# In the proof of Hahn-Banach strict separation theorem.

In the proof of Hahn-Banach strict separation theorem it is written that

"Since $d(A,B)>0$ , there exists $\epsilon >0$ s.t. $A_{\epsilon}\cap B_{\epsilon}=\phi$ where $A_{\epsilon}=A+B(0,\epsilon),B_{\epsilon}=B+B(0,\epsilon)$"

I'm not understanding how this comes?

Here the statement of the Hahn-Banach strict separation theorem:

Let $X$ be a $\mathbb K-$linear space , let $A,B \subseteq X$ be non-empty ,disjoint , convex sets and let $A$ be compact and $B$ be closed . Then there exists $F\in X^*$ and $\alpha,\beta \in \mathbb R$ s.t. $Re(F)(a)<\alpha<\beta<Re(F)(b)$ for all $a\in A,b \in B.$

Please someone help. Thank you.

## 1 Answer

Set $\varepsilon = \frac{1}{2}d(A,B) > 0$. Suppose there exists some $x \in A_\varepsilon \cap B_\varepsilon$. Let $a \in A$ and $b \in B$ be such that $x \in B(a,\varepsilon)$ and $x \in B(b,\varepsilon)$. You get $d(a,b) \leq d(a,x) + d(x,b) < 2 \varepsilon = d(A,B)$ which is a contradiction.