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I have some doubts when it comes to my approach to this problem and I would appreciate any feedback.

Let $C\subset E$ be a closed subset of a topological vector space. I want to show that $C$ is convex if and only if $\frac{1}{2}(x+y) \in C$ for all $x,y \in C$.

Since I'm using the definition that $C$ is convex if $$\forall_{x,y \in C} \forall_{\lambda \in [0,1]} \lambda x+(1-\lambda)y \in C$$ thus one direction we get automatically with $\lambda = \frac{1}{2}$. Now for the other direction my idea is the following: we fix $x,y \in C$ and $\lambda \in [0,1]$. Denote $z=\frac{1}{2}(x+y) \in C$. From the assumption we get that $\frac{1}{2}(z+x) \in C$ so $$\frac{1}{2}(z+x) = \frac{1}{2}(\frac{1}{2}(x+y)+x) = \frac{3}{4}x + \frac{1}{4}y \in C.$$ By taking $\frac{1}{2}(z+y)$ we also get that $\frac{1}{4}x+\frac{3}{4}y \in C$. Continuing this way we can prove that for every $n \in \mathbb{N_0}$ and $k=1,2,...2^n-1$ we have $$\frac{k}{2^n} x + (1-\frac{k}{2^n})y \in C.$$ Now, let's suppose for now that we already have that for every $q \in \mathbb{Q}\cap [0,1]$ $$qx + (1-q)y \in C.$$ Then using closeness of $C$ I can choose a sequence of rational numbers $q_n \rightarrow \lambda$ to get my result.

The step that is missing is how to get from having the result for $\frac{k}{2^n}$ where $n\in \mathbb{N_0}$, $k=1,2,...2^n-1$ to having it for $q \in \mathbb{Q}\cap [0,1]$. My intuition here is that we should be able to approach such $q$ with fractions $\frac{k}{2^n}$. For example $$ \sup\{\frac{k}{2^n}: n\in \mathbb{N_0}, k=1,2,...2^n-1, \frac{k}{2^n}<q \} =q $$ and by the property of supremum we can find a sequence of such elements $\frac{k}{2^n}$ approaching $q$ (or even perhaps any $\lambda \in [0,1]$, if changing $q$ for $\lambda$ in the formula above?). My doubts coming from the fact that usually the existing sequence of fractions is completely abstract, compare to the sequence $\frac{k}{2^n}$ that is very specific.

But here I'm not sure if this idea is rigorous enough, that is why I would love to hear some feedback here.

Thanks!

btfm
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    related: https://math.stackexchange.com/questions/1371147/prove-if-c-is-midpoint-convex-and-closed-then-its-a-convex-set – LinAlg Jan 29 '21 at 18:45
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    You can approach any number, not just rational numbers, in the form you need. Think about writing binary "decimals." Just write out enough digits and you can get an approximation that's as close as you like. More precisely, the closed interval centered on $r$ with diameter $\frac{1}{2^k}$ must have at least one quotient of integers with denominator $2^k$. – Robert Shore Jan 29 '21 at 19:07
  • perfect, thank you! And thanks for the related link, I completely missed that post. Is there a way of me marking my post as a duplicate of the one you posted, @LinAlg ? I'm honestly not sure how it works here exactly. – btfm Jan 29 '21 at 19:20
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    @btfm five people would need to vote on closing this question, I would not worry about it – LinAlg Jan 29 '21 at 19:24

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