A question about convex set

Question

I need to prove the closed set $C\subseteq \mathbb{R}_{+}$ is a convex. And let $x$, $y$ be arbitrary given in $C$, I have proved that $1/2(x+y)\in C$. Then does this means $C$ is convex ?

Answer 1

If $C$ is a closed subset of $\Bbb R_+,$ then it does indeed suffice to prove that $C$ is midpoint-convex. Indeed, suppose that $C$ were not convex, so that there exist $x,y\in C$ such that $tx+(1-t)y\notin C$ for some $0<t<1$. Without loss of generality, suppose that $x<y,$ and put $z=tx+(1-t)y$, so that $x<z<y.$ Since the complement of $C$ is open and $x,y\in C,$ then $z$ lies in some open interval $(a,b)$ disjoint from $C$ with $a,b\in C.$ (Why?) But then $\frac{a+b}2\notin C,$ so $C$ is not midpoint-convex. Thus, midpoint-convex implies convex by contrapositive (convex of course implies midpoint-convex).

Now, it's worth noting that not all closed subsets of $\Bbb R_+$ will be (midpoint-)convex. Consider $\{1,2\},$ for example. Given that, I must assume that one of the following is true:

1. You've been given a particular closed set $C$ that you just haven't specified.
2. You've been told that $C$ is an arbitrary closed subset of $\Bbb R_+,$ asked to prove or disprove that it is convex, and made a mistake in your "proof" of midpoint-convexity.
3. You've been given an arbitrary closed midpoint-convex set, and asked to prove that it is convex.

In the first case, there's no problem, since midpoint-convex implies convex for closed subsets of $\Bbb R_+$ using the proof outlined above. In the second case, just use the above counterexample (and take a good hard look at your "proof" to see where you went wrong). In the third case, you need only complete the proof outlined above.