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The lemma is as follow:

Let $X$ be a normal space, let {$U_1, U_2,...$} be a point- finite indexed open covering of $X$. Then there exists an indexed open covering {$V_1, V_2,...$} of X such that $\overline{V_n} \subset U_n$ for each n.

Proof:

Proceed by induction. First, note the set $A=X- \cup_{i\geqslant 2}U_{i}$ is closed in $X$. Because {$U_1, U_2,...$} covers $X$, the set A is contained in open set $U_1$, by normality there is an open neighborhood $V_1$ of $A$ such that $\overline{V_1} \subset U_1$ and {$V_1, U_2, U_3,...$} covers $X$. In general, given open sets $V_1,...,V_{k-1}$ such that the collection {$V_1,.., V_{k-1}, U_k,...$} covers $X$, let $A = X - (V_1 \cup .. \cup V_{k-1}) - \cup_{i \geqslant k+1}U_{i}$ is a closed subset of $X$ which is contained in open set $U_{k}$. Using normailty, choose $V_{k}$ to be an open set containing $A$ such that $\overline{V_k} \subset U_k$. Then {$V_1,.., V_{k}, U_{k+1},...$} covers X. All we have to do left is to show the collection {$V_1, V_2,..$} covers X. Here is where the condition point-finite indexed family is used. Let $x \in X$, since there are only finite open sets in $U_i$ containing $x$, which means for some $k \in \mathbb{Z}_{+}$ that $x \in X - \cup_{i > k}U_{i}$ therefore $x$ belongs to the set $\cup_{i=1}^{k}V_{i}$ so {$V_1, V_2,..$} covers X as desired.

The fact that the space is normal implies the construction of {$V_1, V_2,...$} such that for each $i \in \mathbb{Z}_{+}$, we have {$V_1,.., V_{k}, U_{k+1},...$} covers $X$, and the condition point-finite indexed open covering is used to show {$V_1, V_2,...$} covers $X$. I've been thinking to construct a normal space which does not have any point-finite indexed open covering to show that this condition is indeed neccessary, but I have not figured out one yet. Concretely, my hope is to find a space such that whatever the sequence of open set $V_n$ which is constructed using the procedure above, there will be a point such that it is not covered. In particular, when we define $V_1$ such that $X-\cup_{i \geqslant 2}U_i \subset V_1 \subset \overline{V_1} \subset \overline{U_1}$, there must be some points in $U_1$ not be covered by $V_1$ and each of them is contained in an infinite number of set $U_i$. Does anyone know how to construct such a space?

Hoang Nguyen
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This paper notes that a space $X$ is countably shrinking (i.e. every countable open cover has a shrinking, as you describe) iff $X$ is normal and countably paracompact. And as there are so-called Dowker spaces (in ZFC) that are normal and not countably paracompact, there are normal spaces in which there are covers without a shrinking, e.g. Mary-Ellen Rudin's Dowker space. The description isn't that hard (read the original paper by Mary Ellen), and one should be able to find such a cover from that.

It is indeed the case that we can have a shrinking for all point-finite covers, a proof of which I wrote down here.

Henno Brandsma
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