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I am having trouble seeing this. I have read and understood the proofs that cl(ri(C))=cl(C) and ri(cl(C))=ri(C). But to conclude that cl(C1)=cl(C2) iff ri(C1)=ri(C2) from the above two equalities? Do I need to show cl(ri(C)=ri(cl(C))? I'm not sure how I'd show this. All sets in consideration are in Rn and convex. Thanks in advance!

RV702
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2 Answers2

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I might be wrong, but try if this works:

We're trying to prove if cl(C1)=cl(C2) then ri(C1)=ri(C2). This is the same as proving $ri(C1)\not=ri(C2)\implies cl(C1)\not=cl(C2)$.

$ri(C1)\not=ri(C2)\implies ri(cl(C1))\not=ri(cl(C2))\implies cl(C1)\not= cl(C2)$

The other direction should be similar.

badatmath
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The if part: $cl(C_1)=cl(ri(C_1))=cl(ri(C_2))=cl(C_2)$.

The only if part: $ri(C_1)=ri(cl(C_1))=ri(cl(C_2))=ri(C_2)$.

See proof of Proposition 1.3.5 in Convex Optimization Theory by Bertsekas.

feixh
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