hcju/proofwikiqa huggingface dataset:
Let $\xi$ be the completed Riemann zeta function :$\displaystyle \xi(s) = \frac 12 s(s-1) \pi^{-s/2} \Gamma\left(\frac s2\right) \zeta(s)$ By Gamma Difference Equation, this gives  :$\displaystyle \xi(s) = (s-1) \pi^{-s/2} \Gamma\left(\frac s2 + 1\right) \zeta(s)$ So taking the logarithm, :$\displaystyle \log \xi(s) = \log (s - 1) - \frac s2 \log\pi + \log\Gamma\left(\frac s2 + 1\right) + \log \zeta(s)$ And taking the derivative we obtain :$\displaystyle \frac{\xi'(s)}{\xi(s)} = \frac1{s-1} -\frac12 \log \pi + \frac{\Gamma'\left(s/2 + 1\right)}{\Gamma\left(s/2 + 1\right)} + \frac{\zeta'(s)}{\zeta(s)} \qquad (1)$ We have that the Completed Riemann Zeta Function has Order One, so by the Hadamard Factorisation Theorem, there exist constants $A$, $B$ such that :$\displaystyle \xi(s) = \exp(A+Bs)\: \prod_\rho \left( 1 - \frac s{\rho} \right) \exp\left( \frac s{\rho} \right)$ where $\rho$ runs over the zeros of $\xi$, that is, the nontrivial zeros of $\zeta$. Taking the logarithm, we obtain :$\displaystyle \log \xi(s) = A+Bs + \sum_\rho \left[ \log \left( 1 - \frac s{\rho} \right) + \frac s{\rho} \right]$ which has derivative :$\displaystyle \frac{\xi'(s)}{\xi(s)} = B + \sum_\rho \left( \frac 1{s - \rho} + \frac 1\rho\right) \qquad (2)$ Combining $(1)$ and $(2)$ we have :$\displaystyle B + \sum_\rho \left( \frac 1{s - \rho} + \frac 1\rho\right) =  \frac1{s-1} -\frac12 \log \pi + \frac{\Gamma'\left(s/2 + 1\right)}{\Gamma\left(s/2 + 1\right)} + \frac{\zeta'(s)}{\zeta(s)}$ from which the result follows by rearranging the terms. {{qed}}
Fix $T \ge 1$ and let $\rho_1, \rho_2, \ldots, \rho_n$ be an enumeration of the zeros of $f$ with modulus less than $T$, counted with multiplicity. By Jensen's Formula: :$\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map f {T e^{i \theta} } } \rd \theta = \ln \cmod {\map f 0} + \sum_{k \mathop = 1}^n \paren {\ln T - \ln \size {\rho_k} }$ Let $\rho_0 = 1$, $\rho_{n + 1} = T$, $r_k = \size {\rho_k}$. Then: {{begin-eqn}} {{eqn | l = \int_0^T \map N t \frac {\d t} t       | r = \sum_{k \mathop = 0}^n \int_{r_k}^{r_{k + 1} } \map N t \frac {\d t} t }} {{eqn | r = \sum_{k \mathop = 0}^n k \, \map \ln {\frac{r_{k + 1} } {r_k} }       | c = as by the definition of $N$, it is constant value $k$ on each interval $\openint {\size {\rho_k} }  {\size {\rho_{k + 1} } }$ }} {{eqn | r = \map \ln {\frac {T^n} {r_1 \dotsm r_n} } }} {{eqn | r = \sum_{k \mathop = 1}^n \paren {\ln T - \ln r_k} }} {{end-eqn}} and {{begin-eqn}} {{eqn | l = \int_0^T \map N t \frac {\d t} t       | r = \int_0^2 \map N {\frac {T \theta} 2} \frac {\d \theta} \theta }} {{eqn | o = \ge       | r = \map N {\frac T 2} \int_1^2 \frac {\d \theta} \theta       | c = Integration by Substitution }} {{eqn | r = \map N {\frac T 2} \ln 2       | c = {{Defof|Logarithm}} }} {{end-eqn}} Moreover, by hypothesis we have that: :$\displaystyle \frac 1 {2 \pi} \int_0^{2 \pi} \ln \size {\map f {T e^{i \theta} } } \rd \theta \le \sup_{\theta \mathop \in \closedint 0 {2 \pi} } \ln \size {\map f {T e^{i \theta} } } \ll \map \alpha T$ Putting these facts into Jensen's Formula we have: :$\map N {\dfrac T 2} \ln 2 + \cmod {\map f 0} \ll \map \alpha T$ which implies: :$\map N T \ll \map \alpha {2 T}$ {{qed}} === Proof of Corollary === Let $\epsilon > 0$, $\map N 0 = 0$, so that: :$\displaystyle \sum_{k \mathop \ge 1} \size {\rho_k}^{-1 - \epsilon} \le \sum_{T \mathop \ge 1} \paren {\map N T - \map N {T - 1} } T^{-1 - \epsilon}$ We have $\map N T \ll 2 T$, so $\map N T - \map N {T - 1}$ is bounded in $T$, say by $C > 0$. Therefore: :$\displaystyle \sum_{k \mathop \ge 1} \size {\rho_k}^{-1 - \epsilon} \le C \ \sum_{T \mathop \ge 1} \frac 1 {T^{1 + \epsilon} }$ and the sum on the {{RHS}} converges absolutely for $\epsilon > 0$. {{qed}} Category:Entire Functions jndqk6h2kh48t5fq924xgeqplpf53wv
Let $T = \struct {S, \tau}$ be strongly locally compact. Let $x \in S$. By definition, there exists an open set $U_x$ of $T$ such that: :$x \in U_x$ :${U_x}^-$ (the closure of $U_x$) is compact. From Set is Subset of its Topological Closure, $U_x \subseteq {U_x}^-$ and so $x \in {U_x}^-$. Thus $x$ is contained in a compact neighborhood. As this holds for all $x$, $T$ is a weakly locally compact. {{qed}}
Let $x \in S$. As $T$ is weakly locally compact, $x$ is contained in a compact neighborhood $N_x$. As $T$ is a $T_2$ (Hausdorff) space, we can use the result Compact Subspace of Hausdorff Space is Closed. Thus the interior of $N_x$ has a closure which is compact. Hence the result, from definition of strongly locally compact space. {{qed}}
=== 1 implies 2 === Let $T = \struct {S, \tau}$ be weakly $\sigma$-locally compact. Then by definition: :$T$ is $\sigma$-compact :$T$ is weakly locally compact. From Sigma-Compact Space is Lindelöf it follows directly that: :$T$ is Lindelöf :$T$ is weakly locally compact. {{qed|lemma}} === 2 implies 1 === Now let $T = \struct {S, \tau}$ be weakly locally compact and Lindelöf. By definition: :$T$ is weakly locally compact {{iff}} every point of $S$ is contained in a compact neighborhood. :$T$ is Lindelöf {{iff}} every open cover of $S$ has a countable subcover. Thus the interiors of the compact neighborhoods are an open cover of $S$. As $T$ is Lindelöf, this cover has a countable subcover. Thus $T$ is the union of countably many compact subspaces. That is, $T$ is $\sigma$-compact, and so weakly $\sigma$-locally compact. {{qed}}
Let $T$ be second-countable. Then by definition its topology has a countable basis. Let $\BB$ be this countable basis. Let $\CC$ be an open cover of $T$. Every set in $\CC$ is the union of a subset of $\BB$. So $\CC$ itself is the union of a subset of $\BB$. This union of a subset of $\BB$ is therefore a countable subcover of $\CC$. That is, $T$ is by definition Lindelöf. {{qed}}
Let $T = \struct {S, \tau}$ be a separable space. Then by definition there exists a countable subset of $X$ which is everywhere dense. We need to show that every disjoint set of open sets of $T$ is countable. {{ProofWanted}}
By Sum of Reciprocals of Powers as Euler Product: :$\displaystyle \map \zeta s = \prod_p \frac 1 {1 - p^{-s} }$ where $p$ ranges over the primes. From Laws of Logarithms: :$\displaystyle \ln \map \zeta s = - \sum_p \map \ln {1 - p^{-s} }$ Taking the derivative: {{begin-eqn}} {{eqn | l = \frac {\map {\zeta'} s} {\map \zeta s}       | r = -\sum_p \frac {\paren {\ln p} p^{-s} } {1 - p^{-s} }       | c = Logarithmic Derivative of Infinite Product of Analytic Functions }} {{eqn | r = -\sum_p \paren {\ln p} \paren {\frac 1 {1 - p^{-s} } - 1}       | c =  }} {{eqn | r = -\sum_p \paren {\ln p} \sum_{n \mathop \ge 1} p^{-n s}       | c = Sum of Infinite Geometric Sequence }} {{end-eqn}} Also, by the definition of $\Lambda$: :$\displaystyle \sum_{n \mathop \ge 1} \map \Lambda n n^{-s} = \sum_p \paren {\ln p} \sum_{n \mathop \ge 1} p^{-n s}$ and the proof is complete. {{qed}} Category:Riemann Zeta Function 7qdyq1f0w6xbecwodhq440lfqs8qgvh
Throughout, the complex variable $s$ is $s = \sigma + it$. We have, for $\sigma > 1$, {{begin-eqn}} {{eqn|l= -\frac{\zeta'(s)}{\zeta(s)}      |r= \sum_{n\geq 1} \Lambda(n) n^{-s}      |c= By Logarithmic Derivative of Riemann Zeta Function }} {{eqn|l=       |r=\sum_{n\geq 1} \Lambda(n) n^{-\sigma} \exp\left( -it\log n\right)      |c=Where $s = \sigma + it$ }} {{eqn|l=       |r=\sum_{n\geq 1} \Lambda(n) n^{-\sigma} \left( \cos(t\log n) - i \sin(t\log n) \right)      |c=By Euler's Formula }} {{end-eqn}} Therefore, :$\displaystyle -\Re\left( \frac{\zeta'(s)}{\zeta(s)} \right) = \sum_{n\geq 1} \Lambda(n) n^{-\sigma} \cos(t\log n) \qquad (1)$ Now observe that :$3 + 4\cos\theta + \cos(2\theta) = 2(1+ \cos\theta)^2 \geq 0$ Because for all $n \geq 1$ we have $\Lambda(n)n^{-\sigma} \geq 0$, we have {{begin-eqn}} {{eqn|l=0      |r=\sum_{n\geq 1} \Lambda(n) n^{-\sigma} \left\{ 3 + 4\cos(t\log n) + \cos(2t\log n) \right\}      |c=      |o=\leq }} {{eqn|l=      |r=-\Re\left( 3\frac{\zeta'(\sigma)}{\zeta(\sigma)} + 4\frac{\zeta'(\sigma + it)}{\zeta(\sigma + it)} + \frac{\zeta'(\sigma + 2it)}{\zeta(\sigma + 2it)} \right)      |c=By $(1)$ }} {{end-eqn}} Now let  :$\eta(s) = \zeta(s)^3\cdot \zeta(s+it)^4\cdot \zeta(s+2it)$ Then the above computation has shown that :$\displaystyle\Re\left( \frac{\eta'(s)}{\eta(s)} \right) \leq 0$ By Poles of Riemann Zeta Function we know that $\zeta$ has a simple pole at $s=1$ with residue $1$. Suppose that $1+it$ is a zero of $\zeta$ of order $d \geq 1$. Therefore, at $s = 1$, $\eta$ has a zero of order $4d - 3 \geq 0$, that is, :$\displaystyle \eta(s) \sim (s-1)^{4d -3}$ as $s \to 1^+$, where $\sim$ indicates asymptotic equality, and superscript $+$ denotes a limit from the right along the real line. Therefore :$\displaystyle \frac{\eta'(s)}{\eta(s)} \sim \frac{4d -3}{s-1}$ as $s \to 1^+$. Since $\displaystyle \Re\left( \frac{4d -3}{s-1}\right) \to + \infty$ as $s \to 1^+$, it follows that :$\displaystyle \Re\left(\frac{\eta'(s)}{\eta(s)}\right) \to \infty$ as $s \to 1^+$. But we have already shown that :$\displaystyle\Re\left( \frac{\eta'(s)}{\eta(s)} \right) \leq 0$ a contradiction. {{qed}} Category:Riemann Zeta Function fo6o84yn30601r2nq1x3org4ioua0l4
Let $T = \struct {S, \tau}$ be a compact space. Then by definition every open cover of $S$ has a finite subcover. So every countable open cover of $S$ has a finite subcover. Hence by definition $T$ is countably compact. {{qed}}
=== Necessary Condition === We have that a compact space is countably compact (whether the space is second-countable or not). {{qed|lemma}} === Sufficient Condition === Let $T$ be countably compact. Since $T$ is second-countable, we can choose a countable (analytic) basis $\BB$ for $\tau$. Let $\CC$ be an open cover for $S$. Define: :$\CC' = \set {B \in \BB: \exists C \in \CC: B \subseteq C}$ Then $\CC' \subseteq \BB$, and so $\CC'$ is countable. By the definition of an analytic basis, we have $\BB \subseteq \tau$; since $\subseteq$ is a transitive relation, it therefore follows that $\CC' \subseteq \tau$. From Equivalence of Definitions of Analytic Basis: :$\forall C \in \CC: \forall x \in C: \exists B \in \CC': x \in B$ That is, by the definition of set union: :$\displaystyle \forall C \in \CC: C \subseteq \bigcup \CC'$ Hence, by Union is Smallest Superset: :$\displaystyle S \subseteq \bigcup \CC \subseteq \bigcup \CC'$ Since $\subseteq$ is a transitive relation, it follows that $\CC'$ is an open cover for $S$. Since $T$ is countably compact, there exists a finite subcover $\FF$ of $\CC'$ for $S$. By the Principle of Finite Choice, there exists an indexed family $\family {C_B}_{B \mathop \in \FF}$ such that: :$\forall B \in \FF: B \subseteq C_B \in \CC$ Note that $\set {C_B: B \in \FF}$ is a finite subcover of $\CC$ for $X$. Hence, $T$ is compact. {{qed}}
We have for $s \notin \Z$ Euler's Reflection Formula: :$\map \Gamma s \, \map \Gamma {1 - s} = \dfrac \pi {\map \sin {\pi s} }$ Replacing $s \mapsto \dfrac {1 + s} 2$ we deduce: {{begin-eqn}} {{eqn | l = \map \Gamma {\frac {1 + s} 2} \, \map \Gamma {\frac {1 - s} 2}       | r = \frac \pi {\map \sin {\pi \paren {1 + s} / 2} }       | c = substituting $s \mapsto \dfrac {1 + s} 2$ }} {{eqn | r = \frac \pi {\map \cos {\pi s / 2} }       | c = Sine and Cosine are Periodic on Reals }} {{end-eqn}} Also, we have Legendre's Duplication Formula for $z \notin -\dfrac 1 2 \N_0$: :$\map \Gamma s \, \map \Gamma {s + \dfrac 1 2} = 2^{1 - 2 s} \sqrt \pi \map \Gamma {2 s}$ Replacing $s \mapsto s / 2$ this yields: :$\map \Gamma {\dfrac s 2} \, \map \Gamma {\dfrac {1 + s } 2} = 2^{1 - s} \sqrt \pi \, \map \Gamma s$ Together these give: :$(1): \quad \dfrac {\map \Gamma {s / 2} } {\map \Gamma {\paren {1 - s} / 2} } = 2^{1 - s} \pi^{-1/2} \, \map \Gamma s \, \map \cos {\pi s / 2}$ Now we take the Functional Equation for Riemann Zeta Function: :$\pi^{-s/2} \, \map \zeta s \, \map \Gamma {s / 2} \, \map \Gamma {\dfrac {1 - s} 2}^{-1} = \pi^{\paren {s - 1} / 2} \, \map \zeta {1 - s}$ and substitute $(1)$ to give: :$\pi^{\paren {s - 1} / 2} \, \map \zeta {1 - s} = \pi^{-\paren {s + 1} / 2} \, \map \zeta s 2^{1 - s} \, \map \Gamma s \, \map \cos {\pi s / 2}$ Multiplying by $\pi^{\paren {s - 1} / 2}$ this becomes: :$\map \zeta {1 - s} = \pi^{-s} 2^{1 - s} \, \map \cos {\pi s / 2} \, \map \Gamma s \, \map \zeta s$ as desired. {{qed}} Category:Riemann Zeta Function e0b4dqr0ckvuouvjabqz7z7qo1p7ry1
By Existence of Laurent Series, there exists a Laurent series: :$\displaystyle f\left({z}\right) = \sum_{n \mathop = -\infty}^\infty c_n \left({z - a}\right)^n$ which is convergent in $D \setminus \left\{a\right\}$, where $\left({c_n}\right)$ is a doubly infinite sequence of complex coefficients.  We are given that $f$ has only a simple pole at $a$. Thus $c_n = 0$ for $n < -1$. So we can write:  :$\displaystyle f\left({z}\right) = \sum_{n \mathop = 0}^\infty c_n \left({z - a}\right)^n + \frac {c_{-1} } {z - a}$ Then: {{begin-eqn}} {{eqn	| l = \lim_{z \mathop \to a} \left({z - a}\right) f\left({z}\right) 	| r = \lim_{z \mathop \to a} \left({z - a}\right) \left({ \sum_{n \mathop = 0}^\infty c_n \left({z - a}\right)^n + \frac {c_{-1} } {z - a} } \right) }} {{eqn	| r = \lim_{z \mathop \to a} \left({ \sum_{n \mathop = 0}^\infty c_n \left({z - a}\right)^{n + 1} + c_{-1} }\right) }} {{eqn	| r = \sum_{n \mathop = 0}^\infty c_n \left({a - a}\right)^{n+1} + c_{-1} }} {{eqn	| r = 0 \sum_{n \mathop = 0}^\infty c_n + c_{-1}  }} {{eqn	| r = c_{-1} }} {{eqn	| r = \operatorname{Res} \left({f, a}\right) 	| c = {{Defof|Residue (Complex Analysis)|Residue}} }} {{end-eqn}} {{qed}} Category:Complex Analysis 0zekccwvg5l1wwblefgy8gxori8tg9q
By Analytic Continuation of Riemann Zeta Function using Mellin Transform of Fractional Part: :$\displaystyle \map \zeta s = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$ is meromorphic for $\map \Re s > 0$, and the integral converges to a finite value for fixed $s$ in this region. Therefore in this region the only pole of $\zeta$ is at $s = 1$, with residue: {{begin-eqn}} {{eqn | l = \Res \zeta 1       | r = \lim_{s \mathop \to 1} \paren {s - 1} \paren {\frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x}       | c = Residue at Simple Pole }} {{eqn | r = 1 }} {{end-eqn}} By Unsymmetric Functional Equation for Riemann Zeta Function: :$\displaystyle \map \zeta {1 - s} = 2^{1 - s} \pi^{-s} \map \cos {\frac 1 2 s \pi} \, \map \Gamma s \, \map \zeta s$ Therefore, for $\map \Re s \le 0$: {{begin-eqn}} {{eqn | l = \map \zeta s       | r = \map \zeta {1 - t}       | c = for some $t$ with $\map \Re t \ge 1$ }} {{eqn | r = 2^{1 - t} \pi^{-t} \map \cos {\frac 1 2 t \pi} ,\ \map \Gamma t \,\map \zeta t }} {{end-eqn}} By Complex Exponential Function is Entire, the factor $\displaystyle 2^{1 - t} \pi^{-t}$ has no poles when $\map \Re t \ge 1$. By Poles of Gamma Function, $\map \Gamma t$ has no poles when $\map \Re t \ge 1$. By Complex Cosine Function is Entire, $\map \cos {\dfrac 1 2 t \pi}$ also has no poles in this region. Therefore, the only possible pole is a simple pole at $t = 1$ from the factor $\map \zeta t$. But at this point: :$\map \cos {\dfrac 1 2 t \pi} = \map \cos {\dfrac \pi 2} = 0$ which cancels the pole of $\zeta$. {{qed}} Category:Riemann Zeta Function jsk325rcv2jct7le98xlkhivv8t9y96
By Structure Induced by Abelian Group Operation is Abelian Group, $\struct {\AA, +}$ is an abelian group. By Properties of Dirichlet Convolution, $*$ is commutative, associative and has a unity. Therefore $\struct {\AA, +, *}$ is a commutative ring with unity. {{finish|link to a proof for distributive}} Category:Arithmetic Functions Category:Examples of Rings jy35b34ylhbf2lw3bll4jlig6smlf56
Follows immediately from Invertibility of Arithmetic Functions and the definition of ring of arithmetic functions. {{qed}} Category:Arithmetic Functions ms9ywwi9foyyqyxreu2xt3go953vc5z
Let $u$ be the unit arithmetic function and $\iota$ the identity arithmetic function. Let $*$ denote Dirichlet convolution. Then equation $(1)$ states that $f = g * u$ and  $(2)$ states that $g = f * \mu$. The proof rests on the following facts: By the lemma to Sum of Möbius Function over Divisors: :$\mu * u = \iota$ By Properties of Dirichlet Convolution, Dirichlet convolution is commutative, associative and $h * \iota = h$ for all $h$. We have: {{begin-eqn}} {{eqn | l = f = g * u       | o = \leadsto       | r = f * \mu = \paren {g * u} * \mu }} {{eqn | o = \leadsto       | r = f * \mu = g * \paren {u * \mu} }} {{eqn | o = \leadsto       | r = f * \mu = g }} {{end-eqn}} Conversely: {{begin-eqn}} {{eqn | l = g = f * \mu       | o = \leadsto       | r = g * u = \paren {f * \mu} * u }} {{eqn | o = \leadsto       | r = g * u = f * \paren {\mu * u} }} {{eqn | o = \leadsto       | r = g * u = f }} {{end-eqn}} Hence the result. {{qed}} {{Namedfor|August Ferdinand Möbius|cat = Möbius}}
From the definition, $T$ is fully normal {{iff}}: :$T$ is fully $T_4$ :$T$ is a $T_1$ (Fréchet) space. We have that a fully $T_4$ space is also a $T_4$ space. So: :$T$ is a $T_4$ Space :$T$ is a $T_1$ (Fréchet) space which is precisely the definition of a normal space. {{qed}}
From the definition, $T$ is paracompact space {{iff}} every open cover of $T$ has an open refinement which is locally finite. This also applies to all countable open covers. So every countable open cover of $T$ has an open refinement which is locally finite. This is precisely the definition for a countably paracompact space. {{qed}}
From the definition, $T$ is metacompact space iff every open cover of $S$ has an open refinement which is point finite. This also applies to all countable open covers. So every countable open cover of $S$ has an open refinement which is point finite. This is precisely the definition for a countably metacompact space. {{qed}}
Recall that $\struct {S, \tau}$ is a $T_4$ space {{iff}}: :$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$  That is, for any two disjoint closed sets $A, B \subseteq X$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively. Consider the open cover $\UU = \set {\map \complement A, \map \complement B}$. Since $T$ is fully $T_4$, there exists a star refinement $\VV$ of $\UU$. Define $\VV_A$ and $\VV_B$ by: :$\VV_A := \set {V \in \VV: A \cap V \ne \O}$ :$\VV_B := \set {V \in \VV: B \cap V \ne \O}$ and subsequently $U_A$ and $U_B$ by: :$U_A := \bigcup \VV_A$ :$U_B := \bigcup \VV_B$ Now given $a \in A$, since $\VV$ is a cover, there exists $V \in \VV$ such that $a \in V$. But then $V \in \VV_A$, so that: :$a \in U_A$ Hence $A \subseteq U_A$, and similarly $B \subseteq U_B$. {{AimForCont}} that $x \in U_A \cap U_B$. Then there exist $V_A \in \VV_A, V_B \in \VV_B$ such that: :$x \in V_A, x \in V_B$ By definition of $\VV_A$ and $\VV_B$, there exist: :$a \in A \cap V_A$ :$b \in B \cap V_B$ Now recall the definition of the star $x^*$ of $x$: :$x^* := \bigcup \set {V \in \VV: x \in V}$ Since $T$ is fully $T_4$, either $x^* \subseteq \map \complement A$ or $x^* \subseteq \map \complement B$. But $a \in V_A \subseteq x^*$, so $x^* \nsubseteq \map \complement A$. Similarly $b \in V_B \subseteq x^*$, so $x^* \nsubseteq \map \complement B$. From this contradiction it is inferred that $x \in U_A \cap U_B$ cannot exist. That is: :$U_A \cap U_B = \O$ and it follows that $T$ is a $T_4$ space. {{qed}}
From the definition, $T$ is fully normal {{iff}}: :$T$ is fully $T_4$ :$T$ is a $T_1$ (Fréchet) space. Then $T$ is fully $T_4$ {{iff}} every open cover of $S$ has a star refinement. Let $\UU$ be an open cover for $T$. Then from the definition, there exists a cover $\VV$ for $T$ such that: :$\displaystyle \forall x \in S: \exists U \in \UU: \paren {\bigcup \set {V \in \VV: x \in V} } \subseteq U$ Recall from the definition of paracompact: :$T$ is paracompact {{iff}} every open cover of $S$ has an open refinement which is locally finite. {{proof wanted|It remains to fill in the gap.}}
From the definition of subcover: : $\mathcal V \subseteq \mathcal C$ That is, every element of $\mathcal V$ is an element of $\mathcal C$. From the definition of subset, every element of $\mathcal V$ is the subset of some element of $\mathcal C$. This is precisely the definition of refinement. {{qed}} Category:Covers jctjtgc73oevd6f4pktjiwyjrpblx12
From the definition, $T$ is compact {{iff}} every open cover of $S$ has a finite subcover. From Subcover is Refinement of Cover, it follows that every open cover of $S$ has an open refinement which is locally finite. This is precisely the definition of paracompact. {{qed}}
From the definition, $T$ is paracompact {{iff}} every open cover of $T$ has an open refinement which is locally finite. Consider some open cover $\UU$ of $T$. Let $x \in S$. Then there exists some neighborhood $\exists N_x$ of $x$ which intersects only finitely many elements of $\UU$. Thus $x$ itself can be in only finitely many elements of $\UU$. Hence $T$ must be, by definition, metacompact. {{qed}}
From the definition, $T$ is countably paracompact {{iff}} every countable open cover of $X$ has an open refinement which is locally finite. Consider some countable open cover $\mathcal U$ of $X$. Let $x \in X$. Then there exists some neighborhood $\N_x$ of $x$ which intersects only finitely many elements of $\mathcal U$. Thus $x$ itself can be in only finitely many elements of $\mathcal U$. Hence $T$ must be, by definition, countably metacompact. {{qed}}
From the definition, $T$ is countably compact {{iff}} every countable open cover of $S$ has a finite subcover. From Subcover is Refinement of Cover, it follows that every countable open cover of $S$ has an open refinement which is locally finite. This is precisely the definition of countably paracompact. {{qed}}
From the definition of countably compact space, every countable open cover of $S$ has a finite subcover. Let $T = \struct {S, \tau}$ be a countably compact space which is also metacompact. Let $\UU_\alpha$ be an open cover of $S$. Then let $\VV_\beta$ be the open refinement which is point finite, guaranteed by its metacompactness. Let $x \in S$. We have that $x$ is an element of only a finite number of the elements of $\VV_\beta$, as $\VV_\beta$ is point finite. Consider all the subcovers of $\VV_\beta$, and order them by subset. Consider any chain $\CC$ of such subcovers. Suppose $x$ is not in the intersection of $\CC$. Then it would fail to be covered by one of the elements of $\CC$. This would be a contradiction of the fact that every element of $\CC$ is a cover of $\VV_\beta$.  So the intersection of a chain of such subcovers is itself a subcover. Hence it must be that $\VV$ has a minimal subcover, $\VV_\gamma$, say. Now each element $V_\gamma$ of $\VV_\gamma$ must itself contain a unique element $x_\gamma \in V_\gamma$ which belongs to no element of $\VV_\gamma$. This is because $\VV_\gamma$ is minimal. If $\VV_\gamma$ were infinite, the set of all $\set {x_\gamma}$ would be an infinite set without an $\omega$-accumulation point. This can not be the case. {{explain|Why can this not be the case?}} Thus $\VV_\gamma$ is a finite cover. So we have constructed a finite subcover for $\UU_\alpha$, demonstrating that $T$ is 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 metacompact space then every countable open cover of $S$ has an open refinement which is point finite. It follows trivially that every open cover of $S$ has an open refinement which is point finite. Hence the result by definition of metacompact. {{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 paracompact space then every countable open cover of $S$ has an open refinement which is locally finite. It follows trivially that every open cover of $S$ has an open refinement which is locally finite. Hence the result by definition of paracompact. {{qed}}