avewright/proofwiki-math huggingface dataset:
Theorem: Riemann Zeta Has No Zeroes With Real Part One

Let $\zeta$ be the Riemann zeta function.
Then for all $t \in \R$:
:$\map \zeta {1 + i t} \ne 0$

Proof:
Throughout, the complex variable $s$ is $s = \sigma + i t$.
We have, for $\sigma > 1$,
{{begin-eqn}}
{{eqn | l = -\frac {\map {\zeta'} s} {\map \zeta s}
      | r = \sum_{n \mathop \ge 1} \map \Lambda n n^{-s}
      | c = Logarithmic Derivative of Riemann Zeta Function
}}
{{eqn | l = 
      | r = \sum_{n \mathop \ge 1} \map \Lambda n n^{-\sigma} \map \exp {-i t \log n}
      | c = where $s = \sigma + i t$
}}
{{eqn | l = 
      | r = \sum_{n \mathop \ge 1} \map \Lambda n n^{-\sigma} \paren {\map \cos {t \log n} - i \map \sin {t \log n} }
      | c = Euler's Formula
}}
{{end-eqn}}
Therefore,
:$(1): \quad \ds -\map \Re {\dfrac {\map {\zeta'} s} {\map \zeta s} } = \sum_{n \mathop \ge 1} \map \Lambda n n^{-\sigma} \map \cos {t \log n}$
Now observe that:
:$3 + 4 \cos \theta + \map \cos {2 \theta} = 2 \paren {1 + \cos \theta}^2 \ge 0$
Because for all $n \ge 1$ we have $\map \Lambda n n^{-\sigma} \ge 0$, we have:
{{begin-eqn}}
{{eqn | l = 0
      | o = \le
      | r = \sum_{n \mathop \ge 1} \map \Lambda n n^{-\sigma} \paren {3 + 4 \map \cos {t \log n} + \map \cos {2 t \log n} }
      | c = 
}}
{{eqn | r = -\map \Re {3 \frac {\map {\zeta'} \sigma} {\map \zeta \sigma} + 4 \frac {\map {\zeta'} {\sigma + i t} } {\map \zeta {\sigma + i t} } + \frac {\map {\zeta'} {\sigma + 2 i t} } {\map \zeta {\sigma + 2 i t} } }
      | c = from $(1)$
}}
{{end-eqn}}
Now let:
:$\map \eta s = \map \zeta s^3 \cdot \map \zeta {s + i t}^4\cdot \map \zeta {s + 2 i t}$
Then the above computation has shown that
:$\map \Re {\dfrac {\map {\eta'} s} {\map \eta s} } \le 0$
By Poles of Riemann Zeta Function we know that $\zeta$ has a simple pole at $s = 1$ with residue $1$.
{{AimForCont}} that $1 + i t$ is a zero of $\zeta$ of order $d \ge 1$.
Therefore, at $s = 1$, $\eta$ has a zero of order $4 d - 3 \ge 0$, that is:
:$\map \eta s \sim (s - 1)^{4 d - 3}$
as $s \to 1^+$, where:
:$\sim$ indicates asymptotic equality
:superscript $+$ denotes a limit from the right along the real line.
Therefore:
:$\dfrac {\map {\eta'} s} {\map \eta s} \sim \dfrac {4 d - 3} {s - 1}$
as $s \to 1^+$.
Since $\map \Re {\dfrac {4 d -3} {s - 1} } \to + \infty$ as $s \to 1^+$, it follows that:
:$\map \Re {\dfrac {\map {\eta'} s} {\map \eta s} } \to \infty$
as $s \to 1^+$.
But we have already shown that:
:$\map \Re {\dfrac {\map {\eta'} s} {\map \eta s} } \le 0$
which is a contradiction.
{{qed}}
Category:Riemann Zeta Function
pzelpennnm76seluzr7ze4t7liolp63
Theorem: Compact Space is Countably Compact

Let $T = \struct {S, \tau}$ be a compact space.
Then $T$ is a countably compact space.

Proof:
{{Recall|Countably Compact Space|countably compact space}}
{{:Definition:Countably Compact Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a compact space.
{{Recall|Compact Topological Space|compact space}}
{{:Definition:Compact Topological Space/Definition 1}}
So every countable open cover of $S$ has a finite subcover.
Hence by definition $T$ is a countably compact space.
{{qed}}
Theorem: Compact Space is Countably Compact

Let $T = \struct {S, \tau}$ be a compact space.
Then $T$ is a countably compact space.

Proof:
{{Recall|Countably Compact Space|countably compact space|index = 3}}
{{:Definition:Countably Compact Space/Definition 3}}
Let $T = \struct {S, \tau}$ be a sequentially compact topological space.
{{Recall|Sequentially Compact Space|sequentially compact space}}
{{:Definition:Sequentially Compact Space}}
The result follows from Limit of Sequence is Accumulation Point.
{{qed}}
{{ACC|Equivalence of Definitions of Countably Compact Space|3}}
Theorem: Compact Space is Countably Compact

Let $T = \struct {S, \tau}$ be a compact space.
Then $T$ is a countably compact space.

Proof:
{{Recall|Countably Compact Space|countably compact space|index = 1}}
{{:Definition:Countably Compact Space/Definition 1}}
Let $T = \struct {S, \tau}$ be a sequentially compact topological space.
{{Recall|Sequentially Compact Space|sequentially compact space}}
{{:Definition:Sequentially Compact Space}}
{{AimForCont}} $T$ is not countably compact.
Then there exists a countable open cover $\family {U_n}_{n \mathop \in \N}$ of $T$ which has no finite subcover.
Let:
:$\ds x_n \in S \setminus \bigcup_{i \mathop < n} U_i \ne \O$
for all $n \in \N$.
Then for arbitrary $p \in S$:
:$p \in U_n$
for at least some $n$.
Hence no infinite subsequence of $\sequence {x_n}$ converges to $p$.
This contradicts the fact that $T$ is sequentially compact.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Second-Countable Space is Compact iff Countably Compact

Let $T = \struct {S, \tau}$ be a second-countable space.
Then $T$ is compact {{iff}} $T$ is countably compact.

Proof:
Let $T = \struct {S, \tau}$ be a second-countable space.
Theorem: Unsymmetric Functional Equation for Riemann Zeta Function

Let $\zeta$ be the Riemann zeta function.
Let $\Gamma$ be the gamma function.
Then for all $s \in \C$:
:$\map \zeta {1 - s} = 2^{1 - s} \pi^{-s} \map \cos {\dfrac {\pi s} 2} \map \Gamma s \map \zeta s$

Proof:
For $s \notin \Z$, we have 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
Category:Reflection Formulas
g2itga0v2etq16x2h3z4emm1qe39ipo
Theorem: Countably Compact First-Countable Space is Sequentially Compact

A countably compact first-countable topological space is also sequentially compact.

Proof:
Follows directly from:
: Infinite Sequence in Countably Compact Space has Accumulation Point
: Accumulation Point of Infinite Sequence in First-Countable Space is Subsequential Limit
{{qed}}
Theorem: Countably Compact First-Countable Space is Sequentially Compact

A countably compact first-countable topological space is also sequentially compact.

Proof:
Let $T = \struct {S, \tau}$ be a countably compact first-countable topological space.
By definition of sequentially compact, it is sufficient to show that every infinite sequence in $S$ has a convergent subsequence.
Let $\sequence {s_n}$ be any sequence in $S$.
By Infinite Sequence in Countably Compact Space has Accumulation Point, $\sequence {s_n}$ has an accumulation point $p \in S$.
As $T$ is first-countable, $p$ has a countable local basis, say:
:$\set {V_n: V_1 \supseteq V_2 \supseteq V_3 \supseteq \cdots}$
Then a subsequence $\sequence {s_{n_i} }$, where $s_{n_i} \in V_i$, converges to $p$.
{{handwaving|This last statement is "by intimidation"; justification needed}}
{{qed}}
Theorem: Residue at Simple Pole

Let $f: \C \to \C$ be a function meromorphic on some region, $D$, containing $a$. 
Let $f$ have a simple pole at $a$. 
Then the residue of $f$ at $a$ is given by:
:$\ds \Res f a = \lim_{z \mathop \to a} \paren {z - a} \map f z$

Proof:
By Existence of Laurent Series, there exists a Laurent series:
:$\ds \map f z = \sum_{n \mathop = -\infty}^\infty c_n \paren {z - a}^n$
which is convergent in $D \setminus \set a$, where $\sequence {c_n}$ 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: 
:$\ds \map f z = \sum_{n \mathop = 0}^\infty c_n \paren {z - a}^n + \frac {c_{-1} } {z - a}$
Then:
{{begin-eqn}}
{{eqn	| l = \lim_{z \mathop \to a} \paren {z - a} \map f z
	| r = \lim_{z \mathop \to a} \paren {z - a} \paren {\sum_{n \mathop = 0}^\infty c_n \paren {z - a}^n + \frac {c_{-1} } {z - a} }
}}
{{eqn	| r = \lim_{z \mathop \to a} \paren {\sum_{n \mathop = 0}^\infty c_n \paren {z - a}^{n + 1} + c_{-1} }
}}
{{eqn	| r = \sum_{n \mathop = 0}^\infty c_n \paren {a - a}^{n + 1} + c_{-1}
}}
{{eqn	| r = 0 \sum_{n \mathop = 0}^\infty c_n + c_{-1} 
}}
{{eqn	| r = c_{-1}
}}
{{eqn	| r = \Res f a
	| c = {{Defof|Residue (Complex Analysis)|Residue}}
}}
{{end-eqn}}
{{qed}}
Category:Complex Analysis
61ofywbwba83iz6m3b4tw3z17pjtg5y
Theorem: Poles of Riemann Zeta Function

Let $\zeta$ be the Riemann zeta function.
Then $\zeta$ has a simple pole at $s = 1$ with residue $1$, and no other poles.

Proof:
By Analytic Continuation of Riemann Zeta Function using Mellin Transform of Fractional Part:
:$\ds \map \zeta s = \frac s {s - 1} - s \int_1^\infty \fractpart x x^{-s - 1} \rd x$
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:
:$\map \zeta {1 - s} = 2^{1 - s} \pi^{-s} \map \cos {\dfrac 1 2 s \pi} \map \Gamma s \map \zeta s$
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 $\ds 2^{1 - t} \pi^{-t}$ 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
66i7gp44gvg79rs6n2v6uhnteb0tio7
Theorem: Ring of Arithmetic Functions is Ring with Unity

Let $\AA$ be the set of all arithmetic functions.
Let $*$ denote Dirichlet convolution, and $+$ the pointwise sum of functions.
The ring of arithmetic functions $\struct {\AA, +, *}$ is a commutative ring with unity.

Proof:
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
7s5729eyq24b4o0avx5jq7bx993weic
Theorem: Units of Ring of Arithmetic Functions

Let $f$ be an arithmetic function.
Then $f$ is a unit in the ring of arithmetic functions {{iff}}:
:$\map f 1 \ne 0$

Proof:
Follows immediately from Invertibility of Arithmetic Functions and the definition of ring of arithmetic functions.
{{qed}}
Category:Arithmetic Functions
ms9ywwi9foyyqyxreu2xt3go953vc5z
Theorem: Möbius Inversion Formula

Let $f$ and $g$ be arithmetic functions.
Then:
:$(1): \quad \ds \map f n = \sum_{d \mathop \divides n} \map g d$
{{iff}}:
:$(2): \quad \ds \map g n = \sum_{d \mathop \divides n} \map f d \, \map \mu {\frac n d}$
where:
:$d \divides n$ denotes that $d$ is a divisor of $n$
:$\mu$ is the Möbius function.

Proof:
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 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}}
Theorem: Fully Normal Space is Normal

Let $T = \struct {S, \tau}$ be a fully normal space.
Then $T$ is a normal space.

Proof:
{{Recall|Fully Normal Space|fully normal space}}
{{:Definition:Fully Normal Space/Definition 2}}
We have that a fully $T_4$ space is also a $T_4$ space.
So:
:$T$ is a $T_4$ Space
:$T$ is a $T_1$ space
which is precisely the definition of a normal space.
{{qed}}
Theorem: Paracompact Space is Countably Paracompact

Let $T = \struct {S, \tau}$ be a paracompact space.
Then $T$ is a countably paracompact space.

Proof:
{{Recall|Countably Paracompact Space|countably paracompact space}}
{{:Definition:Countably Paracompact Space}}
Let $T = \struct {S, \tau}$ be a paracompact space.
{{Recall|Paracompact Space|paracompact space}}
{{:Definition:Paracompact Space}}
This applies in particular 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}}
Theorem: Metacompact Space is Countably Metacompact

Let $T = \struct {S, \tau}$ be a metacompact space.
Then $T$ is a countably metacompact space.

Proof:
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}}
Theorem: Fully T4 Space is T4

Let $T = \struct {S, \tau}$ be a fully $T_4$ space.
Then $T$ is a $T_4$ space.

Proof:
{{Recall|T4 Space|$T_4$ space}}
{{:Definition:T4 Space/Definition 1}}
Consider the open cover $\UU = \set {\relcomp S A, \relcomp S B}$.
Since $T$ is fully $T_4$, there exists a barycentric 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 $x^* \subseteq \relcomp S A$ or $x^* \subseteq \relcomp S B$.
But $a \in V_A \subseteq x^*$, so $x^* \nsubseteq \relcomp S A$.
Similarly $b \in V_B \subseteq x^*$, so $x^* \nsubseteq \relcomp S B$.
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}}
Theorem: Fully Normal Space is Paracompact

Let $T = \struct {S, \tau}$ be a fully normal space.
Then $T$ is paracompact.

Proof:
{{Recall|Paracompact Space|paracompact}}
{{:Definition:Paracompact Space}}
Let $T = \struct {S, \tau}$ be a fully normal space.
{{Recall|Fully Normal Space|fully normal}}
{{:Definition:Fully Normal Space/Definition 2}}
{{Recall|Fully T4 Space|fully $T_4$}}
{{:Definition:Fully T4 Space}}
Let $\UU$ be an open cover for $T$.
Then, from the definition of barycentric refinement, there exists a cover $\VV$ for $T$ such that:
:$\ds \forall x \in S: \exists U \in \UU: \paren {\bigcup \set {V \in \VV: x \in V} } \subseteq U$
{{proof wanted|It remains to fill in the gap. Despite the fact that the stronger result Fully T4 Space is Paracompact exists, it is edifying to have a first-principles proof from before the latter was developed.}}
Theorem: Fully Normal Space is Paracompact

Let $T = \struct {S, \tau}$ be a fully normal space.
Then $T$ is paracompact.

Proof:
Let $T = \struct {S, \tau}$ be a fully normal space.
{{Recall|Fully Normal Space|fully normal}}
{{:Definition:Fully Normal Space/Definition 2}}
Thus $T$ is a fully $T_4$ space.
The result follows from Fully $T_4$ Space is Paracompact.
Theorem: Subcover is Refinement of Cover

Let $S$ be a set.
Let $\CC$ be a cover for $S$.
Let $\VV$ be a subcover of $\CC$.
Then $\VV$ is a refinement of $\CC$.

Proof:
From definition of subcover:
:$\VV \subseteq \CC$
That is, every element of $\VV$ is an element of $\CC$.
From definition of subset, every element of $\VV$ is the subset of some element of $\CC$.
This is precisely the definition of refinement.
{{qed}}
Theorem: Compact Space is Paracompact

Let $T = \struct {S, \tau}$ be a compact space.
Then $T$ is paracompact.

Proof:
{{Recall|Compact Topological Space|compact space}}
{{:Definition:Compact Topological Space}}
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}}
Theorem: Compact Space is Paracompact

Let $T = \struct {S, \tau}$ be a compact space.
Then $T$ is paracompact.

Proof:
Follows directly from:
{{begin-itemize}}
{{item||Compact Space is Strongly Paracompact}}
{{item||Strongly Paracompact Space is Paracompact}}.
{{end-itemize}}
{{qed}}
Theorem: Paracompact Space is Metacompact

Let $T = \struct {S, \tau}$ be a paracompact space.
Then $T$ is also metacompact.

Proof:
{{Recall|Metacompact Space|metacompact space}}
{{:Definition:Metacompact Space}}
Let $T = \struct {S, \tau}$ be a paracompact space.
{{Recall|Paracompact Space|paracompact space}}
{{:Definition:Paracompact Space}}
Consider an arbitrary open cover $\CC$ of $S$.
Then {{apriori}} there exists an open refinement $\UU$ of $\CC$ which is locally finite.
From Locally Finite Refinement is Point Finite it follows that $\UU$ is point finite.
Hence $T$ must be, by definition, metacompact.
{{qed}}
Theorem: Countably Paracompact Space is Countably Metacompact

Let $T = \struct {S, \tau}$ be a countably paracompact space.
Then $T$ is countably metacompact.

Proof:
{{Recall|Countably Paracompact Space|countably paracompact space}}
{{:Definition:Countably Paracompact Space}}
{{Recall|Countably Metacompact Space|countably metacompact space}}
{{:Definition:Countably Metacompact Space}}
The result follows from Locally Finite Refinement is Point Finite.
{{qed}}
Theorem: Countably Compact Space is Countably Paracompact

Let $T = \struct {S, \tau}$ be a countably compact space.
Then $T$ is countably paracompact.

Proof:
{{Recall|Countably Paracompact Space|countably paracompact space}}
{{:Definition:Countably Paracompact Space}}
Let $T = \struct {S, \tau}$ be a countably compact space.
{{Recall|Countably Compact Space|countably compact space}}
{{:Definition:Countably Compact Space/Definition 1}}
Let $\CC$ be an arbitrary countable open cover of $T$.
Then {{apriori}} $\CC$ has a finite subcover $\CC'$.
From Subcover is Refinement of Cover, it follows that $\CC'$ is a refinement of $\CC$.
By Subcover of Open Cover is Open, $\CC'$ is an open cover of $T$.
From Finite Cover is Locally Finite, $\CC'$ is a locally finite cover of $S$.
Thus $\CC'$ is an open refinement of $\CC$ which is locally finite
As $\CC$ is arbitrary, 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}}
Theorem: Metacompact Countably Compact Space is Compact

Let $T = \struct {S, \tau}$ be a countably compact space which is also metacompact.
Then $T$ is compact.

Proof:
Let $T = \struct {S, \tau}$ be a countably compact space which is also metacompact.
{{Recall|Countably Compact Space|countably compact space}}
{{:Definition:Countably Compact Space/Definition 1}}
{{Recall|Metacompact Space|metacompact}}
{{:Definition:Metacompact Space}}
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.
{{AimForCont}} $x$ 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 cannot 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}}
Theorem: Countably Metacompact Lindelöf Space is Metacompact

Let $T = \struct {S, \tau}$ be a Lindelöf space which is also countably metacompact.
Then $T$ is metacompact.

Proof:
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}}
Theorem: Countably Paracompact Lindelöf Space is Paracompact

Let $T = \struct {S, \tau}$ be a Lindelöf space which is also countably paracompact.
Then $T$ is paracompact.

Proof:
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}}