Characterization of Paracompactness in T3 Space/Lemma 14

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Theorem

Let $X$ be a set.


Let $X \times X$ denote the cartesian product of $X$ with itself.


Let $\sequence{V_n}_{n \in \N}$ be a sequence of subsets of $X \times X$ containing the diagonal $\Delta_X$ of $X \times X$:

$\forall n \in \N_{> 0}$ the composite relation $V_n \circ V_n$ is a subset of $V_{n - 1}$, that is, $V_n \circ V_n \subseteq V_{n - 1}$


For all $n \in \N_{> 0}$, let:

$U_n = V_n \circ V_{n - 1}, \circ \cdots \circ V_1$


Then:

$\forall n \in \N_{>0}: U_n \subseteq V_0$

Proof

Proof: $V_n \subseteq V_{n - 1}$

We have:

\(\ds \forall n \in \N_{> 0}: \, \) \(\ds \tuple{x, y} \in V_n\) \(\leadsto\) \(\ds \tuple{x, y}, \tuple{y, y} \in V_n\) as $\Delta_X \subseteq V_n$
\(\ds \) \(\leadsto\) \(\ds \tuple{x, y} \in V_n \circ V_n\) Definition of Composite Relation
\(\ds \leadsto \ \ \) \(\ds \forall n \in \N_{> 0}: \, \) \(\ds V_n\) \(\subseteq\) \(\ds V_n \circ V_n\) Definition of Subset
\(\ds \forall n \in \N_{> 0}: \, \) \(\ds V_n \circ V_n\) \(\subseteq\) \(\ds V_{n - 1}\) by hypothesis
\(\text {(1)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \forall n \in \N_{> 0}: \, \) \(\ds V_n\) \(\subseteq\) \(\ds V_{n - 1}\) From Subset Relation is Transitive

$\Box$


Proof: $U_n \subseteq V_{n - 1} \circ U_{n - 1}$

We have:

\(\ds \forall n \in \N_{> 0}: \, \) \(\ds V_n\) \(\subseteq\) \(\ds V_{n - 1}\) $(1)$ above
\(\ds \leadsto \ \ \) \(\ds \forall n \in \N_{> 0}: \, \) \(\ds V_n \circ U_{n - 1}\) \(\subseteq\) \(\ds V_{n - 1} \circ U_{n - 1}\) Composition of Relations Preserves Subsets
\(\ds \forall n > 1: \, \) \(\ds U_n\) \(=\) \(\ds V_n \circ U_{n - 1}\) Definition of $U_n$
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \forall n \in \N_{> 0}: \, \) \(\ds U_n\) \(\subseteq\) \(\ds V_{n - 1} \circ U_{n - 1}\) Subset Relation is Transitive

$\Box$


Proof: $V_n \circ U_n \subseteq V_0$

Proceeds by induction on $n$.

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$V_n \circ U_n \subseteq V_0$


Basis for the Induction

$\map P 1$ is the case:

\(\ds V_1 \circ U_1\) \(=\) \(\ds V_1 \circ V_1\) Definition of $U_1$
\(\ds \) \(\subseteq\) \(\ds V_0\) by hypothesis

and $\map P 1$ is seen to hold.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P n$ is true, where $n > 0$, then it logically follows that $\map P {n + 1}$ is true.


So this is our induction hypothesis:

$V_n \circ U_n \subseteq V_0$


Then we need to show:

$V_{n + 1} \circ U_{n + 1} \subseteq V_0$


Induction Step

This is our induction step:

\(\ds V_{n + 1} \circ U_{n + 1}\) \(=\) \(\ds V_{n + 1} \circ \paren {V_{n + 1} \circ U_n}\)
\(\ds \) \(=\) \(\ds \paren{V_{n + 1} \circ V_{n + 1} } \circ U_n\) Composition of Relations is Associative
\(\ds \) \(\subseteq\) \(\ds V_n \circ U_n\) by hypothesis and Composition of Relations Preserves Subsets
\(\ds \) \(\subseteq\) \(\ds V_0\) Induction Hypothesis


So $\map P n \implies \map P {n + 1}$ and by the Principle of Mathematical Induction:

$(3):\quad\forall n \in \N_{> 0} : V_n \circ U_n \subseteq V_0$

$\Box$


Proof: $U_n \subseteq V_0$

We have:

\(\ds \forall n \in \N_{> 0}: \, \) \(\ds U_{n + 1}\) \(\subseteq\) \(\ds V_n \circ U_n\) $(2)$ above
\(\ds \forall n \in \N_{> 0}: \, \) \(\ds V_n \circ U_n\) \(\subseteq\) \(\ds V_0\) $(3)$ above
\(\ds \leadsto \ \ \) \(\ds \forall n \in \N_{> 1}: \, \) \(\ds U_n\) \(\subseteq\) \(\ds V_0\) Subset Relation is Transitive
\(\ds U_1\) \(=\) \(\ds V_1\) Definition of $U_1$
\(\ds \) \(\subseteq\) \(\ds V_0\) $(1)$ above
\(\ds \leadsto \ \ \) \(\ds \forall n \in \N_{> 0}: \, \) \(\ds U_n\) \(\subseteq\) \(\ds V_0\)

$\blacksquare$

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