Show that $\mathcal{A}$ is a basis for the topology on $X.$

Question

Having known how to answer the following question (with the help of @Henno Brandsma ):

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

I want to answer this question:

$(b)$ Let $\mathcal{A} = \bigcup_{n \geq 1} \mathcal{A}_{n}.$ Show that $\mathcal{A}$ is a basis for the topology on $X.$

I got a hint that a basis for the topology on $X$ means a set $U \subseteq X$ is open iff it is a union of sets in $\mathcal{A}.$ Still, I am unable to write the solution neatly.

Could anyone help me in proving so, please?

Answer 1

Again (I've remarked it earlier) $\mathcal{A}_n < \mathcal{A}_{n+1}$ must mean that $\mathcal{A}_n$ is strictly refined by $\mathcal{A}_{n+1}$.

As a general background, see Munkres' book §13, or Wikipedia to find the two criteria that a collection of sets $\mathcal{A}$ has to obey in order to be a base for some topology on a set $X$:

1. $\bigcup \mathcal{A} = X$.
2. For all $A_1, A_2 \in \mathcal{A}: \forall x \in A_1 \cap A_2: \exists A_3 \in \mathcal{A}: x \in A_3 \subseteq A_1 \cap A_2$.

And for your collection 1. is obvious as for every $n$ we already have $\bigcup \mathcal{A}_n =X$ so certainly this holds for the larger collection $\mathcal{A}$.

And to see 2.: if we have $A_1, A_2 \in \mathcal{A}$, there are $n,m$ such that $A_1 \in \mathcal{A}_n$ and $A_2 \in \mathcal{A}_m$. WLOG $n < m$ so $\mathcal{A}_n$ is strictly refined by $\mathcal{A}_m$, so $A_1$ is a (strict) superset of some $A'_1 \in \mathcal{A}_m$. So either $A'_1 \cap A_2$ is empty or $A_1'= A_2$ (because they both live in $\mathcal{A}_m$) In the former case $A_1 \cap A_2$ is empty and 2. is voidly fulfilled, in the latter case $A_2 \subsetneq A_1$ and we can take $A_3 = A_2$ to fulfill it.

This shows that $\mathcal{A}$ is a base for some topology $\mathcal{T}(\mathcal{A})$ which is the smallest topology that contains $\mathcal{A}$ as a subset and consists of all unions of subfamilies of $\mathcal{A}$. So $\mathcal{T}(\mathcal{A})$ is a coarser topology than $\mathcal{T}_X$ because all members of $\mathcal{A}$ are open in $\mathcal{T}_X$.

In fact the topology indeed equals $\mathcal{T}_X$, because it is Hausdorff: If $x \neq y$, find $n$ so large that $\frac{1}{n} < d(x,y)$. Then $x \in A_1 \in \mathcal{A}_n, y \in A_2 \in \mathcal{A}_n$ for some $A_1, A_2$ and we cannot have $A_1 =A_2$ or else $\operatorname{diam}(A_1)\ge d(x,y) > \frac{1}{n}$, contradicting property 2., so $A_1 \cap A_2 = \emptyset$ and so $(X,\mathcal{T}(\mathcal{A}))$ is Hausdorff. Then the identity map $1_X$ from $(X,\mathcal{T}_X)$ to $(X,\mathcal{T}(\mathcal{A}))$ is continuous (as all members of $\mathcal{A}$ are open in $\mathcal{T}_X$) and a bijection and closed (as a continuous map from a compact to a Hausdorff space) and hence a homeomorphism, hence open, and this implies $$\mathcal{T}_X = \mathcal{T}(\mathcal{A})$$

which shows that $\mathcal{A}$ is a base for $X$.