I found in Brezis' Analyse Fonctionelle the Hahn-Banach theorem ("second geometric form") and there is a passage I can't understand. In newest versions of this book the proof has been modified.
Theorem. Let $A \subset E$ and $B \subset E$ be two nonempty convex subsets such that $A \cap B = \emptyset$. Assume that $A$ is closed and
$B$ is compact. Then there exists a closed hyperplane that strictly separates $A$ and $B$.
Proof. For $\epsilon > 0$ let $A_\epsilon = A+B(0,\epsilon)$ and $B_\epsilon = B+B(0,\epsilon)$; those sets are convex and nonempty. They are disjoint too (otherwise, we can find a sequence $\epsilon_n \to 0$ and $x_n \in A$, $y_n \in B$ such that $\|x_n – y_n \|< 2\epsilon_n$; so we can extract a subsequence $y_n \to y \in A \cap B$). $\quad$ [CUT]
My question is:
How can we extract such a subsequence when we don't know if $A_\epsilon \cap B_\epsilon$ has elements in common with $A \cap B$?
Best Answer
The passage is worded somewhat ambiguously, but I think what is claimed is that there exists some $\epsilon > 0$ such that $A_\epsilon$ and $B_\epsilon$ are disjoint. For suppose the contrary: that for every $\epsilon > 0$ there exists $z \in A_\epsilon \cap B_\epsilon$. Choose a sequence $\epsilon_n \to 0$ and $z_n$ with $z_n \in A_{\epsilon_n} \cap B_{\epsilon_n}$. By definition of $A_\epsilon, B_\epsilon$ there exists $x_n \in A$ and $y_n \in B$ such that $\lVert x_n - z_n \rVert < \epsilon_n$, $\lVert y_n - z_n \rVert < \epsilon_n$, so by the triangle inequality $\lVert x_n - y_n \rVert < 2 \epsilon_n$. Since $B$ is compact, we can pass to a subsequence and assume $y_n$ converges to some $y \in B$. Then it follows from the triangle inequality that $x_n \to y$ as well. $A$ is closed, so $y \in A$, and thus $A,B$ are not disjoint, a contradiction.