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Could someone help me with the following question?

Let $X$ be a locally compact Hausdorff space (LCHS) and $K_1,K_2\subset{X}$ two disjoint compact sets. Then there exist $V_1,V_2$ disjoint open sets such that $K_1\subset{V_1},K_2\subset{V_2}$ and $\overline{V_1},\overline{V_2}$ are compact.

I have found several proofs of this fact (see Lemma 5.1 here), but they use sophisticated arguments.

Thanks.

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Let K,L be two disjoint compact sets.

Pick y in L.
For all x in K, exists open disjoint U$_x$, V$_x$ with x in U$_x$, y in V$_x$.
C = { U$_x$ : x in K } covers K.
Let Cf be a finite subcover of C.
U = $\cup${ U$_x$ : U$_x$ in Cf }, V = $\cap${ V$_x$ : V$_x$ in Cf } are open.
K subset U, y in V and U and V are disjoint.

Using the above, for all y in L,
exists open disjoint U$_y$, V$_y$ with K subset U$_y$, y in V$_y$.
C = { V$_y$ : y in L } covers L.
Let Cf be a finite subcover of C.
U = $\cap${ U$_x$ : U$_x$ in Cf }, V = $\cup${ V$_x$ : V$_x$ in Cf } are open.
K subset U, L subset V and U and V are disjoint.

Thus two disjoint compact set of a Hausdorff space can be separated by disjoint open sets.

Assume K is compact and U is open within a locally compact space Hausdorff space.
If K subset U, then for all x in K,
exists open U$_x$ with x in U$_x$ subset U, compact $\overline U_x$.
C = { U$_x$ : x in K } covers K.
Let Cf be a finite subcover of C.
U' = $\cup${ U$_x$ : U$_x$ in Cf } is open. K subset U' subset U.
Finally show $\overline {U'}$ is compact.

Do the same for L and pull together the loose ends for the conclusion you wish to prove. Lengthy but mostly simple stuff.

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  • $\begingroup$ I don't undertsand this sentence of the second part: "If $K$ subset $U$, then for all $x$ in $K$ exists open $U_x$ with $x$ in $U_x$ subset U, compact $\overline{U_x}$ " $\endgroup$ – mathlife Mar 17 at 17:33
  • $\begingroup$ @mathlife. Space is locally compact. $\endgroup$ – William Elliot Mar 17 at 22:03

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