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Theoren 6.3 from Convex analysis, by Rockafellar, says that if $C\subset \mathbb R^n$ is a convex set, then $cl(ri C)=cl C$ and $ri(cl C)=ri C$.

(Here $cl C$ denotes the close of $C$ and $ri C$- a relative interior of $C$).

I wish to give a proof of the following Corollary 6.3.2 from the above book.

If $C\subset \mathbb R^n$ is convex and $V\subset \mathbb R^n$ is open and $V\cap cl C \neq \emptyset$, then $V\cap ri C \neq \emptyset$.

Thanks

Richard
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1 Answers1

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Assume $V$ and $C$ are as you state.
Take $\bar{x}\in V\cap \rm{cl}\,C = V\cap \rm{cl}\,\rm{ri}\,C$.
Then there exists a sequence $(x_n)$ in $\rm{ri}\,C$ converging to $\bar{x}$.
Hence, for all $n$ sufficiently large, $x_n\in V$ (because $V$ is open).
These $x_n$ will do the job.

max_zorn
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