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Let $A\subset E$ and $B\subset E$ two convexe sets, non empty and disjoint. Suppose $A$ open. There is a closed hyperplan that separate $A$ and $B$.

In the book "Analyse fonctionnelle : théorie et application" (sorry it's in french), page 6 (demonstration théorème I.6) they do as follow : Set $C=A-B$. In particular $C$ is convexe (easy to prove), open (since $C=\bigcup_{x\in B} A-y$) and $0\notin C$ (because $A\cap B=\emptyset$.)

Question : What's the thing with this $C$ ? Since $A$ and $B$ is disjoint, $C=A-B=A$. So nothing to prove about the convexity of $C$ and the fact that $C$ is open... I don't understand what they try to do. Did they made a mistake somewhere ?

Peter
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1 Answers1

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You are confusing $$A-B=\left\{x-y:x\in A,y\in B\right\} $$ with $$A\setminus B=\left\{x\in A : x\notin B\right\} $$ which is sometimes also denoted with $A-B$.

Lorenzo Q
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  • Waou... I would never think about $A-B={x-y\mid x\in A, y\in B}$... Why didn't they precise it ? But anyway, thanks vvveerrrryyyyy much ! :) – Peter Aug 28 '18 at 13:38
  • @Peter I agree that the author should have specified the meaning, although this notation is rather common in this context. – Lorenzo Q Aug 28 '18 at 13:54