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\begin{document}

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\section*{shrinking lemma}

\hypertarget{context}{}\subsubsection*{{Context}}\label{context}

\hypertarget{topology}{}\paragraph*{{Topology}}\label{topology}

[[!include topology - contents]]

\hypertarget{contents}{}\section*{{Contents}}\label{contents}

\noindent\hyperlink{idea}{Idea}\dotfill \pageref*{idea} \linebreak
\noindent\hyperlink{statement}{Statement}\dotfill \pageref*{statement} \linebreak
\noindent\hyperlink{references}{References}\dotfill \pageref*{references} \linebreak


\hypertarget{idea}{}\subsection*{{Idea}}\label{idea}

In [[topology]], the ``shrinking lemma'' (lemma \ref{PatchesOfOpenCoverOfNormalSpaceMayBeMadeSmallerSoThatTheirClosuresAreContained} below) states that on a [[normal topological space]] the patches of every [[locally finite cover]] may be replaced by smaller patches which still cover the space, but such that their [[topological closures]] are contained in the original patches.

If there is more than a [[countable set]] of elements in the original cover, then this conclusion requires [[excluded middle]] and [[Zorn's lemma]], hence the [[axiom of choice]].

The shrinking lemma is needed in the proof that [[ paracompact Hausdorff spaces equivalently admit subordinate partitions of unity]].

\hypertarget{statement}{}\subsection*{{Statement}}\label{statement}

\begin{lemma}
\label{PatchesOfOpenCoverOfNormalSpaceMayBeMadeSmallerSoThatTheirClosuresAreContained}\hypertarget{PatchesOfOpenCoverOfNormalSpaceMayBeMadeSmallerSoThatTheirClosuresAreContained}{}
\textbf{(shrinking lemma for locally finite covers)}

Assuming the [[axiom of choice]] then:

Let $X$ be a [[topological space]] which is [[normal topological space|normal]] and let $\{U_i \subset X\}_{i \in I}$ be a [[locally finite cover|locally finite]] [[open cover]].

Then there exists another open cover $\{V_i \subset X\}_{i \in I}$ such that the [[topological closure]] $Cl(V_i)$ of its elements is contained in the original patches:

\begin{displaymath}
\underset{i \in I}{\forall}
  \left(
     V_i \subset Cl(V_i) \subset U_i
  \right)
  \,.
\end{displaymath}
\end{lemma}
We now \textbf{prove} this in increasing generality, first for binary open covers (lemma \ref{ShrinkingLemmaForBinaryCover} below), then for finite covers (lemma \ref{ShrinkinglemmaForFiniteCovers}), then for locally finite countable covers (lemma \ref{ShrinkingLemmaForLocallyFiniteCountableCovers}), and finally for general locally finite covers (lemma \ref{PatchesOfOpenCoverOfNormalSpaceMayBeMadeSmallerSoThatTheirClosuresAreContained}, proof \hyperlink{PatchesOfOpenCoverOfNormalSpaceMayBeMadeSmallerSoThatTheirClosuresAreContained}{below}). It is only the last statement that needs the [[axiom of choice]].

\begin{lemma}
\label{ShrinkingLemmaForBinaryCover}\hypertarget{ShrinkingLemmaForBinaryCover}{}
\textbf{(shrinking lemma for binary covers)}

Let $(X,\tau)$ be a [[normal topological space]] and let $\{U_i \subset X\}_{i \in \{1,2\}}$ an [[open cover]] by two [[open subsets]].

Then there exists an open set $V_1 \subset X$ whose [[topological closure]] is contained in $U_1$

\begin{displaymath}
V_1 \subset Cl(V_1) \subset U_1
\end{displaymath}
and such that $\{V_1,U_2\}$ is still an open cover of $X$.

\end{lemma}
\begin{proof}
Since $U_1 \cup U_2 = X$ it follows (by [[de Morgan's law]]) that their [[complements]] $X \backslash U_i$ are [[disjoint subset|disjoint]] [[closed subsets]]. Hence by normality of $(X,\tau)$ there exist disjoint open subsets

\begin{displaymath}
V_1 \supset X \backslash U_2
  \phantom{AAA}
  V_2 \supset X \backslash U_1
  \,.
\end{displaymath}
By their disjointness, we have the following inclusions:

\begin{displaymath}
V_1 \subset X \backslash V_2 \subset U_1
  \,.
\end{displaymath}
In particular, since $X \backslash V_2$ is closed, this means that $Cl(V_1) \subset X \backslash V_2$.

Hence it only remains to observe that $V_1 \cup U_2 = X$, by definition of $V_1$.

\end{proof}
\begin{lemma}
\label{ShrinkinglemmaForFiniteCovers}\hypertarget{ShrinkinglemmaForFiniteCovers}{}
\textbf{(shrinking lemma for finite covers)}

Let $(X,\tau)$ be a [[normal topological space]], and let $\{U_i \subset X\}_{i \in \{1, \cdots, n\}}$ be an [[open cover]] with a [[finite number]] $n \in \mathbb{N}$ of patches. Then there exists another open cover $\{V_i \subset X\}_{i \in I}$ such that $Cl(V_i) \subset U_i$ for all $i \in I$.

\end{lemma}
\begin{proof}
By [[induction]] using lemma \ref{ShrinkingLemmaForBinaryCover}.

To begin with, consider $\{ U_1, \underoverset{i = 2}{n}{\cup} U_i\}$. This is a binary open cover, and hence lemma \ref{ShrinkingLemmaForBinaryCover} gives an open subset $V_1 \subset X$ with $V_1 \subset Cl(V_1) \subset U_1$ such that $\{V_1, \underoverset{i = 2}{n}{\cup} U_i\}$ is still an open cover, and accordingly so is

\begin{displaymath}
\{ V_1  \} \cup \left\{ U_i \right\}_{i \in \{2, \cdots, n\}}
  \,.
\end{displaymath}
Similarly we next find an open subset $V_2 \subset X$ with $V_2 \subset Cl(V_2) \subset U_2$ and such that

\begin{displaymath}
\{ V_1, ,V_2  \} \cup \left\{ U_i \right\}_{i \in \{3, \cdots, n\}}
\end{displaymath}
is an open cover. After $n$ such steps we are left with an open cover $\{V_i \subset X\}_{i \in \{1, \cdots, n\}}$ as required.

\end{proof}
\begin{remark}
\label{}\hypertarget{}{}
Beware that the [[induction]] in lemma \ref{ShrinkinglemmaForFiniteCovers} does \emph{not} give the statement for infinite [[countable covers]]. The issue is that it is not guaranteed that $\underset{i \in \mathbb{N}}{\cup} V_i$ is a cover.

And in fact, assuming the [[axiom of choice]], then there exists a counter-example of a countable cover on a normal spaces for which the shrinking lemma fails (a [[Dowker space]] due to \hyperlink{Beslagic85}{Beslagic 85}).

\end{remark}
This issue is evaded if we consider [[locally finite covers]]:

\begin{lemma}
\label{ShrinkingLemmaForLocallyFiniteCountableCovers}\hypertarget{ShrinkingLemmaForLocallyFiniteCountableCovers}{}
\textbf{(shrinking lemma for locally finite countable covers)}

Let $(X,\tau)$ be a [[normal topological space]] and $\{U_i \subset X\}_{i \in \mathbb{N}}$ a [[locally finite cover|locally finite]] [[countable cover]]. Then there exists [[open subsets]] $V_i \subset X$ for $i \in \mathbb{N}$ such that $V_i \subset Cl(V_i) \subset U_i$ and such that $\{V_i \subset X\}_{i \in \mathbb{N}}$ is still a cover.

\end{lemma}
\begin{proof}
As in the proof of lemma \ref{ShrinkinglemmaForFiniteCovers}, there exist $V_i$ for $i \in \mathbb{N}$ such that $V_i \subset Cl(V_i) \subset U_i$ and such that for every finite number, hence every $n \in \mathbb{N}$, then

\begin{displaymath}
\underoverset{i = 0}{n}{\cup} V_i
  \;=\; 
  \underoverset{i = 0}{n}{\cup} U_i
  \,.
\end{displaymath}
Now the extra assumption that $\{U_i \subset X\}_{i \in I}$ is [[locally finite cover|locally finite]] implies that every $x \in X$ is contained in only finitely many of the $U_i$, hence that for every $x \in X$ there exists $n_x \in \mathbb{N}$ such that

\begin{displaymath}
x \in \underoverset{i = 0}{n_x}{\cup} U_i
  \,.
\end{displaymath}
This implies that for every $x$ then

\begin{displaymath}
x \in \underoverset{i = 0}{n_x}{\cup} V_i 
  \subset \underset{i \in \mathbb{N}}{\cup} V_i
\end{displaymath}
hence that $\{V_i \subset X\}_{i \in \mathbb{N}}$ is indeed a cover of $X$.

\end{proof}
We now invoke [[Zorn's lemma]] to generalize the shrinking lemma for finitely many patches (lemma \ref{ShrinkinglemmaForFiniteCovers}) to arbitrary sets of patches:

\begin{proof}
of the general shrinking lemma \ref{PatchesOfOpenCoverOfNormalSpaceMayBeMadeSmallerSoThatTheirClosuresAreContained}

Let $\{U_i \subset X\}_{i \in I}$ be the given locally finite cover of the normal space $(X,\tau)$. Consider the set $S$ of [[pairs]] $(J, \mathcal{V})$ consisting of

\begin{enumerate}%
\item a [[subset]] $J \subset I$;


\item an $I$-indexed set of open subsets $\mathcal{V} = \{V_i \subset X\}_{i \in I}$



\end{enumerate}
with the property that

\begin{enumerate}%
\item $(i \in J \subset I) \Rightarrow ( Cl(V_i) \subset U_i )$;


\item $(i \in I \backslash J) \Rightarrow ( V_i = U_i )$.


\item $\{V_i \subset X\}_{i \in I}$ is an open cover of $X$.



\end{enumerate}
Equip the set $S$ with a [[partial order]] by setting

\begin{displaymath}
\left(
    (J_1, \mathcal{V})
    \leq
    (J_2, \mathcal{W})
  \right)
    \Leftrightarrow
  \left(
    \left(
      J_1 \subset J_2
    \right)
    \,\text{and}\,
    \left(
      \underset{i \in J_1}{\forall}
      \left(
         V_i = W_i
      \right)
    \right)
  \right)  
  \,.
\end{displaymath}
By definition, an element of $S$ with $J = I$ is an open cover of the required form.

We claim now that a [[maximal element]] $(J, \mathcal{V})$ of $(S,\leq)$ has $J = I$.

For assume on the contrary that there were $i \in I \backslash J$. Then we could apply the construction in lemma \ref{ShrinkingLemmaForBinaryCover} to replace that single $V_i$ with a smaller open subset $V'_i$ to obtain $\mathcal{V}'$ such that $Cl(V'_i) \subset V_i$ and such that $\mathcal{V}'$ is still an open cover. But that would mean that $(J,\mathcal{V}) \lt (J \cup \{i\}, \mathcal{V}')$, contradicting the assumption that $(J,\mathcal{V})$ is maximal. This [[proof by contradiction|proves by contradiction]] that a maximal element of $(S,\leq)$ has $J = I$ and hence is an open cover as required.

We are reduced now to showing that a maximal element of $(S,\leq)$ exists. To achieve this we invoke [[Zorn's lemma]]. Hence we have to check that every [[chain]] in $(S,\leq)$, hence every [[total order|totally ordered]] [[subset]] has an [[upper bound]].

So let $T \subset S$ be a [[total order|totally ordered]] subset. Consider the union of all the index sets appearing in the pairs in this subset:

\begin{displaymath}
K
    \;\coloneqq\;
  \underset{(J,\mathcal{V}) \in T  }{\cup} J
  \,.
\end{displaymath}
Now define open subsets $W_i$ for $i \in K$ picking any $(J,\mathcal{V})$ in $T$ with $i \in J$ and setting

\begin{displaymath}
W_i \coloneqq V_i \phantom{AAA} i \in K
  \,.
\end{displaymath}
This is independent of the choice of $(J,\mathcal{V})$, hence well defined, by the assumption that $(T,\leq)$ is totally ordered.

Moreover, for $i \in I\backslash K$ define

\begin{displaymath}
W_i \coloneqq U_i  \phantom{AAA} i \in I \backslash K
  \,.
\end{displaymath}
We claim now that $\{W_i \subset X\}_{i \in I}$ thus defined is a cover of $X$. Take an arbitrary point $x \in X$. If $x \in U_i$ for some $i \notin K$, we have $U_i = W_i$ and therefore $x$ is in $\underset{i \in I}{\cup} W_i$. Otherwise, combining with the assumption that $\{U_i \subset X\}_{i \in I}$ is locally finite, the set $J_x$ of indices $i \in I$ such that $x \in U_i$ is finite and $J_x \subset K$. Since $(T,\leq)$ is a total order, it must contain an element $(J, \mathcal{V})$ such that $J_x \subset J$. And since that $\mathcal{V}$ is a cover and $x$ cannot belong to any $U_i$ with $i$ outside of $J_x$, it must be that $x \in \underset{i \in J_x}{\cup} V_i \subset \underset{i \in J}{\cup} V_i$, and hence $x$ is in $\underset{i \in I}{\cup} W_i$.

This shows that $(K,\mathcal{W})$ is indeed an element of $S$. It is clear by construction that it is an upper bound for $(T ,\leq )$. Hence we have shown that every [[chain]] in $(S,\leq)$ has an upper bound, and so Zorn's lemma implies the claim.

\end{proof}
\hypertarget{references}{}\subsection*{{References}}\label{references}

The above account follows

\begin{itemize}%
\item [[Matt Rosenzweig]], \emph{\href{https://matthewhr.wordpress.com/2014/06/11/general-shrinking-lemma-for-normal-spaces/}{General shrinking lemma for normal spaces}}

\end{itemize}
The example (a [[Dowker space]]) of a normal space with a (not locally-finite) countable cover to which the shrinking lemma does not apply is given in

\begin{itemize}%
\item Amer Beslagic, \emph{A Dowker product}, Transactions of the AMS, vol 292, number 2 (1985) (\href{http://www.ams.org/journals/tran/1985-292-02/S0002-9947-1985-0808735-X/S0002-9947-1985-0808735-X.pdf}{pdf})

\end{itemize}


\end{document}