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If the "closed convex hull" of A is the intersection of all closed convex sets containing A, is this the same as the closure of the convex hull of A?

Many have asked whether the closure of the convex hull is the same as the convex hull of the closure (answer: no), but I think this is a bit different.

I feel like this should have a simple answer, either based on set logic (if it's true) or a simple counterexample (if it's false).

Measure Theory Penguin
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  • I put "normed spaces" tag, because the original ones (topological/metric spaces) seemed not directly relevant to the concept of convex hull. –  May 06 '16 at 17:56

2 Answers2

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They are equal.

Let $C$ be the closed convex hull of $A$, defined as the intersection of all closed convex sets containing $A$. (Closed half-spaces are enough here.)

Let $B$ be the convex hull of $A$, defined as the intersection of all convex sets containing $A$. (Half-spaces, either open or closed, are not enough here.)

Clearly $B\subset C$. Since $C$ is closed, it follows that $\overline{B}\subset C$.

Conversely, $\overline{B}$ is both closed and convex, hence $C\subset \overline{B}$.

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While user147263's answer is correct, the second part of his proof is not fully detailed. Let me elaborate on it a little bit.

First, it can easily be seen that the closure of a convex set is too a convex set.

Let $B$ be the convex hull of A. In symbols, $B = \cap \{M,\, M \text{ is convex, } A \subset M\}$.

Let $C$ be the closed convex hull of $A$. Again in symbols, $C = \cap \{M,\, M \text{ is convex and closed, } A \subset M\}$.

Then clearly, $\overline{B}$ is closed, convex and contains A. It follows from here that $\overline{B}$ is one of the intersected sets in the definition of C, hence $C \subset \overline{B}$.

Conversely, $B \subset C$. Then also $ \overline{B} \subset \overline{C}$. Because $C$ is closed, then $C = \overline{C}$ and $\overline{B} \subset C$.

tahotoxic
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