I am studying Michael's article "Topologies on Spaces of Subsets" and I just can't figure out the proof of lemma 2.2.2' below
I have already proven that if $[X,U]$ is $T_1$ (which we are assuming for this), then it is $T_3$. Using this and with an assumption I'm not achieving to prove, I have proven this lemma as I write next. Here I am using the uniform structure induced by entourages, and by neighborhood of a subset I mean a set that contains an open that contains the subset. (More information about notation is in the remark.)
It suffices to prove that its complement is open, which is the set of all closed sets that are contained in the complement of $A$. Since the space $[X,U]$ is $T_3$ under the uniform topology, given $B$ disjoint from $A$ and closed, there are for each $a \in A$ a neighborhood $V_a(a)$ disjoint from a neighborhood $U_a$ which I don't know how to write in terms of $B$, but suppose that it is possible to have $V_a(a) \cap V_a(B)=\emptyset$. We find a covering of $A$, and then a finite sub-covering is found since $A$ is compact. Because uniformities are closed by finite intersections, if the neighborhoods of this finite covering are indexed by $\{a_1,\ldots,a_n\}$, we can take $W = \cap_{i=1}^n V_{a_i}$ and this will be an entourage of $[X,U]$. Since the $V_{a_i}(a_i)$'s covers $A$, we end up having $W(B) \cap A = \emptyset$, it is to say $W(B) \subset A^c$. Then we have $\widehat{W}(B)$ a neighborhood of $B$ in $2^X$ such that it is contained in the complement of the set which we want to prove it is closed. Since $B$ was arbitrary, we conclude that the complement is open.
The huge problem is that I don't know how to prove that there is one entourage satisfying $V(a)\cap V(B) = \emptyset$.
Remark:
- $2^X$ is the collection of all closed subsets of $X$ in the uniform topology
- $2^{\mathtt{U}}$ is the uniformity in $2^X$ generated by the basis of the entourages
$$\widehat{V} = \{(E,F) \in 2^X \times 2^X : E \subset V(F) \text{ and } F \subset V(E)\}$$
where $V \in \mathtt{U}$ - $|2^\mathtt{U}|$ is the topology induced by the uniformity $2^\mathtt{U}$.
Best Answer
$A$ and $B$ should, of course, be assumed to be disjoint; otherwise the conclusion obviously fails.
For each $a\in A$, we have an open neighborhood of $a$ disjoint from $B$, because $B$ is closed, and this neighborhood can be taken to be of the form $U_a(a)$for some entourage $U_a$. Choose symmetric entourages $V_a$ with $V_a\circ V_a\subseteq U_a$. Since $A$ is compact and is covered by the open sets $V_a(a)$ (for $a\in A$), it is covered by finitely many of these, say the ones corresponding to $a\in F$, where $F$ is a certain finite subset of $A$.
Now let $V=\bigcap_{a\in F}V_a$; since $F$ is finite, this is an entourage. I'll show that $V(A)$ is disjoint from $B$. It suffices to show that $V(x)$ is disjoint from $B$ for all $x\in A$, since $V(A)$ is the union of these. So consider any $x\in A$ and note that, by our choice of $F$, $x\in V_a(a)$ for some $a\in F$. Fix such an $a$. Since $V$ is symmetric and $\subseteq V_a$, we have $V(x)\subseteq V_a(x)\subseteq V_a(V_a(a))\subseteq U_a(a)$, which is disjoint from $B$ as required.