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In the proof of the Hahn-Banach separation theorem my notes claim the following: Let $X$ be a normed $\mathbf{R}$ vector space, $A,B\subset X$ be nonempty, disjoint, convex, $A$ compact and $B$ closed. Then there exists $r>0$ such that $N_r(A)\cap B=\emptyset$, where $N_r(A):=\bigcup_{x\in A} B_r(x)$. Then there comes a proof of several lines. Can't we just say let $r=dist(A,B)$ (using this result). $\square$?

I mean if $y\in N_r(A)\cap B$ then $d(x,y)<r=dist(A,B)$ with $x\in A$ and $y\in B$!

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    $\begingroup$ Yes. But, the question is: was that result proved before? The proof of several lines in your notes basically proves that result. (But, yes, with a bit of background, the proof is a one-liner.) $\endgroup$ – Daniel Fischer Jul 25 '14 at 13:03
  • $\begingroup$ @DanielFischer: First of all thank you for the comment. Yes it is basically a proof of this, but I am of the opinion that one should use these facts in full generality instead of reproving them all the time. (I don't believe that it was proven in the same course.) By the way I would accept it as an answer, otherwise this will most likely be left unanswered and don't want to delete the question, as it might be useful for other people $\endgroup$ – HAMMER Jul 25 '14 at 13:21
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Yes, in the given situation, one could just use $r = \operatorname{dist}(A,B)$, and then have

$$N_r(A) = \bigcup_{x\in A} B_r(x) = \{ y\in X : \operatorname{dist}(y,A) < r\}$$

as a convex open neighbourhood of $A$ disjoint from $B$.

But of course, to use that result, $\operatorname{dist}(A,B) > 0$ must have been proved. In the setting of a normed space, that is a one-liner by stating that the continuous function $y \mapsto \operatorname{dist}(y,B)$ attains its minimum on the compact set $A$, and $\{x\in X : \operatorname{dist}(x,B) = 0\} = \overline{B} = B$.

I suspect that the several-lines proof in your notes proves the more general proposition that in a locally convex space, there is a convex open neighbourhood $U$ of $0$ with $(A + U) \cap B = \varnothing$ when $A$ and $B$ are disjoint convex sets with $A$ compact and $B$ closed. The proof in the more general setting is a bit more complicated than for normed spaces (or for locally convex spaces with a translation-invariant metric whose balls are convex, where the same argument as for normed spaces works).

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    $\begingroup$ Thanks, very helpful! In particular the proof for normed spaces, actually the fact that $dist(-,B)$ is continuous was proven in an earlier course, so I am allowed to use it. (But the proof for metric spaces is also very simple.) $\endgroup$ – HAMMER Jul 25 '14 at 13:58

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