A question about convex open set in a topological vector space.

Question

Supose $E$ is a topological vector space(may not be Hausdorff). $U\subset E$ is an open set such that $U+U=2U$.

How to show $U$ is convex?

I can see if $E$ is $T_1$，then $E$ should be Hausdorff. and every line segment in $U$ is compact, then closed. and more,then I can show the line segment is in $U$.

What about $E$ not $T_1$?

Answer 1

Let $x,y\in U$ and $\lambda_0\in (0,1)$. We need to show that $\lambda_0 x+(1-\lambda_0)y\in U$, as well.

At first we observe that $x+y\in U+U=2U$, and hence $\frac{1}{2}(x+y)\in U$. Similarly $$ \frac{1}{4}x+\frac{3}{4}y=\frac{1}{2}\left(\frac{1}{2}(x+y)+y\right)\in U, $$ and in general it can be shown that $$ \frac{m}{2^n}x+\left(1-\frac{m}{2^n}\right)y\in U, \quad\text{for all $m,n\in\mathbb N$ and $x,y\in U$.} \tag{$\star$} $$ But as $\lambda\mapsto \lambda x+(1-\lambda)y$ is continuous from $\mathbb R\to X$, and as $U$ is open and $x\in U$, there exists an $\varepsilon>0$, such that $$ \{(1-t)x+ty: \lvert t\rvert<\varepsilon\}\subset U. $$ We shall now show that $\lambda_0 x+(1-\lambda_0)y\in U$. Pick $\mu,\nu\in\mathbb N$, so that $$ \frac{\mu}{2^\nu}\in (\lambda_0,\lambda_0+\varepsilon). $$ This is possible, as this type of rationals are dense in $\mathbb R$.

Then $$ \lambda_0 x+(1-\lambda_0)y= \left(\lambda_0 x+\Big(\frac{\mu}{2^\nu}-\lambda_0\Big)y\right)+\Big(1-\frac{\mu}{2^\nu}\Big)y\\ =\frac{\mu}{2^\nu}\left(\frac{2^\nu\lambda_0}{\mu} x+\Big(1-\frac{2^\nu\lambda_0}{\mu}\Big)y\right)+\Big(1-\frac{\mu}{2^\nu}\Big)y. $$ But $\mu,\nu$ can be chosen so that $\,\,\big\lvert\frac{2^\nu\lambda_0}{\mu}-1\big\rvert<\varepsilon,\,$ as well, and hence $$ x_1=\frac{2^\nu\lambda_0}{\mu} x+\Big(1-\frac{2^\nu\lambda_0}{\mu}\Big)y \in \{(1-t)x+ty: \lvert t\rvert<\varepsilon\}\subset U, $$ and thus, so does $$ \lambda_0 x+(1-\lambda_0)y=\frac{\mu}{2^\nu}x_1+\left(1-\frac{\mu}{2^\nu}\right)y $$ as in $(\star)$.

Note. The fact that $U$ is open can not be omitted, as shown in the following example. Let $E=\mathbb R$, then $\mathbb Q+\mathbb Q=2\mathbb Q$, but $\mathbb Q$ is not a convex subset of $\mathbb R$.

Answer 2

Take any $x,y\in X$ and let $l=\{\lambda x +(1-\lambda) y: \lambda \in \mathbb{R}\} .$ Th e set $A=l\cap U$ is open in $l$ ( with induced topology) and $A+A =(l\cap U) +(l\cap U)\subset l\cap (U+U) =2(l\cap U)=2A.$ Hence the set $A$ is convex.