Proof: Every point-finite open cover in a normal space has an open shrinkage

Question

I try to unterstand the proof of the statement:

Each point-finite open cover $\mathcal{U}$ of a normal topological space $(X,\mathcal{T})$ has an open refinement $\{V_{U}\mid U\in\mathcal{U}\}$ such that $\overline{V_{U}}\subset U$ for every $U\in\mathcal{U}$.

I have found the following proof in the book "Geometric Aspects of General Topology" by Katsuro Sakai, page 51/52:

I have two questions about this proof:

(1) Where exactly in the proof does the author use that our topological space is normal?

(2) Is the map $\varphi$ in his proof well-defined? Because he define $\varphi$ as $$\varphi(U)=\bigcap_{\psi\in\Psi}\psi(U)$$ But $\varphi$ should be a function with values in $\mathcal{T}$ and the intersection of arbitrary many open sets is in general not open.....So why is $\varphi(U)$ open?

Thanks in advance!

Answer 1

Let me answer your second question first. The key is that $\varPsi$ is linearly ordered by $\le$. Suppose that $U\in\mathscr{U}$ and $\psi_0,\psi_1\in\varPsi$ are such that $\psi_0(U)\ne U\ne\psi_1(U)$. Without loss of generality assume that $\psi_0\le\psi_1$. Then the definition of $\le$ ensures that $\psi_0(U)=\psi_1(U)$, and it follows that $\psi(U)\in\{U,\psi_0(U)\}$ for all $\psi\in\varPsi$. In particular, $\bigcap_{\psi\in\varPsi}\psi(U)$ is actually an intersection of at most two different open sets.

Now for your first question. The proof is incomplete precisely because the author did not use normality: more argument is required in order to justify the assertion that if $\varphi_0$ is a maximal element of $\varPhi$, then $\{\varphi_0(U):U\in\mathscr{U}\}$ is the desired refinement of $\mathscr{U}$, and that argument uses the normality of $X$.

Suppose that $X$ is not normal, and let $H$ and $K$ in $X$ that cannot be separated by disjoint open sets. Let $U_H=X\setminus H$ and $U_K=X\setminus K$, and let $\mathscr{U}=\{U_H,U_K\}$. Suppose that $\{V_H,V_K\}$ is an open cover of $X$ such that $\operatorname{cl}V_H\subseteq U_H$ and $\operatorname{cl}V_K\subseteq U_K$. Then $V_K\cap K\subseteq U_K\cap K=\varnothing$, so $K\subseteq V_H$. Moreover, $X\setminus\operatorname{cl}V_H\supseteq X\setminus U_H=H$, so $X\setminus\operatorname{cl}V_H$ and $V_H$ are disjoint open sets containing $H$ and $K$, respectively. This is impossible, so $\mathscr{U}$ has no open shrinkage. Any maximal element $\varphi_0$ in $\varPhi$ will be such that $\varphi_0(U)=U\ne\operatorname{cl}U$ for some $U\in\mathscr{U}$.

The author needs to show further that if $\varphi(U)=U$ for some $\varphi\in\varPhi$ and $U\in\mathscr{U}$, then there is a $\psi\in\varPhi$ such that $\varphi\le\psi$ and $\operatorname{cl}\psi(U)\subseteq U$, i.e., $\psi$ does shrink $U$ and also still shrinks every member of $\mathscr{U}$ that was shrunk by $\varphi$.

Let

$$F=X\setminus\bigcup\left\{\varphi(V):V\in\mathscr{U}\setminus\{U\}\right\}\;;$$

then $F$ is a closed subset of $U$, and by normality of $X$ there is an open $W$ such that $F\subseteq W\subseteq\operatorname{cl}W\subseteq U$. Let $\psi(U)=W$, and let $\psi(V)=\varphi(V)$ for $V\in\mathscr{U}\setminus\{U\}$; then $\varphi\le\psi$ and $\operatorname{cl}\psi(U)\subseteq U$, as desired.

This shows that any maximal element of $\varPhi$ must in fact shrink every member of $\mathscr{U}$.

Answer 2

As an alternative proof: let $\{U_\alpha: \alpha < \gamma\}$, where $\gamma$ is an ordinal number, be a point-finite cover of the normal space $X$ by open sets. Then we will construct by transfinite recursion family of open sets $\{V_\alpha: \alpha < \gamma \}$ such that $\forall \alpha: \overline{V_\alpha} \subseteq U_\alpha$ such that at each stage we obey the condition

$$X=\bigcup \left(\{V_\alpha: \alpha < \beta\} \cup \{U_\alpha: \alpha \ge \beta\}\right)\tag{$\ast_\beta$}$$

for each $\beta < \gamma+1$.

For $\beta=0$ we have to do nothing, because the $U_\alpha$ indeed form an open cover.

Successor stage: Suppose that we have the $V_\alpha$ for all $\alpha < \beta$, obeying $(\ast_\beta)$; we then have to construct $V_\beta$ to obey $(\ast_{\beta+1})$. To this end define $A=\left(\bigcup \{V_\alpha: \alpha < \beta\} \cup \{U_\alpha: \alpha \ge \beta+1\} \right)^\complement$, which is closed and obeys $A \subseteq U_\beta$ (if $x \in A$ it must be covered by the cover from $(\ast_\beta)$ and being in $A$, only $x \in U_\beta$ is possible.) Now we apply normality and find an open $V_\beta$ such that $A \subseteq V_\beta \subseteq \overline{V_\beta} \subseteq U_\beta$. Now with the new $V_\beta$ covering $A$, we see that $(\ast_{\beta+1})$ is now satisfied. (nitpick: note that $A=\emptyset$ would not be a problem: pick $V_\beta = \emptyset$ and the recursion goes on anyway.)

Limit stage: Suppose that $(\ast_{\delta})$ is satisfied for all $\delta < \beta$ and $\beta$ is a limit ordinal. Then we already know that $(\ast_{\beta})$ is satisfied too; this uses the point-finiteness of the cover: let $x \in X$ and $x$ is not covered by the cover from $(\ast_\beta)$. But $x \in U_{\alpha_1},\ldots, U_{\alpha_n}$, for finite but non-zero number of indices $\alpha_1 < \ldots < \alpha_n$. By assumption, these $\alpha_i < \beta$ (or the cover from $(\ast_\beta)$ would have sufficed) and as $\beta$ is a limit we find $\delta$ with $\alpha_n < \delta < \beta$. But then $x$ must covered by one of the $V_\alpha$ with $\alpha < \delta$ by the recursion assumption $(\ast_\delta)$ (one of the $\alpha_i$, of course) and $(\ast_\beta)$ is satisfied after all.

Now, if $\mathcal{U}$ is a finite cover we have a closed shrinking, using only the recursion step for the successor in the previous proof. Otherwise we index $\mathcal{U}$ by an ordinal $\gamma$ (using AC, as Zorn does too) and do this construction. At the end, by $(\ast_\gamma)$ we have our shrinking $(V_\alpha)_{\alpha < \gamma}$, as required.