Proving existence of finite, clopen,disjoint, strict refinement cover with diameter at most $r.$

Question

I want to solve part(b) of this problem:

Let $X$ be a totally disconnected compact metric space, and let $\mathcal{A}$ be a clopen cover of $X,$ and fix $r>0$.

(a) Show that there is a cover $\mathcal{B}$ having all five of the properties:

I. each member of $\mathcal{B}$ is clopen,

2. each member of $\mathcal{B}$ has diameter at most $r,$

3. $\mathcal{B}$ refines $\mathcal{A},$

4. $\mathcal{B}$ is a finite cover, and

5. $\mathcal{B}$ is a disjoint cover.

$(b)$ Show that if $X$ is perfect, then $(3)$ can be improved to $\mathcal{A} < \mathcal{B}.$

And also I proved (with the very great help of @Henno Brandesma on this site) part(a) of the previous question and the following question :

Let $X$ be a compact metric space that is totally disconnected, and let $\epsilon > 0.$\ (a) Show that $X$ has a finite cover $\mathcal{A}$ clopen sets with diameter at most $\epsilon.$\ (b) Show that there is a clopen cover $\mathcal{B}$ such that $\mathcal{A}$ refines $\mathcal{B}$ and distinct numbers of $\mathcal{B}$ are disjoint.

And we know the following definition:

DEFINITION.

Suppose $\mathcal{A}$ and $\mathcal{B}$ are two covers of $X.$ We say that $\mathcal{B}$ refines $\mathcal{A}$ if each member of $\mathcal{B}$ is contained in some member of $\mathcal{A}.$ We say that $\mathcal{B}$ strictly refines $\mathcal{A}$ if each member of $\mathcal{B}$ is a proper subset of some member of $\mathcal{A}.$ if $\mathcal{B}$ strictly refines $\mathcal{A},$ we write $\mathcal{A} < \mathcal{B}.$

Definition:

A topological space $X$ is totally disconnected if for any two distinct points $x,y \in X,$ there is a separation $X = U \cup V$ of $X$ with $x \in U $ and $y \in V.$

Definition:

A separation of a space is a cts. function $f: X \rightarrow \{0,1\}$ with $\{0,1\}$ has the discrete topology.

In terms of open sets, a separation of $X$ is an expression $X = U \cup V$ where $U \cap V = \emptyset $ and $U,V$ are both open in $X.$

So, how can all of the above help me in answering the part $(b)$ in the first problem I stated above? Could anyone help me, please?

Answer 1

To show $(b)$ we use the perfectness to show a simple "shrinking lemma":

Let $X$ be perfect too. Suppose that $U$ is non-empty and open. Then there is a clopen subset $V$ such that $\emptyset \neq V \subsetneq U$.

Proof: let $x \neq y$ be two distinct points of $U$. (as $U$ is non-empty and cannot be a singleton, or $X$ would not be perfect). Let $B(x,r) \subseteq U$ such that $y \notin B(x,r)$ (just take any $r \le d(x,y)$ such that the ball also sits inside $U$, by openness of $U$). The earlier lemma on totally disconnected compact spaces then tells you that we have a clopen subset $V$ such that $x \in V \subseteq B(x,r)$. This $V$ is as required.

Now, after (a) you have some clopen refinement $\mathcal{B}$ of $\mathcal{A}$ that obeys 1-5. If for some $B \in \mathcal{B}$ we have that its refining element from $\mathcal{A}$ is just $B$ again, so a contradiction to strictness, we apply the previous lemma to this $U=B$ and use $V$ and $B \setminus V$ (so we split $B$ into two disjoint clopen pieces) and remove $B$. This fixes $\mathcal{B}$ in at most finitely many steps to a strict refinement, if this were not already the case.

The splitting preserves all 1-5 properties: both new sets are clopen, mutually disjoint, refined by the same element of $\mathcal{A}$ and the diameter only gets smaller, and the covers stay finite.

Answer 2

You can just take the resulting cover from part (a) and modify it to get strictness while preserving all the other properties.