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Here is Prob. 5, Sec. 26, in the book Topology by James R. Munkres, 2nd edition:

Let $A$ and $B$ be disjoint compact subspaces of the Hausdorff space $X$. Show that there exist disjoint open sets $U$ and $V$ containing $A$ and $B$, respectively.

First of all, here is Lemma 26.1:

Let $Y$ be a subspace of (a topological space) X. Then $Y$ is compact (relative to the subspace topology that $Y$ inherits from $X$) if and only if every covering of $Y$ by sets open in $X$ contains a finite subcollection covering $Y$.

And, here is Lemma 26.4:

If $Y$ is a compact subspace of the Hausdorff space $X$ and $x_0$ is not in $Y$, then there exist disjoint open sets $U$ and $V$ of $X$ containing $x_0$ and $Y$, respectively.

I think I am clear on the proof of either of Lemmas 26.1 and 26.4. So I'll be using these in my proof of Prob. 5, Sec. 26, which is as follows:

Since $A$ and $B$ are disjoint, therefore for each point $a \in A$, there exist disjoint open sets $U_a$ and $V_a$ of $X$ containing $a$ and $B$, respectively, by Lemma 26.4.

As the collection $$ \left\{ \ U_a \colon \ a \in A \ \right\}$$ is a covering of $A$ by sets open in $X$, so by Lemma 26.1 there is a finite subcollection of this collection that also covers $A$; let $U_{a_1}, \ldots, U_{a_n}$ be this finite subcollection.

Now let us put $$ U \colon= \bigcup_{i=1}^n U_{a_i} \qquad \mbox{ and } \qquad V \colon= \bigcap_{i=1}^n V_{a_i}. \tag{1} $$ Here $V_{a_1}, \ldots, V_{a_n}$ are the open sets that correspond to the sets $U_{a_1}, \ldots, U_{a_n}$, respectively, as in the first paragraph of this proof.

Then both the sets $U$ and $V$ as defined in (1) here are open sets of $X$; moreover the set $U$ contains $A$ by our choice of the sets that $U$ is composed of.

As each set $V_a$ in the first paragraph contains $B$, so does each of the sets $V_{a_i}$ in (1) above; therefore $B$ is contained in $V$.

Finally, if $u \in U$, then $u \in U_{a_k}$ for some $k = 1, \ldots, n$, and thus this point $u$ would not be in the corresponding set $V_{a_k}$, and hence $u$ would not be in the set $V$ in (1) above.

Conversely, if $v \in V$, then $v$ is in each of the sets $V_{a_1}, \ldots, V_{a_n}$, and therefore $v$ is in none of the sets $U_{a_1}, \ldots, U_{a_n}$, which implies that $v$ is not in $U$.

Thus the sets $U$ and $V$ are disjoint.

Is this proof correct? If so, then is it clear in each and every one of its steps? If not, then where are the issues?

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    $\begingroup$ I cannot find any mistakes in it. Redundancy: if you have proved that an element of $U$ cannot be an element of $V$ then you are ready. It is not necessary anymore to prove also that an element of $V$ cannot be be an element of $U$. $\endgroup$ – drhab Sep 10 '18 at 13:46
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Your proof is correct.

Note that the proof idea is exactly the same as for lemma 26.4 where we can replace one point in a Hausdorff space by a compact set and we separate by a union-and-intersection-trick because we get a finite union (from the finite subcover) and a finite corresponding intersection (so that makes this intersection open, as we need; an infinite intersection of open sets can have empty interior).

The proof of the empty intersection could be shorter: suppose we'd have $x \in U = \cup _{i=1}^n U_{a_i}$ and also $x \in V= \cap_{i=1}^n V_{a_i}$.

Then $x \in U_{a_j}$ for some fixed $j \in \{1,\ldots,n\}$ (definition of union). But then by being in the intersection $V$ we also know that $x \in V_{a_j}$ for that same $j$, and this immediately contradicts $U_{a_j} \cap V_{a_j} =\emptyset$ which is how we chose the corresponding pairs of $U_a$'s and $V_a$'s in the first place.
So immediately $U \cap V = \emptyset$ is clear.

You should also note the similarity to the proof of the tube lemma and its generalisations. It's a very common technique in compactness proofs.

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  • $\begingroup$ Following is the link to a post of mine on the generalisation of the tube lemma. You've contributed to that post too. math.stackexchange.com/questions/1278681/… $\endgroup$ – Saaqib Mahmood Sep 11 '18 at 8:38
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    $\begingroup$ @SaaqibMahmood that generalisation is called Wallace's theorem (in Engelking's book) and it also holds for any number of factors: any product of compacts in a product of spaces that sits in an open set has a basic open set sitting between it and that open set. So products of compacts behave like points. Just as compact sets behave like points wr.t. separation axioms: in a Hausdorff space we can separate two disjoint compact sets as well (but not two closed disjoint ones in general). $\endgroup$ – Henno Brandsma Sep 11 '18 at 8:45

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