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August 20, 2021

The Set Equality for Convex Set

math

Follow the idea of user04651 in this post to prove this statement.

Lemma. Let be a normed space, open and . If , then the line segment

Proof. For every , there is a such that . Let be such that .

Now for every (we have discussed the intuition of this radius in this post), we have for some ,

As this is true for every , we conclude .

Proposition. Let be a normed space and a convex set, then .

Proof. The direction is clear.

Let , for the sake of contradiction suppose , then there is an open neighborhood of such that . Since , there is an such that .

Now being an element in , , fix an , by the lemma above the segment

We can pick an that is close enough to , so that , but then , a contradiction.