nyuuzyou/wikis huggingface dataset:
Characterization of Null Sets of Variation of Complex Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be measurable space.
Let $\mu$ be a complex measure on $\struct {X, \Sigma}$. 
Let $\size \mu$ be the variation of $\mu$. 
Then $A \in \Sigma$ is such that $\map {\size \mu} A = 0$ :
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$.
== Proof ==
Let $A \in \Sigma$.
Let $\map P A$ be the set of finite partitions of $A$ into $\Sigma$-measurable sets.
From the definition of variation, we have:
:$\ds \map {\cmod \mu} A = \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$
Suppose that: 
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$.
Then, for any: 
:$\set {A_1, A_2, \ldots, A_n} \in \map P A$
we have: 
:$\map \mu {A_i} = 0$ for each $i$
so:
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} } = 0$
This gives: 
:$\ds \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} = \set 0$
So, from the definition of supremum, we have: 
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} = 0$
so:
:$\map {\cmod \mu} A = 0$ 
Suppose that:
:$\map {\cmod \mu} A = 0$
Then: 
:$\ds \sup \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} = 0$
Note that since each element of: 
:$\ds \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A}$ 
is non-negative, we must have: 
:$\ds \set {\sum_{j \mathop = 1}^n \cmod {\map \mu {A_j} } : \set {A_1, A_2, \ldots, A_n} \in \map P A} = \set 0$
So, for all:
:$\set {A_1, A_2, \ldots, A_n} \in \map P A$
we have: 
:$\ds \sum_{i \mathop = 1}^n \cmod {\map \mu {A_i} } = 0$
Let $B \subseteq A$ be $\Sigma$-measurable.
From Sigma-Algebra Closed under Set Difference, we have: 
:$A \setminus B$ is $\Sigma$-measurable.
We also have: 
:$A = B \cup \paren {A \setminus B}$ 
so:
:$\set {B, A \setminus B}$ is a finite partition of $A$ into $\Sigma$-measurable sets.
That is: 
:$\set {B, A \setminus B} \in \map P A$
We therefore have: 
:$\cmod {\map \mu B} + \cmod {\map \mu {A \setminus B} } = 0$
Since: 
:$\cmod {\map \mu {A \setminus B} } \ge 0$
and:
:$\cmod {\map \mu B} \ge 0$ 
we must have: 
:$\cmod {\map \mu B} = \cmod {\map \mu {A \setminus B} } = 0$
and in particular: 
:$\map \mu B = 0$
So:
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$.
Category:Variation of Complex Measure
Characterization of Null Sets of Variation of Signed Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Let $\size \mu$ be the variation of $\mu$. 
Then $A \in \Sigma$ is such that $\map {\size \mu} A = 0$ :
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$.
== Proof ==
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then, from the definition of variation: 
:$\size \mu = \mu^+ + \mu^-$
Suppose that $A \in \Sigma$ is such that $\map {\size \mu} A = 0$.
Then: 
:$\map {\mu^+} A + \map {\mu^-} A = 0$
Since $\mu^+ \ge 0$ and $\mu^- \ge 0$ we have: 
:$\map {\mu^+} A = 0 $ and $\map {\mu^-} A = 0$
From Null Sets Closed under Subset, we then have: 
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map {\mu^+} B = 0$
and:
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map {\mu^-} B = 0$
So, for each $\Sigma$-measurable set $B \subseteq A$, we have: 
:$\map {\mu^+} B - \map {\mu^-} B = 0$
From the definition of the Jordan decomposition, we therefore have: 
:$\map \mu B = 0$
for each $\Sigma$-measurable set $B \subseteq A$.
Now suppose that: 
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$.
Let $\tuple {P, N}$ be a Hahn decomposition of $\mu$. 
From the definition of Jordan decomposition, and Uniqueness of Jordan Decomposition, we have: 
:$\map {\mu^+} A = \map \mu {P \cap A}$
and:
:$\map {\mu^+} A = -\map \mu {N \cap A}$
From Intersection is Subset, we have: 
:$P \cap A \subseteq A$
and:
:$N \cap A \subseteq A$
So, we have: 
:$\map \mu {P \cap A} = 0$
and:
:$\map \mu {N \cap A} = 0$
from the assumption.
That is: 
:$\map {\mu^+} A = 0$
and:
:$\map {\mu^-} A = 0$
So:
:$\map {\size \mu} A = \map {\mu^+} A + \map {\mu^-} A = 0$
So, if: 
:for each $\Sigma$-measurable set $B \subseteq A$, we have $\map \mu B = 0$
then:
:$\map {\size \mu} A = 0$
Category:Variation of Signed Measure
Characterization of Open Ball in P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
For any $\epsilon \in \R_{>0}$ and $a \in \Q_p$ let $\map {B_\epsilon} a$ denote the open ball of center $a$ of radius $\epsilon$. 
Let $n \in \Z$.
Let $x, y \in \Q_p$.
:$(1): \quad x \in \map {B_{p^{-n} } } y$
:$(2): \quad \norm{x - y}_p < p^{-n}$
:$(3): \quad \map {B_{p^{-n} } } x = \map {B_{p^{-n} } } y$
:$(4): \quad x - y \in p^{n + 1} \Z_p$
:$(5): \quad x + p^{n + 1} \Z_p = y + p^{n + 1} \Z_p$
== Proof ==
From P-adic Numbers form Non-Archimedean Valued Field:
:$\norm {\,\cdot\,}_p$ is a non-Archimedean norm.
=== Condition $(1)$ iff Condition $(2)$ ===
This follows directly from the definition of a open ball in the $p$-adic numbers.
=== Condition $(1)$ iff Condition $(3)$ ===
By definition, $\map {B_{p^{-n}}} y$ is an open ball in a non-Archimedean norm $\norm {\,\cdot\,}_p$.
From Centers of Open Balls in Non-Archimedean Division Rings:
:$x \in \map {B_{p^{-n}}} y \leadsto \map {B_{p^{-n}}} x = \map {B_{p^{-n}}} y$
From Center is Element of Open Ball in P-adic Numbers:
:$\map {B_{p^{-n}}} x = \map {B_{p^{-n}}} y \leadsto x \in \map {B_{p^{-n}}} x = \map {B_{p^{-n}}} y$
=== Condition $(2)$ iff Condition $(4)$ ===
=== Condition $(3)$ iff Condition $(5)$ ===
From Open Balls of P-adic Number,
:$\map {B_{p^{-n} } } x = x + p^{n + 1} \Z_p$
and 
:$\map {B_{p^{-n} } } y = y + p^{n + 1} \Z_p$
Hence:
:$\map {B_{p^{-n} } } x = \map {B_{p^{-n} } } y$ $x + p^{n + 1} \Z_p = y + p^{n + 1} \Z_p$
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Lemma $3.3.5$
Category:Topology of P-adic Numbers
Characterization of Open Linear Transformation between Normed Vector Spaces

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,}_X}$ and $\struct {Y, \norm {\, \cdot \,}_Y}$ be normed vector spaces over $\GF$. 
Let $T : X \to Y$ be a linear transformation.
Let $B_X^-$ and $B_Y^-$ be the closed unit balls of $X$ and $Y$ respectively. 
:$(1) \quad$ $T$ is an open map
:$(2) \quad$ there exists $\delta > 0$ such that $\delta B_Y^- \subseteq T \sqbrk {B_X^-}$
:$(3) \quad$ there exists $M > 0$ such that for all $y \in Y$ there exists $x \in T^{-1} \sqbrk {\set y}$ such that $\norm x_X \le M \norm y_Y$
== Proof ==
Let $B_X^O$ and $B_Y^O$ be the open unit balls of $X$ and $Y$ respectively.
=== $(1)$ implies $(2)$ ===
Suppose that $T$ is an open map.
Then $T \sqbrk {B_X^O}$ is open in $\struct {Y, \norm {\, \cdot \,}_Y}$.
Hence there exists $\delta > 0$ such that: 
:$\delta B_Y^O \subseteq T \sqbrk {B_X^O}$
We have that: 
:$\ds \frac \delta 2 B_Y^- \subseteq \delta B_Y^O$
We also have that: 
:$T \sqbrk {B_X^O} \subseteq T \sqbrk {B_X^-}$
from Image of Subset under Mapping is Subset of Image.
Hence we obtain: 
:$\ds \frac \delta 2 B_Y^- \subseteq T \sqbrk {B_X^-}$
which is the demand of $(2)$.
=== $(2)$ implies $(1)$ ===
Suppose that there exists $\delta > 0$ such that $\delta B_Y^- \subseteq T \sqbrk {B_X^-}$.
We have: 
From Linear Transformation between Normed Vector Spaces is Open iff Image of Open Unit Ball is Open, we have that $T$ is an open map.
Hence we have $(1)$. 
=== $(2)$ implies $(3)$ ===
Suppose that there exists $\delta > 0$ such that $\delta B_Y^- \subseteq T \sqbrk {B_X^-}$.
If $y = {\mathbf 0}_Y$, then $x = {\mathbf 0}_X$ has: 
:$\norm x_X \le M \norm y_Y$
for any $M > 0$.
It therefore suffices to pick an $M > 0$ that works for $y \in Y \setminus \set { {\mathbf 0}_Y}$.
Let $y \in Y \setminus \set { {\mathbf 0}_Y}$. 
Then: 
:$\ds \frac {\delta y} {\norm y_Y} \in \delta B_Y^-$
so:
:$\ds \frac {\delta y} {\norm y_Y} \in T \sqbrk {B_X^-}$
Then there exists $x \in B_X^-$ such that: 
:$\ds \frac {\delta y} {\norm y_Y} = T x$
so that: 
:$\ds \map T {\frac {\norm y_Y} \delta x} = y$
Setting:
:$\ds x' = \frac {\norm y_Y} \delta x$
we have that $x' \in T^{-1} \sqbrk {\set y}$ with:
:$\ds \norm {x'}_X = \frac {\norm y_Y} \delta \norm x_X \le \frac {\norm y_Y} \delta$
Hence setting: 
:$\ds M = \frac 1 \delta$
we have $(3)$. 
=== $(3)$ implies $(2)$ ===
Suppose that there exists $M > 0$ such that for all $y \in Y$ there exists $x \in T^{-1} \sqbrk {\set y}$ such that $\norm x_X \le M \norm y_Y$.
Let $y \in B_Y^-$.
Then there exists $x \in T^{-1} \sqbrk {\set y}$ such that: 
:$\norm x_X \le M \norm y_Y \le M$
Hence, we have: 
:$B_Y^- \subseteq T \sqbrk {M B_X^-}$
Hence, from Image of Dilation of Set under Linear Transformation is Dilation of Image:
:$\ds \frac 1 M B_Y^- \subseteq T \sqbrk {B_X^-}$
== Sources ==
* : Exercise $2.29$
Category:Open Mappings
Category:Linear Transformations
Category:Normed Vector Spaces
Characterization of Open Set by Open Cover

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $\UU$ be an open cover of $T$.
For each $U \in \UU$, let $\tau_U$ denote the subspace topology on $U$.
Let $W \subseteq S$.
Then $W$ is open in $T$ :
:$\forall U \in \UU: W \cap U$ is open in $\struct{U, \tau_U}$
== Proof ==
=== Necessary Condition ===
This follows immediately from the definition of subspace topology.
=== Sufficient Condition ===
Let:
:$W \cap U$ be open in $\struct{U, \tau_U}$ for each $U \in \UU$
From Open Set in Open Subspace:
:$\forall U \in \UU : W \cap U$ is open in $T$
We have:
By :
:$W$ is open in $T$
Category:Open Sets
Category:Covers
Characterization of Openness in terms of Nets

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $U \subseteq X$. 
Then $U \in \tau$ for each:
:$x \in U$
:directed set $\tuple {\Lambda, \preceq}$
:net $\family {x_\lambda}_{\lambda \in \Lambda}$ converging to $x$
there exists $\lambda \in \Lambda$ with $x_\lambda \in U$.
== Proof ==
=== Necessary Condition ===
Suppose that $U \in \tau$. 
Take $x \in U$, a directed set $\tuple {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ converging to $x$.
By the definition of convergence for net, we have: 
:there exists $\lambda_0 \in \Lambda$ such that for all $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$ we have $x_\lambda \in U$.
In particular, $x_{\lambda_0} \in U$.
=== Sufficient Condition ===
Suppose that for each:
:$x \in U$
:directed set $\tuple {\Lambda, \preceq}$
:net $\family {x_\lambda}_{\lambda \in \Lambda}$ converging to $x$
there exists $\lambda \in \Lambda$ with $x_\lambda \in U$.
 $U \not \in \tau$.
Then $X \setminus U$ is not closed.
From Set is Closed iff Equals Topological Closure, we have $\map \cl {X \setminus U} \ne X \setminus U$. 
Let $x \in \map \cl {X \setminus U} \setminus \paren {X \setminus U}$. 
That is, $x \in \map \cl {X \setminus U} \cap U$. 
From Point in Set Closure iff Limit of Net:
:there exists directed set $\struct {\Lambda, \preceq}$ and a net $\family {x_\lambda}_{\lambda \in \Lambda}$ in $X \setminus U$ converging to $x$.
By hypothesis, since $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$, there exists $\lambda \in \Lambda$ such that $x_\lambda \in U$.
This contradicts that $x_\lambda \in X \setminus U$ for each $\lambda \in \Lambda$. 
Category:Nets (Set Theory)
Characterization of P-adic Unit has Square Root in P-adic Units

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $Z_p^\times$ be the set of $p$-adic units. 
Let $u \in Z_p^\times$ be a $p$-adic unit.
Let $u = c_0 + c_1p + c_2p^2 + \ldots$ be the $p$-adic expansion of $u$.
:
:$(1)\quad \exists x \in \Z_p^\times : x^2 = u$
:$(2)\quad c_0$ is a quadratic residue of $p$
:$(3)\quad \exists y \in \Z_p : y^2 \equiv u \pmod{p\Z_p}$
== Proof ==
From Partial Sum Congruent to P-adic Integer Modulo Power of p:
:$u \equiv c_0 \pmod {p\Z_p}$
=== $(1)$ implies $(2)$ ===
=== $(2)$ implies $(3)$ ===
=== $(3)$ implies $(1)$ ===
== Sources ==
* : $\S 3.4$ Hensel's Lemma $\Q_p$, Theorem $3.4.3$
* : $\S 1.7$ Hensel's Lemma and Congruences: Exercise $36$
Category:P-adic Units
Category:Characterization of P-adic Unit has Square Root in P-adic Units
Characterization of P-adic Unit has Square Root in P-adic Units/Condition 1 implies Condition 2

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $Z_p^\times$ be the set of $p$-adic units. 
Let $u \in Z_p^\times$ be a $p$-adic unit.
Let $u = c_0 + c_1p + c_2p^2 + \ldots$ be the $p$-adic expansion of $u$.
Let there exist $x \in \Z_p^\times$ such that $x^2 = u$.
Then:
:$c_0$ is a quadratic residue of $p$
== Proof ==
From Partial Sum Congruent to P-adic Integer Modulo Power of p:
:$u \equiv c_0 \pmod {p\Z_p}$
Let there exist $x \in \Z_p^\times$ such that $x^2 = u$
Then:
$x^2 \equiv c_0 \pmod {p\Z_p}$
Let $x = x_0 + x_1p + x_2p^2 + x_3p^3 + \ldots$ be the $p$-adic expansion of $x$.
From Partial Sum Congruent to P-adic Integer Modulo Power of p:
:$x \equiv x_0 \pmod {p\Z_p}$
Then:
:$x_0^2 \equiv x^2 \equiv c_0 \pmod {p\Z_p}$
By definition of quadratic residue:
:$c_0$ is a quadratic residue of $p$
Category:Characterization of P-adic Unit has Square Root in P-adic Units
Characterization of P-adic Unit has Square Root in P-adic Units/Condition 2 implies Condition 3

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $Z_p^\times$ be the set of $p$-adic units. 
Let $u \in Z_p^\times$ be a $p$-adic unit.
Let $u = c_0 + c_1p + c_2p^2 + \ldots$ be the $p$-adic expansion of $u$.
Let $c_0$ be a quadratic residue of $p$.
Then:
:$\exists y \in \Z_p : y^2 \equiv u \pmod{p\Z_p}$
== Proof ==
From Partial Sum Congruent to P-adic Integer Modulo Power of p:
:$u \equiv c_0 \pmod {p\Z_p}$
Let $c_0$ be a quadratic residue of $p$.
By definition of quadratic residue:
:$\exists x_0 \in \Z : x_0^2 \equiv c_0 \pmod {p\Z_p}$
Then:
:$x_0^2 \equiv u \pmod {p\Z_p}$
Category:Characterization of P-adic Unit has Square Root in P-adic Units
Characterization of P-adic Unit has Square Root in P-adic Units/Condition 3 implies Condition 1

== Theorem ==
Let $\Z_p$ be the $p$-adic integers for some prime $p \ne 2$.
Let $Z_p^\times$ be the set of $p$-adic units. 
Let $u \in Z_p^\times$ be a $p$-adic unit.
Let there exist $y \in \Z_p$ such that $y^2 \equiv u \pmod {p \Z_p}$.
Then:
:$\exists x \in \Z_p^\times : x^2 = u$
== Proof ==
Let there exist $y \in \Z_p$ such that $y^2 \equiv u \pmod {p\Z_p}$
Let $y = y_0 + y_1 p + y_2 p^2 + y_3 p^3 + \ldots$ be the $p$-adic expansion of $y$.
By definition of $p$-adic expansion:
:$y_0 \in {0, 1, \ldots, p - 1}$
From P-adic Expansion of P-adic Unit:
:$y_0 \ne 0$.
Hence:
:$y_0 \in {1, 2, \ldots, p - 1}$
It follows that:
:$p \nmid y_0$
From Partial Sum Congruent to P-adic Integer Modulo Power of p:
:$y \equiv y_0 \pmod {p \Z_p}$
Then:
:$y_0^2 \equiv y^2 \equiv u \pmod {p \Z_p}$
Let $F \in \Z \sqbrk X$ be the polynomial:
:$\map F X = X^2 - u$
By definition of formal derivative the formal derivative of $F$ is:
:$\map {F'} X = 2X$
We have:
:$\map F {y_0} \equiv 0 \pmod p$
and
:$\map {F'} {y_0} = 2 y_0$
By hypothesis:
:$p \nmid 2$
From the contrapositive statement of Euclid's Lemma for Prime Divisors:
:$p \nmid 2 y_0$
Hence:
:$\map {F'} {y_0} = 2 y_0 \not \equiv 0 \pmod p$ 
From Congruence Modulo Equivalence for Integers in P-adic Integers:
:$\map F {y_0} \equiv 0 \pmod {p \Z}$
and
:$\map {F'} {y_0} \not \equiv 0 \pmod {p \Z}$
From Hensel's Lemma for P-adic Integers:
:$\exists x \in \Z_p : \map F x = 0$
That is:
:$\exists x \in \Z_p : x^2 = u$
It remains to show that $x \in \Z_p^\times$.
We have:
Category:Characterization of P-adic Unit has Square Root in P-adic Units
Characterization of P-adic Valuation on Integers

== Theorem ==
Let $p \in \N$ be a prime number.
Let $\nu_p^\Z: \Z \to \N \cup \set {+\infty}$ be the $p$-adic valuation on $\Z$.
Let $n \in \Z_{\ne 0}$.
Then $\map {\nu_p^\Z} n$ is the unique $r \in \N$ such that:
:$\exists k \in \Z: n = p^r k : p \nmid k$
== Proof ==
=== Uniqueness of $r$ ===
Let $r, r'$ be such that:
:$\exists k \in \Z: n = p^r k : p \nmid k$
and:
:$\exists k' \in \Z: n = p^{r'} k' : p \nmid k'$
, suppose $r \ge r'$.
By subtracting the above equations:
:$0 = p^r k - p^{r'} k' = p^{r'} \paren {p^{r - r'} k - k'} = 0$
Therefore:
:$p ^{r - r'} k = k'$
Thus because $p \nmid k'$:
:$r - r' = 0$
Let $n \in \Z_{\ne 0}$.
In the following, we shall find a $k\in\Z$ such that:
:$n = p^{\map {\nu_p^\Z} n} k$
:$p \nmid k$
By definition of $p$-adic valuation:
:$\map {\nu_p^\Z} n = \sup S$
where
:$S := \set {v \in \N: p^v \divides n}$
$S$ is a non-empty finite set, since:
:$0 \in S \subseteq \closedint 0 {\floor {\log _p n} }$
Hence, the supremum of $S$ is adopted, that is:
:$\map {\nu_p^\Z} n \in S$
:$\forall v > \map {\nu_p^\Z} n : v \notin S$
In particular:
:$p^{\map {\nu_p^\Z} n} \divides n$
:$p^{\map {\nu_p^\Z} n + 1} \nmid n$
Therefore, let:
:$k := \dfrac n {p^{\map {\nu_p^\Z} n} }$
so that:
:$n = p^{\map {\nu_p^\Z} n} k$
Finally, $p \nmid k$, as $p^{\map {\nu_p^\Z} n + 1} \nmid n$.
Category:P-adic Valuations
Characterization of Paracompactness in T3 Space

== Theorem ==
Let $T = \struct {X, \tau}$ be a $T_3$ space.
:$(1): \quad T$ is paracompact
:$(2): \quad$ every open cover of $T$ has a locally finite refinement
:$(3): \quad$ every open cover of $T$ has a closed locally finite refinement
:$(4): \quad$ every open cover of $T$ is even
:$(5): \quad$ every open cover of $T$ has an open $\sigma$-discrete refinement
:$(6): \quad$ every open cover of $T$ has an open $\sigma$-locally finite refinement
== Proof ==
=== Statement $(1)$ implies Statement $(2)$ ===
=== Statement $(1)$ implies Statement $(6)$ ===
=== Statement $(2)$ implies Statement $(3)$ ===
=== Statement $(3)$ implies Statement $(1)$ ===
=== Statement $(3)$ implies Statement $(4)$ ===
=== Statement $(4)$ implies Statement $(5)$ ===
=== Statement $(5)$ implies Statement $(6)$ ===
=== Statement $(6)$ implies Statement $(2)$ ===
== Sources ==
* : Chapter $5$: Compact Spaces: $\S$Paracompactness: Theorem $28$
* : Chapter $6$: Compactness: $\S20$: Paracompactness: Theorem $20.7$
Category:Characterization of Paracompactness in T3 Space
Category:T3 Spaces
Category:Paracompact Spaces
Characterization of Paracompactness in T3 Space/Lemma 1

== Theorem ==
Let $T = \struct{X, \tau}$ be a $T_3$ space.
Let $\UU$ be an open cover of $T$.
Let:
:$\VV = \set{V \in \tau : \exists U \in \UU : V^- \subseteq U}$ 
where $V^-$ denotes the closure of $V$ in $T$.
Then:
:$\VV$ is an open cover of $T$
== Proof ==
Let $x \in S$.
By definition of open cover:
:$\exists U \in \UU : x \in U$
From Characterization of T3 Space:
:$\exists V \in \tau : x \in V : V^- \subseteq U$
Hence:
:$V \in \VV$
Since $x$ was arbitrary, $\VV$ is an open cover by definition.
Category:Characterization of Paracompactness in T3 Space
Characterization of Paracompactness in T3 Space/Lemma 11

== Theorem ==
Let $X$ be a set.
Let $\AA$ and $\VV$ be sets of subsets of $X$.
For each $V \in \VV$, let:
:$V^* = X \setminus \ds \bigcup \set{A \in \AA | A \cap V = \O}$
Let:
:$\VV^* = \set{V^* : V \in \VV}$.
Then:
:$\forall A \in \AA, V^* \in \VV^* : A \cap V^* \ne \O \implies A \cap V \ne \O$
== Proof ==
We prove the contrapositive statement:
:$\forall A \in \AA, V \in \VV^* : A \cap V = \O \implies A \cap V^* = \O$
Let $B \in \AA, V^* \in \VV^* : B \cap V = \O$.
Hence:
:$B \in \set{A \in \AA : A \cap V = \O }$
From Set is Subset of Union:
:$B \subseteq \ds \bigcup \set{A \in \AA : A \cap V = \O }$
We have:
From Subset of Set Difference iff Disjoint Set:
:$V^* \cap B = \O$
Since $B$ and $V^*$ were arbitrary:
:$\forall A \in \AA, V^* \in \VV^* : A \cap V = \O \implies A \cap V^* = \O$
The result follows from Rule of Transposition.
Category:Characterization of Paracompactness in T3 Space
Characterization of Paracompactness in T3 Space/Lemma 12

== Theorem ==
Let $T = \struct{X, \tau}$ be a topological space.
Let $\UU$ be an open cover of $T$.
Let $\VV$ be a closed locally finite refinement of $\UU$.
For all $x \in X$, let:
:$W_x \in \tau: x \in W_x$ and $\set{V \in \VV : V \cap W \ne \O}$ be finite
Let $\WW = \set{W_x : x \in X}$ be an open cover of $T$.
Let $\AA$ be a closed locally finite refinement of $\WW$.
For each $V \in \VV$, let:
:$V^* = X \setminus \ds \bigcup \set{A \in \AA | A \cap V = \O}$
Let $\VV^* = \set{V^* : V \in \VV}$.
For each $V \in \VV$, let:
:$U_V \in \UU : V \subseteq U_V$
Let:
:$\UU^* = \set{V^* \cap U_V : V \in \VV}$
Then:
:$\forall A \in \AA : \set{U^* \in \UU^* : U^* \cap A \ne \O}$ is finite
== Proof ==
===== Lemma 4 =====
===== Lemma 11 =====
Let $A \in \AA$.
From Lemma 4:
:$\set{V \in \VV : V \cap A_0 \ne \O}$ is finite
Consider the surjection:
:$f: \set{V \in \VV : V \cap A \ne \O} \to \set{V^* \cap U_V : V \in \VV : V \cap A \ne \O}$ defined by:
::$\forall V \in \VV: \map f V = V^* \cap U_V$
From Cardinality of Codomain of Surjection:
:$\set{V^* \cap U_V : V \in \VV : V \cap A \ne \O}$ is finite
From Lemma 11:
:$\set{V^* \cap U_V : V \in \VV : V^* \cap A \ne \O} \subseteq \set{V^* \cap U_V : V \in \VV : V \cap A \ne \O}$
From Subsets of Disjoint Sets are Disjoint:
:$\set{V^* \cap U_V : V \in \VV : V^* \cap U_V \cap A \ne \O} \subseteq \set{V^* \cap U_V : V \in \VV : V^* \cap A \ne \O}$
From Subset of Finite Set is Finite:
:$\set{V^* \cap U_V : V \in \VV : V^* \cap U_V \cap A \ne \O}$ is finite
That is:
:$\set{U^* \in \UU^* : U^* \cap A \ne \O}$ is finite
Since $A$ was arbitrary, it follows that:
:$\forall A \in \AA : \set{U^* \in \UU^* : U^* \cap A \ne \O}$ is finite
Category:Characterization of Paracompactness in T3 Space
Characterization of Paracompactness in T3 Space/Lemma 13

== Theorem ==
Let $T = \struct{X, \tau}$ be a topological space.
Let every open cover of $T$ be even.
Let $\tau_{X \times X}$ denote the product topology on the cartesian product $X \times X$.
Let $V$ be a neighborhood of the diagonal $\Delta_X$ of $X \times X$ in the product space $\struct {X \times X, \tau_{X \times X} }$.
Then:
:there exists a sequence $\sequence{V_n}_{n \in \N}$ of neighborhoods of the diagonal $\Delta_X$ in $\struct {X \times X, \tau_{X \times X} }$:
::$V_0 = V$
::$\forall n \in \N_{> 0} : V_n$ is symmetric as a relation on $X \times X$
::$\forall n \in \N_{> 0}$ the composite relation $V_n \circ V_n$ is a subset of $V_{n - 1}$, that is, $V_n \circ V_n \subseteq V_{n - 1}$
== Proof ==
==== Lemma 20 ====
The sequence $\sequence{V_n}_{n \in \N}$ is now constructed using the Principle of Recursive Definition and Zermelo's Well-Ordering Theorem.
Let:
:$\NN = \leftset{U \subseteq X \times X : U }$ is a neighborhood of the diagonal $\Delta_X$ in $\struct {X \times X, \tau_{X \times X} } \rightset{}$
From Zermelo's Well-Ordering Theorem, let:
:$\preccurlyeq$ be a well-ordering on $\NN$
Let $g: \NN \to \NN$ be the mapping defined by:
:$\forall U \in \NN : \map g U = \min \leftset{ V \in \NN : V \circ V \subseteq U}$ and $V$ is symmetric $\rightset{}$
From Lemma 20:
:$\forall U \in \NN : \leftset{ V \in \NN : V \circ V \subseteq U}$ and $V$ is symmetric $\rightset{} \ne \O$
By definition of well-ordering:
:$g$ is well-defined
From Principle of Recursive Definition, there exists exactly one sequence $\sequence{V_n}_{n \in \N}$ of neighborhoods of the diagonal $\Delta_X$ in $\struct {X \times X, \tau_{X \times X} }$:
:$V_0 = V$
:$\forall n \in \N_{> 0} : V_n$ is symmetric as a relation on $X \times X$
:$\forall n \in \N_{> 0}$ the composite relation $V_n \circ V_n$ is a subset of $V_{n - 1}$, that is, $V_n \circ V_n \subseteq V_{n - 1}$
Category:Characterization of Paracompactness in T3 Space