Theorem 13.1.25.label Let $C\suf\R^{n}$ be closed convex and $x_{0}\nin C$ then $C$ and $\curl{x_0}$ can be strongly separated
Proof. WLOG $0\nin C$ then by Lemma 13.1.23 then $\exists z\in C$ such that $\inner{z}{x}\geq \norm{z}{}^{2}$ for all $x\in C$. Let $\al=\frac{\norm{z}{}^{2}}{z}$ then
\begin{align*}\inner{z}{x}>\frac{\norm{z}{}^{2}}{z}=\al&&\fall x\in C\end{align*}
Since $x_{0}\nin C$ we have $\inner{z}{x_0}<\norm{z}{}^{2}\implies \inner{z}{x_0}<\frac{\norm{z}{}^{2}}{z}$ taking $\e\in (0,\al)$ we are done. If $0\in C$ and $z=0$ then consider $c-\de e$ for $\de>0$ and $\til{x_0}=x_{0}-\de e$.$\square$
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