Barycentric Refinement of a Barycentric Refinement is a Star Refinement

Question

I'm reading Willard's topology, and the following came up as an 'easy' exercise in the text, though I am having trouble seeing why it is true. Let $X$ be a set, $\mathscr U$ some cover of $X$, and let $\mathscr W$ be a barycentric refinement of a barycentric refinement $\mathscr V$ of $\mathscr U$. Then, $\mathscr W$ is a star refinement of $\mathscr U$. (See https://en.wikipedia.org/wiki/Star_refinement for the definitions). Here are some of my thoughts on this:

First, by assumption, to every $x \in X$, there is some $y_x \in X$ such that $\text{st}(x, \mathscr W) \subseteq \text{st}(y_x, \mathscr V)$. That is to say, $\bigcup\{ W \in \mathscr W \vert x \in W \} \subseteq \bigcup \{V \in \mathscr V \vert y_x \in V\}$. Moreover, we also have by assumption that for every $z \in X$, there is $U \in \mathscr U$ such that $\text{st}(z, \mathscr V) \subseteq U$. Let $A\subseteq X$; one needs to show $\text{st}(A, \mathscr W) \subseteq U$ for some $U \in \mathscr U$. Notice that $\text{st}(A, \mathscr W) = \bigcup_{x \in A} \text{st}(x, \mathscr W)$.

With notation as above, we then have $\text{st}(A, \mathscr W) \subseteq \bigcup_{x \in A} \text{st}(y_x, \mathscr V)$. I do not see why this union should be contained in a single $U \in \mathscr U$; for each $y_x$, there is some $U_{y_x} \in \mathscr U$ for which $\text{st}(y_x, \mathscr V) \subseteq U_{y_x}$. Is there a way to find a single $U$ containing each of the $\text{st}(y_x, \mathscr V)$'s?

Thanks in advance.

Answer 1

So we have $\mathcal{W} \prec_b \mathcal{V} \prec_b \mathcal{U}$ and we want to show $\mathcal{W} \prec_\ast \mathcal{U}$.

Let $W_0$ be some element of $\mathcal{W}$ and consider $x_0 \in W_0$.

As $\mathcal{V} \prec_b \mathcal{U}$ we know that $\mathrm{st}(x_0, \mathcal{V}) \subseteq U_0$ for some $U_0 \in \mathcal{U}$.

Now my claim is that in fact $\mathrm{st}(W_0,\mathcal{W}) \subseteq U_0$, finishing the proof of star-refinement:

To this end, let $W \in \mathcal{W}$ be any member that intersects $W_0$. Say $x \in W \cap W_0$.

As $\mathcal{W} \prec_b \mathcal{V}$ there is some $V_0 \in \mathcal{V}$ such that $\mathrm{st}(x, \mathcal{W}) \subseteq V_0$.

Then $W \cup W_0 \subseteq \mathrm{st}(x,\mathcal{W}) \subseteq V_0$ so that in particular $x_0 \in V_0$ and so $V_0 \subseteq \mathrm{st}(x_0, \mathcal{V}) \subseteq U_0$ and as $W \subseteq V_0$, also $W \subseteq U_0$.

As $W$ was an arbitrary member of the sets whose union is $\mathrm{st}(W_0, \mathcal{W})$ we see that $\mathrm{st}(W_0, \mathcal{W}) \subseteq U_0$ as required.

Magical isn't it? A search yielded that this argument was also hidden in the third paragraph of this answer but only as a lemma of sorts.