# About convex hull and closed sets

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?

• regardless your proof is correct or not (which is obviously incorrect ), when you prove something (instead of giving example) you are infact showing that thing is always true for all case..... so you showed that always $conv(S)$ is open!! do you think it is correct ? – Red shoes Oct 28 '17 at 5:27
• Where are we? $\mathbb R^n$? An arbitrary Banach space? An arbitrary topological vector space? – Jack M Oct 28 '17 at 8:58
• @JackM in $\mathbb R^n$ – user441848 Oct 28 '17 at 16:12
• @Redshoes 😭😭 I did the proof based on the fact that $conv(S)$ is the intersection of all sets X such that $S\subset X$ – user441848 Oct 28 '17 at 16:17

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.

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.

• It is mot the X axis that is the problem, but the parallel to it through the pt 1,1 . Otherwise a very good example. – orangeskid Oct 28 '17 at 1:02
• @orangeskid The horizontal line through $(1,1)$ is a support line of the convex hull that contains one point of it. It has nothing to do with the convex hull being not closed. – alden Oct 28 '17 at 1:10
• Perhaps a line with a point outside it is an easier example. – A.Γ. Oct 28 '17 at 1:16
• @A.Γ. There is no difference, and it has more points. So, "easier" is moot. – alden Oct 28 '17 at 1:20