avewright/proofwiki-math huggingface dataset:
Theorem: Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods

Let $T = \struct {S, \tau}$ be a Hausdorff space.
Let $V_1$ and $V_2$ be compact sets in $T$ which are disjoint:
:$V_1 \cap V_2 = \O$
Then $V_1$ and $V_2$ have disjoint neighborhoods.

Proof:
We first suppose that $V_1 = \set x$ and $x \not \in V_2$. 
Since $T$ is Hausdorff:
:for each $y \in V_2$ there exists an open neighborhood $O^1_y$ of $x$ and an open neighborhood $O^2_y$ of $y$ such that $O^1_y \cap O^2_y = \O$.
Hence we have:
:$\ds V_2 \subseteq \bigcup_{y \mathop \in V_2} O_y^2$ 
Since $V_2$ is compact, there exists $y_1, \ldots, y_n \in V_2$ such that:
:$\ds V_2 \subseteq \bigcup_{i \mathop = 1}^n O_{y_i}^2$
Now let:
:$\ds U_2 = \bigcup_{i \mathop = 1}^n O_{y_i}^2$
and:
:$\ds U_1 = \bigcap_{i \mathop = 1}^n O_{y_i}^1$
We have that $U_1$ and $U_2$ are open, $x \in U_1$ and $V_2 \subseteq U_2$. 
Further, if $x \in U_2$ then $x \in O_{y_i}^2$ for some $i$. 
Since $O_{y_i}^2 \cap O_{y_i}^1 = \O$, it follows that $x \not \in O_{y_i}^1$. 
Hence $x \not \in U_1$. 
So $U_1 \cap U_2 = \O$.
Hence $U_1$ and $U_2$ are disjoint open sets with $V_1 \subseteq U_1$ and $V_2 \subseteq U_2$. 
Hence we have proven the claim in the case $V_1 = \set x$. 
Now take $V_1$ to be general.
For each $x \in V_1$, there exists open sets $U_{1, x}$ and $U_{2, x}$ such that $U_{1, x} \cap U_{2, x} = \O$, $x \in U_{1, x}$ and $V_2 \subseteq U_{2, x}$.
In particular:
:$\ds V_1 \subseteq \bigcup_{x \in V_1} U_{1, x}$
Since $V_1$ is compact, there exists $x_1, \ldots, x_n \in V_1$ such that:
:$\ds V_1 \subseteq \bigcup_{i \mathop = 1}^n U_{1, x_i}$
Now set:
:$\ds U_1 = \bigcup_{i \mathop = 1}^n U_{1, x_i}$
and:
:$\ds U_2 = \bigcap_{i \mathop = 1}^n U_{2, x_i}$
We have $U_1 \cap U_2 = \O$ by an identical argument to above. 
Further, $U_1$ and $U_2$ are both open.
Since $V_2 \subseteq U_{2, x_i}$ for each $i$ it follows that $V_2 \subseteq U_2$.
Hence $U_1$ and $U_2$ are the desired open sets.
{{qed}}
Theorem: Compact Subsets of T3 Spaces

Let $T = \struct {S, \tau}$ be a $T_3$ space.
Let $A \subseteq S$ be compact in $T$.
Then for each $U \in \tau$ such that $A \subseteq U$:
:$\exists V \in \tau: A \subseteq V \subseteq V^- \subseteq U$
where $V^-$ denotes the closure of $V$.

Proof:
Let $A \subseteq S$ be compact in $T$.
Let $U \in \tau$ such that $A \subseteq U$.
Since $T$ is $T_3$:
:Each open set contains a closed neighborhood around each of its points:
::$\forall x \in U: \exists N_x: \relcomp S {N_x} \in \tau: \exists V_x \in \tau: x \in V_x \subseteq N_x \subseteq U$
Note that $\set {V_x: x \in A}$ forms an open cover for $A$.
By compactness of $A$, $\set {V_x: x \in A}$ has a finite subcover.
Therefore there is a finite subset $I \subseteq A$ such that: 
:$\set {V_x: x \in I}$ is an open cover for $A$.
Let $V = \ds \bigcup_{i \mathop \in I} V_x$, $N = \ds \bigcup_{i \mathop \in I} N_x$.
By definition of a topology, the set $\ds \bigcup_{i \mathop \in I} V_x$ is open since it is a  union of open sets.
By Finite Union of Closed Sets is Closed in Topological Space, the set $\ds \bigcup_{i \mathop \in I} N_x$ is closed since it is a finite union of closed sets.
By Set Union Preserves Subsets, since $V_x \subseteq N_x \; \forall x \in I \subseteq U$:
:$V \subseteq N$
By Set Closure is Smallest Closed Set in Topological Space:
:$V \subseteq V^- \subseteq N$
Thus:
{{begin-eqn}}
{{eqn | l = A
      | o = \subseteq
      | r = V
      | c = {{Defof|Open Cover}}
}}
{{eqn | o = \subseteq
      | r = V^-
}}
{{eqn | o = \subseteq
      | r = N
}}
{{eqn | o = \subseteq
      | r = U
      | c = Union of Subsets is Subset: $\forall x \in I \subseteq U: N_x \subseteq U$
}}
{{end-eqn}}
satisfies our claim.
{{qed}}
Theorem: Compact Hausdorff Space is T4

Let $T = \struct {S, \tau}$ be a compact $T_2$ (Hausdorff) space.
Then $T$ is a $T_4$ space.

Proof:
{{Recall|T4 Space|$T_4$ space|index = 1}}
{{:Definition:T4 Space/Definition 1}}
We have that a Compact Subspace of Hausdorff Space is Closed.
We also have that a Closed Subspace of Compact Space is Compact.
We also have that Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods.
But by the definition above, $T$ is a $T_4$ space when any two disjoint closed subsets of $S$ are separated by neighborhoods.
Hence the result.
{{qed}}
Theorem: Compact Hausdorff Topology is Minimal Hausdorff

Let $T = \struct {S, \tau}$ be a Hausdorff space which is compact.
Then $\tau$ is the minimal subset of the power set $\powerset S$ such that $T$ is a Hausdorff space.

Proof:
{{AimForCont}} there exists a topology $\tau'$ on $S$ such that:
:$\tau' \subseteq \tau$ but $\tau' \ne \tau$
:$\tau'$ is a Hausdorff space.
From Equivalence of Definitions of Finer Topology:
:the identity mapping $I_S: \struct {S, \tau} \to \struct {S, \tau'}$ is continuous.
Let $A \in \tau$.
Then $S \setminus A \subseteq S$ is closed in $\struct {S, \tau}$.
By Closed Subspace of Compact Space is Compact, $S \setminus A$ is compact in $\struct {S, \tau}$.
From Continuous Image of Compact Space is Compact:
:$I_S \sqbrk {S \setminus A}$ is also compact.
By hypothesis, $\struct {S, \tau'}$ is a Hausdorff space.
From Compact Subspace of Hausdorff Space is Closed:
:$I_S \sqbrk {S \setminus A}$ is closed in $\struct {S, \tau'}$.
Hence:
:$A = S \setminus I_S \sqbrk {S \setminus A} \in \tau'$
By definition of subset:
:$\tau \subseteq \tau'$
Thus we have that $\tau \subseteq \tau'$ and $\tau' \subseteq \tau$.
Hence by definition of set equality:
:$\tau' = \tau$
But this contradicts our hypothesis that $\tau' \ne \tau$.
By Proof by Contradiction, it follows that no topology on $S$ which is strictly coarser than $\tau$ can be Hausdorff.
{{qed}}
Theorem: Compact Hausdorff Topology is Maximally Compact

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

Proof:
Let $\tau'$ be a topology on $S$ such that $\tau \subseteq \tau'$ but that $\tau \ne \tau'$.
Consider the identity mapping $I_S: \struct {S, \tau'} \to \struct {S, \tau}$.
From Separation Properties Preserved in Subspace, $I_S$ is a continuous bijection from a Hausdorff space to a compact Hausdorff space.
{{AimForCont}} $\struct {S, \tau'}$ is compact.
From Continuous Bijection from Compact to Hausdorff is Homeomorphism, $I_S$ is also open.
Hence:
:$\tau' \subseteq \tau$
Thus we have that $\tau \subseteq \tau'$ and $\tau' \subseteq \tau$.
Hence by definition of set equality:
:$\tau' = \tau$
But this contradicts our hypothesis that $\tau' \ne \tau$.
By Proof by Contradiction, it follows that no topology on $S$ which is strictly finer than $\tau$ can be compact.
{{qed}}
Theorem: Maximum Cardinality of Separable Hausdorff Space

Let $T = \struct {S, \tau}$ be a Hausdorff space which is separable.
Then $S$ can have a cardinality no greater than $2^{2^{\aleph_0} }$.

Proof:
Let $D$ be an everywhere dense subset of $S$ which is countable, as is guaranteed as $T$ is separable.
Consider the mapping $\Phi: S \to 2^{\powerset D}$ defined as:
:$\forall x \in S: \map {\map \Phi x} A = 1 \iff A = D \cap U_x$ for some neighborhood $U_x$ of $x$
{{explain|It is not clear in Steen & Seeabch what is meant by $\Phi: S \to 2^{\powerset D}$ -- presumably $2^{\powerset D}$ is the ordinal which is the power set of the power set of $D$. It is also not clear what the notation $\map {\map \Phi x} A$ means -- in fact is may be the case that a transcription error has been committed.  Hence the proof cannot be attempted until these points have been cleared up.}}
It is seen that if $T$ is a Hausdorff space, then $\Phi$ is an injection.
It follows that:
:$\card S \le \card {2^{\powerset D} } = 2^{2^{\aleph_0} }$
{{explain|the chain of reasoning leading to the above}}
Theorem: Cluster Point of Ultrafilter is Unique

Let $S$ be a set.
Let $\FF$ be an ultrafilter on $S$.
Let $x \in S$ be a cluster point of $\FF$.
Then there is no point $y \in S: y \ne x$ such that $y$ is also a cluster point of $\FF$.

Proof:
Let $x, y$ both be cluster points of an ultrafilter $\FF$ on a set $S$.
Then by definition:
:$\forall U \in \FF: \set {x, y} \subseteq U$
But then we can create a finer filter $\FF \cup \set {\set x}$ on $S$.
So $\FF$ is not an ultrafilter.
{{qed}}
Theorem: Sequence of Implications of Global Compactness Properties

Let $P_1$ and $P_2$ be topological compactness properties.
Let:
:$P_1 \implies P_2$
mean:
:If a topological space $T$ satisfies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
{|
|-
| align="center" | ||
| align="center" | ||
| align="center" | Sequentially Compact ||
| align="center" | ||
| align="center" | ||
|-
| align="center" | ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | ||
|-
| align="center" | Compact ||
| align="center" | $\implies$ ||
| align="center" | Countably Compact ||
| align="center" | $\implies$ ||
| align="center" | Pseudocompact ||
|-
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | ||
|-
| align="center" | $\sigma$-Compact ||
| align="center" | ||
| align="center" | Weakly Countably Compact ||
| align="center" | ||
| align="center" | ||
|-
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | ||
| align="center" | ||
| align="center" | ||
|-
| align="center" | Lindelöf Space ||
| align="center" | ||
| align="center" | ||
| align="center" | ||
| align="center" | ||
|}
If there is no implication indicated, then there exists a counterexample demonstrating that no such implication exists.

Proof:
The relevant justifications are listed as follows:
* Sequentially Compact Space is Countably Compact
* Compact Space is Countably Compact
* Compact Space is $\sigma$-Compact
* $\sigma$-Compact Space is Lindelöf
* Countably Compact Space is Weakly Countably Compact
* Countably Compact Space is Pseudocompact
Then we have:
* Compact Space is not necessarily Sequentially Compact
* Sequentially Compact Space is not necessarily Compact
* $\sigma$-Compact Space is not necessarily Compact
* Lindelöf Space is not necessarily $\sigma$-Compact
* Countably Compact Space is not necessarily Sequentially Compact
* Countably Compact Space is not necessarily Compact
* Pseudocompact Space is not necessarily Countably Compact
* Weakly Countably Compact Space is not necessarily Countably Compact
{{qed}}
Theorem: Sequence of Implications of Local Compactness Properties

Let $P_1$ and $P_2$ be topological compactness properties.
Let:
:$P_1 \implies P_2$
mean:
:If a topological space $T$ satisfies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
{|
|-
| align="center" | Compact ||
| align="center" | $\implies$ ||
| align="center" | Strongly Locally Compact ||
|-
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
|-
| align="center" | Weakly $\sigma$-Locally Compact ||
| align="center" | $\implies$ ||
| align="center" | Weakly Locally Compact ||
| align="center" | $\Longleftarrow$ ||
| align="center" | Locally Compact ||
|-
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | ||
|-
| align="center" | $\sigma$-Compact ||
| align="center" | ||
| align="center" | ||
|-
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | ||
|-
| align="center" | Lindelöf Space ||
| align="center" | ||
| align="center" | ||
|}
If there is no implication indicated, then there exists a counterexample demonstrating that no such implication exists.

Proof:
The relevant justifications are listed as follows:
* Compact Space is Strongly Locally Compact.
* Strongly Locally Compact Space is Weakly Locally Compact.
* Locally Compact Space is Weakly Locally Compact.
* Compact Space is Weakly $\sigma$-Locally Compact.
* A weakly $\sigma$-locally compact is both weakly locally compact and $\sigma$-compact by definition.
* $\sigma$-Compact Space is Lindelöf Space.
Then we have:
* Strongly Locally Compact Space is not necessarily Compact
* Strongly Locally Compact Space is not necessarily Weakly Sigma-Locally Compact
* Strongly Locally Compact Space is not necessarily Sigma-Compact
* Strongly Locally Compact Space is not necessarily Lindelöf
* Weakly Sigma-Locally Compact Space is not necessarily Strongly Locally Compact
* Weakly Sigma-Locally Compact Space is not necessarily Compact
* Weakly Locally Compact Space is not necessarily Strongly Locally Compact
* Weakly Locally Compact Space is not necessarily Weakly Sigma-Locally Compact
* Weakly Locally Compact Space is not necessarily Sigma-Compact
* Weakly Locally Compact Space is not necessarily Lindelöf
* Lindelöf Space is not necessarily Sigma-Compact
* Sigma-Compact Space is not necessarily Weakly Sigma-Locally Compact
{{qed}}
Theorem: Equivalence of Definitions of Integral Dependence

Let $A$ be an extension of a commutative ring with unity $R$.
For $x \in A$, the following are equivalent:
{{begin-eqn}}
{{eqn | n = 1
      | o =
      | r = \) $x$ is integral over $R$ \(
}}
{{eqn | n = 2
      | o =
      | r = \) The $R$-module $R \sqbrk x$ is finitely generated \(
}}
{{eqn | n = 3
      | o =
      | r = \) There exists a subring $B$ of $A$ such that $x \in B$, $R \subseteq B$ and $B$ is a finitely generated $R$-module \(
}}
{{eqn | n = 4
      | o =
      | r = \) There exists a subring $B$ of $A$ such that $x B \subseteq B$ and $B$ is finitely generated over $R$ \(
}}
{{eqn | n = 5
      | o =
      | r = \) There exists a faithful $R \sqbrk x$-module $B$ that is finitely generated as an $R$-module \(
}}
{{end-eqn}}

Proof:
=== $(1) \implies (2)$ ===
By hypothesis, there exist $r_0, \ldots, r_{n - 1} \in R$ such that:
:$x^n + r_{n - 1} x^{n - 1} + \cdots + r_1 x + r_0 = 0$
So the powers $x^k$, $k \ge n$ can be written as an $R$-linear combination of:
:$\set {1, \ldots, x^{n - 1} }$
Therefore this set generates $R \sqbrk x$.
{{qed|lemma}}
Theorem: Transitivity of Finite Generation

Let $\struct { C, +_C, \circ_C}$ be a ring with unity, where $1_C$ denotes the unity of $C$.
Let $\struct { B, +_B, \circ_B}$ be a subring of $C$, such that $1_C \in B$.
Let $\struct { A, +_A, \circ_A}$ be a subring of $B$, such that $1_C \in A$.
Let $B' = \struct { B, +_B, \circ_B}_A$ be a finitely generated $A$-module, where the scalar multiplication $\circ_B$ is defined as the ring product $\circ_B$ in $B$.
Let $C' = \struct { C, +_C, \circ_C}_B$ be a finitely generated $B$-module, where the scalar multiplication $\circ_C$ is defined as the ring product $\circ_C$ in $C$.
Then $C' ' = \struct { C, +_C, \circ_C}_A$ is a finitely generated $A$-module.

Proof:
As $1_C \in B$, and $1_C \in A$, it follows that $A$ and $B$ are rings with unity.
Let $b_1, \ldots, b_n \in B$ generate $B'$ over $A$.
Let $c_1, \ldots, c_m \in C$ generate $C'$ over $B$.
Then for any $x \in C$, there exist $\beta_k \in B$, $k = 1, \ldots, m$ such that:
:$\ds x = \sum_{k \mathop = 1}^m \beta_k \circ_C c_k$
For any $k \in \set {1, \ldots, m}$, there exist $\alpha_{k j} \in A$, $j = 1, \ldots, n$ such that:
:$\ds \beta_k = \sum_{j \mathop = 1}^n \alpha_{k j} \circ_B b_j$
Then:
{{begin-eqn}}
{{eqn | l = x
      | r = \sum_{k \mathop = 1}^m \paren { \sum_{j \mathop = 1}^n \alpha_{k j} \circ_B b_j } \circ_C c_k
      | c = combining the two equations above
}}
{{eqn | r = \sum_{k \mathop = 1}^m \paren { \sum_{j \mathop = 1}^n \paren { \alpha_{k j} \circ_B b_j } \circ_C c_k }
      | c = {{Ring-axiom|D}}
}}
{{eqn | r  = \sum_{k \mathop = 1}^m \sum_{j \mathop = 1}^n \paren { \paren{ \alpha_{k j} \circ_B b_j } \circ_C c_k }
      | c =  {{Ring-axiom|A2}}
}}
{{eqn | r  = \sum_{k \mathop = 1}^m \sum_{j \mathop = 1}^n \paren { \paren{ \alpha_{k j} \circ_C b_j } \circ_C c_k }
      | c =  as $B$ is a subring of $C$, it follows that $\circ_B$ and $\circ_C$ are identical ring products
}}
{{eqn | r  = \sum_{k \mathop = 1}^m \sum_{j \mathop = 1}^n \alpha_{k j}  \circ_C \paren { b_j  \circ_C c_k }
      | c =  {{Ring-axiom|M1}}
}}
{{end-eqn}}
which shows that $x$ is a linear combination of $\set {b_j \circ_C c_k : j = 1, \ldots, n, k = 1, \ldots, m}$, with scalars $\alpha_{k j} \in A$ for all $k \in \set {1, \ldots, m}, j \in \set{1, \ldots, n}$.
Therefore, $\set {b_j \circ_C c_k : j = 1, \ldots, n, k = 1, \ldots, m}$ generates $C' '$ over $A$.
{{qed}}
Category:Commutative Algebra
Category:Algebraic Number Theory
56wg9pv3grruuckmiqmoyjy3s600kt6
Theorem: Integral Closure is Subring

Let $A$ be an extension of a commutative ring with unity $\struct {R, +, \circ}$.
Let $C$ be the integral closure of $R$ in $A$.
Then $C$ is a subring of $A$.

Proof:
For each $r \in R$, we have $r$ is a zero of $a - r = 0$.
{{MissingLinks|zero}}
Hence $R \subseteq C$, and in particular, $C$ is non-empty.
By Subring Test it is sufficient to prove that if $x, y \in C$ then $x + \paren {-y}, x \circ y \in C$.
Let $x, y \in C$.
By Equivalent Definitions of Integral Dependence: $(1) \implies (2)$, $R \sqbrk x$ and $R \sqbrk y$ are finitely generated over $R$.
Because $R \subseteq R \sqbrk x$, $R \sqbrk x \sqbrk y$ is also finitely generated over $R \sqbrk x$.
Therefore, we have finitely generated extensions:
:$R \hookrightarrow R \sqbrk x \hookrightarrow R \sqbrk x \sqbrk y = R \sqbrk {x, y}$
Hence, by Transitivity of Finite Generation, $R \sqbrk {x, y}$ is finitely generated over $R$.
So by Equivalent Definition of Integral Dependence: $(2) \implies (1)$, every element of $R \sqbrk {x, y}$ is integral over $R$.
In particular, $x + \paren {-y}$ and $x \circ y$ are integral over $R$.
{{qed}}
Category:Commutative Algebra
Category:Algebraic Number Theory
a4tpxtnv2uundr184ymd3rw6xox1tln
Theorem: Localization of Ring is Unique

Let $A$ be a commutative ring with unity.
Let $S \subseteq A$ be a multiplicatively closed subset.
Let $\struct {A_S, \iota}$ and $\struct {\tilde A_S, \tilde \iota}$ both satisfy the definition of the localization of $A$ at $S$.
Then there is a canonical isomorphism $\phi: A_S \to \tilde A_S$.

Proof:
By the definition of localization, there exist unique homomorphisms:
:$g : A_S \to \tilde A_S$
:$h : \tilde A_S \to A_S$
such that:
:$h \circ \iota = \tilde \iota$
and:
:$g \circ \tilde \iota = \iota$
Therefore:
:$\tilde \iota = h \circ \iota = h \circ \paren {g \circ \tilde \iota} = \paren {h \circ g} \circ \tilde \iota$
The identity mapping also satisfies this equality (that is, $\tilde \iota = \paren {h \circ g} \circ \tilde \iota$).
Therefore by the uniqueness of $h$ and $g$ we have:
:$h \circ g = I_{\tilde A_S}$
where $I_{\tilde A_S}$ denotes the identity mapping on $\tilde A_S$.
Similarly we have that:
:$g \circ h = I_{A_S}$
where $I_{A_S}$ denotes the identity mapping on $A_S$.
Therefore by Bijection iff Left and Right Inverse $g = \phi$ is a bijective homomorphisms, that is, an isomorphism.
{{qed}}
Category:Localization of Rings
tm0hgp68kzwqwo5aop27b0jfw60iby2
Theorem: Spectrum of Ring is Nonempty

Let $A$ be a non-trivial commutative ring with unity.
Then its prime spectrum is non-empty:
:$\Spec A \ne \O$

Proof:
This is a reformulation of Ring with Unity has Prime Ideal.
{{qed}}
Theorem: Compact Space is Strongly Locally Compact

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

Proof:
{{Recall|Strongly Locally Compact Space|strongly locally compact space}}
{{:Definition:Strongly Locally Compact Space/Definition 1}}
Let $T = \struct {S, \tau}$ be compact.
From Underlying Set of Topological Space is Clopen, $S$ is clopen in $T$.
From Closed Set Equals its Closure, $S = S^-$.
So every point of $S$ is contained in an open set (that is, $S$) whose closure (that is, $S$ again) is compact (as  $T = \struct {S, \tau}$ itself is compact).
That is precisely the definition of a strongly locally compact space.
{{qed}}
Theorem: Compact Space is Weakly Sigma-Locally Compact

Let $T = \struct {S, \tau}$ be a compact space.
Then $T$ is a weakly $\sigma$-locally compact space.

Proof:
{{Recall|Weakly Sigma-Locally Compact Space|weakly $\sigma$-locally compact space}}
{{:Definition:Weakly Sigma-Locally Compact Space}}
Let $T = \struct {S, \tau}$ be a compact space.
We have:
{{begin-itemize}}
{{item||Compact Space is $\sigma$-Compact}}
{{item||Compact Space is Weakly Locally Compact}}
{{end-itemize}}
Hence by definition $T$ is weakly $\sigma$-locally compact space.
{{qed}}
Theorem: Sequence of Implications of Paracompactness Properties

Let $P_1$ and $P_2$ be paracompactness properties.
Let:
:$P_1 \implies P_2$
mean:
:If a topological space $T$ satsifies property $P_1$, then $T$ also satisfies property $P_2$.
Then the following sequence of implications holds:
{|
|-
| align="center" | Fully Normal ||
| align="center" | ||
| align="center" | Compact ||
| align="center" | $\implies$ ||
| align="center" | Countably Compact ||
|-
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
|-
| align="center" | Fully $T_4$ ||
| align="center" | $\implies$ ||
| align="center" | Paracompact ||
| align="center" | $\implies$ ||
| align="center" | Countably Paracompact ||
|-
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
| align="center" | ||
| align="center" | $\Big \Downarrow$ ||
|-
| align="center" | $T_4$ ||
| align="center" | ||
| align="center" | Metacompact ||
| align="center" | $\implies$ ||
| align="center" | Countably Metacompact ||
|}
If there is no implication indicated, then there exists a counterexample demonstrating that no such implication exists.

Proof:
The relevant justifications are listed as follows:
* Compact Space is Countably Compact
* Compact Space is Paracompact
* Fully $T_4$ Space is Paracompact
* Countably Compact Space is Countably Paracompact
* Paracompact Space is Countably Paracompact
* Fully Normal Space is Fully $T_4$ by definition
* Fully $T_4$ Space is $T_4$
* Paracompact Space is Metacompact
* Metacompact Space is Countably Metacompact
* Countably Paracompact Space is Countably Metacompact
{{qed|lemma}}
Then we have:
* Countably Compact Space is not necessarily Compact
* Countably Compact Space is not necessarily Metacompact
* Metacompact Space is not necessarily Paracompact
* Metacompact Space is not necessarily Countably Paracompact
* Countably Paracompact Space is not necessarily Paracompact
* Countably Paracompact Space is not necessarily Metacompact
* Countably Paracompact Space is not necessarily Countably Compact
* Countably Metacompact Space is not necessarily Countably Paracompact
* Countably Metacompact Space is not necessarily Metacompact
* Paracompact Space is not necessarily Compact
* Paracompact Space is not necessarily Countably Compact
* Paracompact Space is not necessarily Fully $T_4$
* Compact Space is not necessarily Fully Normal
* Fully Normal Space is not necessarily Compact
* Fully $T_4$ Space is not necessarily Fully Normal
* $T_4$ Space is not necessarily Fully $T_4$
{{qed}}