Cfrgtkky

55 31 $begingroup$ Claim: Let $X$ be a metric space. If $A,Bin X$ are disjoint, if A is compact, and if B is closed, then $exists delta>0: |alpha-beta|geqdelta;;;forallalphain A,betain B$. Proof. Assume the contrary. Let $alpha_nin A,beta_nin B$ be chosen such that $|alpha_n-beta_n|rightarrow0$ as $nrightarrow infty$. Since A is compact, there exists a convergent subsequence of $alpha_n;(ninmathbb{N})$, $alpha_{n_m};(minmathbb{N})$, which converges to $alphain A$. We have $$|alpha-beta_{n_m}|leq|alpha-alpha_{n_m}|+|alpha_{n_m}-beta_{n_m}|rightarrow0 ;;;as;;mrightarrowinfty.$$ Hence $alpha$ is a limit point of B and since B is closed $alphain B$, contradiction. Is my proof correct? I feel as though I am missing something simple which would trivialize the proof.