $U$ and $V$ respectively :$\Img g$ denotes the image set of a mapping $g$. From Constant Mapping is Continuous, both $f {\restriction_U}$ and $f {\restriction_V}$ are continuous on $U$ and $V$ respectively. From Continuous Mapping of Separation, it follows that $f$ is continuous on $S$ itself. By definition, $f$ is then an Urysohn function on $x$ and $y$. Hence $T$ is Urysohn. Having proved that $T$ is an Urysohn space, it follows from Urysohn Space is Completely Hausdorff Space that $T$ is also completely Hausdorff. {{qed}}
Let $\left({X,Y}\right), \left({X',Y'}\right) \in \mathbf C \times \mathbf D$. Let $\left({f,g}\right) : \left({X,Y}\right) \to \left({X',Y'}\right)$ and $\left({h,k}\right) : \left({X',Y'}\right) \to \left({X,Y}\right)$ be morphisms. Let $\operatorname{id}_X$, $\operatorname{id}_Y$ be the identity morphisms for the objects $X$ and $Y$ respectively. Then: {{begin-eqn}} {{eqn|l=\left({f, g}\right) \circ \left({\operatorname{id}_X, \operatorname{id}_Y}\right)      |r=\left({f \circ \operatorname{id}_X, g \circ \operatorname{id}_Y}\right)      |c=By the definition of composition in the product category. }} {{eqn|l=      |r=\left({f, g}\right)      |c=By the definition of the identity morphisms }} {{end-eqn}} Similarly, {{begin-eqn}} {{eqn|l=\left({\operatorname{id}_X, \operatorname{id}_Y}\right) \circ \left({h, k}\right)      |r=\left({\operatorname{id}_X \circ h, \operatorname{id}_Y \circ k}\right)      |c=By the definition of composition in the product category. }} {{eqn|r=\left({h, k}\right)      |c=By the definition of the identity morphisms }} {{end-eqn}} Therefore $\left({\operatorname{id}_X, \operatorname{id}_Y}\right)$ satisfies the property of an identity morphism. Now let $\left({f, g}\right)$, $\left({h, k}\right)$ and $\left({\ell, m}\right)$ be composable morphisms of $\mathbf C \times \mathbf D$. We have: {{begin-eqn}} {{eqn|l=\left({ \left({f, g}\right) \circ \left({h, k}\right) }\right) \circ \left({\ell, m}\right)      |r= \left({f \circ h, g \circ k}\right) \circ \left({\ell, m}\right)      |c=By the definition of composition in the product category. }} {{eqn|r=\left({ \left({f \circ h}\right) \circ \ell, \left({g \circ k}\right) \circ m}\right)      |c= }} {{eqn|r=\left({f \circ \left({h \circ \ell}\right), g \circ \left({k \circ m}\right) }\right)      |c=By associativity of morphisms of $\mathbf C$ and $\mathbf D$. }} {{eqn|l=      |r=\left({f, g}\right) \circ \left({ \left({h, k}\right) \circ \left({\ell, m}\right) }\right)      |c= }} {{end-eqn}} Therefore composition of morphisms in $\mathbf C \times \mathbf D$ is also associative. {{qed}} Category:Category Theory dhcnr0or9t8cmmt0o8r8c1lf4voy08r
Let us verify the axioms $(\text C 1)$ up to $(\text C 3)$ for a metacategory. For any two mappings their composition (in the usual set theoretic sense) is again a mapping by Composite Mapping is Mapping. For any set $X$, we have the identity mapping $\operatorname{id}_X$. By Identity Mapping is Left Identity and Identity Mapping is Right Identity, this is the identity morphism for $X$. Finally by Composition of Mappings is Associative, the associative property is satisfied. Hence $\mathbf{Set}$ is a metacategory. {{qed}} Category:Examples of Categories Category:Category of Sets rpd0zesyfpvfk4le8e9ow3t0xrsnjyl
Let $T = \struct {S, \tau}$ be a $T_2$ (Hausdorff) topological space such that the closure of every open set of $T$ is open. Let $V \subseteq S$ be closed in $T$. Then $S \setminus V$ is open by definition. Then its closure $\paren {S \setminus V}^-$ is open by hypothesis. By Complement of Interior equals Closure of Complement we have that: :$\paren {S \setminus V}^- = S \setminus V^\circ$ where $V^\circ$ is the interior of $V$. As $S \setminus V^\circ$ is open in $T$, it follows that $V^-$ is closed. So the interior of every closed set of $T$ is closed. {{qed|lemma}} By a similar argument we see that if the interior of every closed set of $T$ is closed in $T$, then the closure of every open set of $T$ is open. Hence the result.
=== Definition $1$ {{iff}} Definition $2$ === Follows directly from Complement of Interior equals Closure of Complement. {{qed|lemma}} === Definition $1$ implies Definition $3$ === Let $A, B \subseteq S$ be disjoint open sets. Then: {{begin-eqn}} {{eqn | l = A \cap B       | r = \O       | c = {{Defof|Disjoint Sets}} }} {{eqn | ll= \leadsto       | l = A^- \cap B       | r = \O       | c = Disjoint Open Sets remain Disjoint with one Closure }} {{eqn | ll= \leadsto       | l = A^- \cap B^-       | r = \O       | c = Disjoint Open Sets remain Disjoint with one Closure; $A^-$ is open }} {{end-eqn}} As $A, B$ are arbitrary: :the closures of every pair of open sets which are disjoint are also disjoint. {{qed|lemma}} === Definition $3$ implies Definition $1$ === Let $A \subseteq S$ be an open set. By Topological Closure is Closed, $A^-$ is closed. Hence $\relcomp S {A^-}$ is open. We have: {{begin-eqn}} {{eqn | l = A       | o = \subseteq       | r = A^-       | c = Set is Subset of its Topological Closure }} {{eqn | ll= \leadsto       | l = A \cap \relcomp S {A^-}       | r = \O       | c = Empty Intersection iff Subset of Complement }} {{eqn | ll= \leadsto       | l = A^- \cap \paren {\relcomp S {A^-} }^-       | r = \O       | c = Definition $3$ }} {{eqn | ll= \leadsto       | l = \paren {\relcomp S {A^-} }^-       | o = \subseteq       | r = \relcomp S {A^-}       | c = Empty Intersection iff Subset of Complement }} {{eqn | ll= \leadsto       | l = \paren {\relcomp S {A^-} }^-       | r = \relcomp S {A^-}       | c = Set is Subset of its Topological Closure }} {{end-eqn}} By Set is Closed iff Equals Topological Closure, $\relcomp S {A^-}$ is closed. Thus $\relcomp S {\relcomp S {A^-} } = A^-$ is open. As $A$ is arbitrary: :the closure of every open set is open.
Let $T = \struct {S, \tau}$ be an extremally disconnected topological space. Then by definition $T$ is a $T_2$ (Hausdorff) space such that the closure of every open set of $T$ is open. Let $x, y \in S: x \ne y$. As $T$ is a $T_2$ (Hausdorff) space, there exist disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively. Let $U^-$ denote the closure of $U$. From Topological Closure is Closed, $U^-$ is closed. But $U$ is open, so from the definition of extremally disconnected space, then $U^-$ is also open. So $U^-$ is clopen. From Disjoint Open Sets remain Disjoint with one Closure, $U^- \cap V = \O$, so $y \notin U^-$. Thus $\set {U^- \mid S \setminus U^-}$ is a partition where $x \in U^-$ and $y \in X \setminus U^-$. Thus, by definition, $T$ is totally separated. {{qed}}
Let $T = \struct {S, \tau}$ be a zero dimensional space. Let $F \subseteq S$ be closed in $T$. Let also $y \notin F$. Then by definition of closed, $\relcomp S F$ is open in $T$, where $\relcomp S F$ is the complement of $F$ in $S$. As $T$ is zero dimensional, it has a basis $\BB$ which consists entirely of clopen sets. As $\BB$ is a basis for $T$, it follows that: :$\displaystyle \exists \UU \subseteq \BB: \relcomp S F = \bigcup \UU$ that is, $\relcomp S F$ is the union of a subset of elements of $\BB$. Thus, there is a set $U \in \UU$ such that $y \in U$ and $F \subseteq \relcomp S U$. But the elements of $\UU$ are clopen sets, so $U$ is itself clopen. Thus, by definition, $\relcomp S U$  is also clopen. So we have that $U$ and $\relcomp S U$ are open sets in $T$ such that: :$\exists W, R \in \tau: F \subseteq W, y \in R: R \cap W = \O$ by setting $R = U$ and $W = \relcomp S U$. That is, $T$ is a $T_3$ space. {{qed}}
Let $T = \struct {S, \tau}$ be a zero dimensional space which is also a $T_0$ (Kolmogorov) space. As $T$ is zero dimensional, there exists a basis $\BB$ whose sets are all clopen. Let $x, y \in S$. As $T$ is a $T_0$ space: :$\exists U \in \tau: x \in U, y \notin U$ or: :$\exists U \in \tau: y \in U, x \notin U$ {{WLOG}}, suppose that $\exists U \in \tau: x \in U, y \notin U$. Then: :$\exists V \in \BB: x \in V$ and $V \subseteq U$ by definition of basis. The set $V$ is clopen by the definition of $\BB$. But then $x \in V$ which is open and $y \in S \setminus V$ which is also open. $\set {V \mid S \setminus V}$ is a partition and hence $T$ is totally separated. {{qed}}
Let $T = \struct {S, \tau}$ be a scattered space which is also a $T_1$ (Fréchet) space. We have that every Non-Trivial Connected Set in $T_1$ Space is Dense-in-itself. As $T$ is scattered, every $H \subseteq S$ contains at least one point which is isolated in $H$. So $H$ is not dense-in-itself and so if $H$ has more than one element it can not be connected. As $H$ is arbitrary, it follows that $T$ is totally disconnected. {{qed}}
{{AimForCont}} $H$ is not dense-in-itself. Then $\exists x \in H$ such that $x$ is isolated in $H$. That is, $\exists U \in \tau: U \cap H = \set x$. Since $H$ has more than one element, we can find $y \in H$ with $y \ne x$. Since $T$ is $T_1$: :$\forall y \in H: y \ne x \implies \paren {\exists V_y \in \tau: y \in V_y, x \notin V_y}$ Define $V = \displaystyle \bigcup_{y \in H\\ y \ne x} V_y$. Then $\forall z \in H$: :$z = x \implies z \in U$ :$z \ne x \implies z \in V_z \subseteq V$ Showing that $H \subseteq U \cup V$. This also shows that both $U \cap H$ and $V \cap H$ are non-empty. Since $x \notin V_y$ for any $y \in H$,  we have $x \notin V$.  This implies $\set x \cap V = \O$ So $H \cap U \cap V = \O$. Thus we have: :$H \subseteq U \cup V$ :$H \cap U \cap V = \O$ :$U \cap H \ne \O$ :$V \cap H \ne \O$ Showing that $H$ is disconnected. This is a contradiction. Hence we must have $H$ is dense-in-itself.  {{qed}}
The relevant justifications are listed as follows: * Discrete Space is $T_0$ and Discrete Space is Zero Dimensional. * Discrete Space is $T_1$ and Discrete Space is Scattered. * Zero Dimensional Space is T3, and by definition of a regular space as being both $T_3$ and $T_0$. * Zero Dimensional $T_0$ Space is Totally Separated. * Discrete Space is Extremally Disconnected. * Extremally Disconnected Space is Totally Separated. * Totally Separated Space is Completely Hausdorff and Urysohn. * Totally Separated Space is Totally Disconnected. * Scattered $T_1$ Space is Totally Disconnected. * Totally Disconnected Space is Totally Pathwise Disconnected. * Totally Disconnected Space is $T_1$. {{qed}}
Suppose $T$ is not a $T_0$ (Kolmogorov) space. From Equivalence of Definitions of $T_0$ Space, there exist $x, y \in S$ such that $x$ and $y$ are both limit points of each other. So by definition of isolated point, neither $x$ nor $y$ are isolated in $\set {x, y}$. Thus we have found a subset $\set {x, y} \subseteq T$ such that $\set {x, y}$ is by definition dense-in-itself. So $T$ is not scattered. Hence the result by Rule of Transposition. {{qed}}
{{AimForCont}} $H$ is not biconnected. Then by definition there exist disjoint non-degenerate connected sets $U, V$ such that $H = U \cup V$. {{WLOG}}, let $p \in U$. Then $V \subset H \setminus \set p$. As $p$ is a dispersion point of $H$, $H \setminus \set p$ is totally disconnected. Thus, by definition, $H \setminus \set p$ contains no non-degenerate connected sets. But this contradicts our definition of $V$, as $V \subset H \setminus \set p$. It follows by Proof by Contradiction that $H$ is biconnected. {{qed}}
Let $T = \struct {S, \tau}$ be totally disconnected. Then by definition its components are singletons. Thus by definition each of its connected sets are degenerate. {{qed}}
By definition of pseudometric, we have that: {{begin-axiom}} {{axiom | n = \text M 1         | q = \forall x \in A         | m = \map d {x, x} = 0 }} {{axiom | n = \text M 2         | q = \forall x, y, z \in A         | m = \map d {x, y} + \map d {y, z} \ge \map d {x, z} }} {{axiom | n = \text M 3         | q = \forall x, y \in A         | m = \map d {x, y} = \map d {y, x} }} {{end-axiom}} Hence: {{begin-eqn}} {{eqn | lo= \forall x, y \in X:       | l = \map d {x, y} + \map d {y, x}       | o = \ge       | r = \map d {x, x}       | c = $\text M 2$ }} {{eqn | ll= \leadsto       | l = 2 \map d {x, y}       | o = \ge       | r = 0       | c = $\text M 1$ and $\text M 3$ }} {{eqn | ll= \leadsto       | l = \map d {x, y}       | o = \ge       | r = 0 }} {{end-eqn}} {{qed}} Category:Pseudometric Spaces hhxteeliri2ilwxojg011vsuv8hloiz
Suppose first that $d$ satisfies the hypotheses 1. and 2.. Since $d(u,v) \geq 0$ for all $u,v \in V$, $\| u \| = d(u,0) \geq 0$ for all $u \in V$. Moreover if $\| u \| = 0$ then $d(u,0) = 0$, so $u = 0$. Now let $\lambda \in K$, $u \in V$. Then, using the enlargement property of $d$: {{begin-eqn}} {{eqn|l=\norm{\lambda v}      |r=d(\lambda v, 0 )      |c=By the definition of $\norm{\,\cdot\,}$ }} {{eqn|l=      |r=d(\lambda v, \lambda \cdot 0)      |c= }} {{eqn|l=      |r=\norm{\lambda}_k d(v,0)      |c=By the enlargement property }} {{eqn|l=      |r=\norm{\lambda}_k \norm{v}      |c=By the definition of $\norm{\,\cdot\,}$ }} {{end-eqn}} Finally if $u,v \in V$, then we have {{begin-eqn}} {{eqn|l= \Vert u + v \Vert      |r= d(u + v, 0)      |c=By the definition of $\Vert \cdot \Vert$ }} {{eqn|l=      |r= d(u, -v)      |c=By translation invariance of $d$ }} {{eqn|l=      |r=d(u, 0) + d(0,-v)      |o=\leq      |c=By the triangle inequality for metrics. }} {{eqn|l=      |r=d(u, 0) + d(v,0)      |c=By the enlargement property }} {{eqn|l=      |r=\Vert u \Vert + \Vert v \Vert      |c=By the definition of $\Vert \cdot \Vert$ }} {{end-eqn}} {{stub|converse to add}} {{qed}} Category:Metric Spaces 18ys98ewi06rum6rtjtl9lio10uxcsm
Let $\SS$ be the set of all sets. Then $\SS$ must be an element of itself, in symbols, $\SS \owns \SS$. Thus we have an infinite descending sequence of membership: :$\SS \owns \SS \owns \SS \owns \cdots$ But by the axiom of foundation, no such sequence exists, a contradiction. {{qed}} {{questionable|This is circular: We need to demonstrate the contradiction and then demonstrate why ZF (specifically the AoF) prevents it happening.}} Category:Antinomies 87mh65z8bqgp2d2r28fe681ukov2elw
By definition, a perfect set is a set which equals its set of limit points. Let $x \in \R$. Consider the sequence: :$\left \langle{y_k}\right \rangle = x + \dfrac 1 k$ Then as $\left \langle{z_k}\right \rangle = \dfrac 1 k$ is a basic null sequence it follows that: :$\displaystyle \lim_{n \to \infty} \left \langle{y_k}\right \rangle = x$ Thus we see that $x$ is a limit point of $S$. {{qed}} Category:Real Numbers Category:Perfect Sets p7sv1yxcmud2uc77hsudekcha3vuuoi
First suppose that $a, b \in \R$. We have that the unit interval is uncountable. {{MissingLinks|Unit Interval is Uncountable (page does not exist at time of editing)}} Let $f: \closedint 0 1 \to \closedint a b$ such that $\map f x = a + \paren {b - a} x$. Then if $\map f {x_1} = \map f {x_2}$, we have: :$a + \paren {b - a} x_1 = a + \paren {b - a} x_2$ Because $b > a$: :$b - a > 0$ so this implies: :$x_1 = x_2$ Therefore $f$ is injective. Now if $\closedint a b$ were countable, there would be an injective function $g : \closedint a b \to \N$. By Composite of Injections is Injection, this implies that $g \circ f$ is an injection, so $\closedint 0 1$ is countable, a contradiction. If one or both of $a, b$ is (are) not real, then we can pick a closed interval $\closedint c d \subseteq \closedint a b$ with $c, d \in \R, c < d$. We know by the above that $\closedint c d$ is uncountable. Let $S = \set {x \in \R : a \le x \le b}$.  The identity function $I_\R : \closedint c d \to S$ is trivially injective. If $S$ were uncountable, we would have an injective function $g : S \to \N$. Then by Composite of Injections is Injection we would have an injection $\closedint c d \to \N$ given by $g \circ I_\R$. This was shown above to be false, so $S$ is uncountable. {{qed}}
Let $S, T \subseteq A$ such that $S$ and $T$ are separated in $A$. Then: :each point $x \in S$ has an open $\epsilon$-ball $\map {B_{\epsilon_x} } x$ which is disjoint from $T$ :each point $y \in T$ has an open $\epsilon$-ball $\map {B_{\epsilon_y} } y$ which is disjoint from $S$. Then: :$U_S = \displaystyle \bigcup_{x \mathop \in S} \map {B_{\epsilon_x / 2} } x$ :$U_T = \displaystyle \bigcup_{y \mathop \in T} \map {B_{\epsilon_y / 2} } y$ are disjoint open neighborhoods of $S$ and $T$ respectively. {{finish|that's cutting the corner}} Hence the result by the definition of $T_5$ space. {{qed}}
We have that a metric space is $T_4$. We also have that every closed set in a metric space is a $G_\delta$ set. Hence the result, by definition of a perfectly $T_4$ space. {{qed}}
We have that: :A metric space is a $T_2$ (Hausdorff) space. :A metric space is a perfectly $T_4$ space. :A metric space is a $T_5$ space. From Sequence of Implications of Separation Axioms we then have: :$T_5$ space is $T_4$ space. :$T_2$ (Hausdorff) Space is $T_1$ (Fréchet) Space. :$T_1$ (Fréchet) Space is $T_0$ (Kolmogorov) Space. By definition, a perfectly normal space is: :a perfectly $T_4$ space :a $T_1$ (Fréchet) space So $M$ is a perfectly normal space. By definition, a completely normal space is: :a $T_5$ space :a $T_1$ (Fréchet) space So $M$ is a completely normal space. Then from Sequence of Implications of Separation Axioms we can complete the chain: :Completely Normal Space is Normal Space. :Normal Space is Tychonoff (Completely Regular) Space. :Completely Regular (Tychonoff) implies $T_{3 \frac 1 2}$ by definition. :Completely Regular (Tychonoff) Space is Regular Space. :Completely Regular (Tychonoff) Space is Urysohn Space. :Regular implies $T_3$ by definition. :Regular Space is Completely Hausdorff Space. :Regular Space is Semiregular Space. {{qed}} There are other chains of implications which can be used, but the above are sufficient to prove the hypothesis.
We have that a metric space is fully $T_4$. We also have that a metric space is a $T_1$ (Fréchet) space. Hence the result, by definition of a fully normal space. {{qed}}
We have that a Metric Space is Fully Normal. Then we have that a Fully Normal Space is Paracompact. {{qed}}