About convex hull and closed sets

Question

Let S be a closed set. Show with an example that $conv(S)$ is not necessarily closed. Also show that if S is compact then $conv(S)$ is always closed.

Here $conv(S)$ denotes the hull of S.

Proof: (I didn't show an example I did the proof)

Recall that $conv(S)$ is the intersection of all the sets X such that $S\subset X$.

w.l.o.g we can suppose that all the sets X such that $S\subset X$ are open and also suppose their intersection is finite.

Thus $conv(S)$ can't be closed, is open.

Is my proof correct?

Answer 1

There are a couple problems with your proof.

The biggest one is simply that you have the definition of convex hull wrong.   $\DeclareMathOperator{conv}{conv} \conv{(S)}$ is not simply $\bigcap{\{X:S\subseteq X\}}$; in fact, $$\bigcap{\{X:S\subseteq X\}}=S$$ Instead, $$\conv{(S)}=\bigcap{\{X:S\subseteq X\text{ and }X\text{ is convex}\}}$$

Another biggie that alden already pointed out is that sets can be both open closed. Consider $\mathbb{Q}$ as a subset of $\mathbb{R}$: we have $\conv{(\mathbb{Q})}=\mathbb{R}$, which is both open and closed. This is a common mistake beginning math students make; the mantra I've heard people use to try and teach this is that "sets are not doors." Hopefully you find that helpful too.

And finally, I'm troubled by your notion of "proof." You are correct that, a priori, the sets that contain $S$ and $X$ could be open, and there could be only finitely many of them. If that is so, then $\conv{(S)}$ would be open, for the reason you illustrate. If closed sets are not open, anyone trying to prove that $\conv{(S)}$ were closed would need to overcome this obstruction; you very well could have identified a gap in their proof.

But that doesn't mean that they can't overcome the obstruction; perhaps the gap you identified simply does not occur. Consider any infinite set $U$ with the indiscrete topology (only $\emptyset$ and $U$ are open). Then any coinfinite set $S$ has infinitely many supersets in $U$ and exactly one ($U$) is open. (On the other hand, proof may well be necessary for this question. Once you find a counterexample $S$ with convex hull $T$, you will need to prove that $\conv{(S)}=T$ and that $T$ is not closed.)

alden already gave a hint, so I'll direct you to his answer if you're still looking for a counterexample.

Answer 2

Take the points of the plane $(0,0)$ and $(n,1/n)$ for $n=1,2,...$. Its convex hull gets arbitrarily close to the all the positive part of the X axis, but it is not going to include any of its points except $(0,0)$.

Your argument is not correct, since the convex hull doesn't have to be open. For example the convex hull of $\{(0,0)\}$ is itself, which is closed.

Moreover, proving that a set is open doesn't necessarily imply that it is not closed. A set can be simultaneously open and closed.