Proving existence of sequence of finite, clopen and disjoint cover each member with diameter at most $1/n.$

Question

I want to answer the following question:

Let $X$ be a totally disconnected, compact, and perfect metric space.

(a) Show that there is a sequence of covers $\mathcal{A}_{1}, \mathcal{A}_{2}, ... $ such that:

1- Each $\mathcal{A}$ is a finite disjoint clopen cover,

2- Each member of $\mathcal{A}_{n}$ has diameter at most $\frac{1}{n},$ and

3- $\mathcal{A}_{1} < \mathcal{A}_{2} < ... < \mathcal{A}_{n} < \mathcal{A}_{n + 1} < ...\, .$

Could anyone help me in proving so please?

Answer 1

By the proposition $(b)$ in this problem, we have $\mathcal{A}_1$ of sets of diameter $< 1$.

If we have constructed $\mathcal{A}_i, i \le n$ already, obeying the demands, we just pick by the same problem, a disjoint clopen finite cover $\mathcal{A}_{n+1} \prec \mathcal{A}_n$ such that all its sets have diameter $< \frac{1}{n+1}$ and this preserves all properties ( when we modify the refinement relation as I said)

So the sequence exists by the recursion theorem.