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Let $(X,\tau)$ be a topological vector space, $D \subset X$ and $(D, \tau_{D})$ a topological space. We are working on this subspace topology.

Suppose we have a closed convex set $C \subset D$, which has a non-empty interior, $C^{\mathrm{o}}$.

If $x \in \partial C$ and $y \in C^{\mathrm{o}}$, then for any $\lambda \in (0,1)$, is $\lambda x + (1-\lambda)y \in C^{\mathrm{o}}$?

I think this is true, but I am not sure how to prove it. Both answers to this (especially the answer which is not accepted) Closure of interior of closed convex set may be helpful.

Namch96
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The point $y$ is in the interior of $C$, hence for sufficiently small $\varepsilon>0$, the $\varepsilon$-ball around $y$ is also in $C$. As $C$ is convex, the line from any point in the $\varepsilon$-ball to $x$ is also in $C$. The union of all these lines forms a kind of cone that is part of $C$. For any point on the line from $x$ to $y$ that is not $x$ itself a small ball around this point is inside this cone and thus in $C$. This proves that any point on this line apart from $x$ is not on the boundary of $C$.

quarague
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  • This certainly seems like the correct way of going about proving it. Thank you. – Namch96 Jan 10 '20 at 16:29
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    Is there a more rigorous way to show this. This kind of cone doesn't seem mathematically very solid to me, even though I understand the argument. – Felix Crazzolara Mar 08 '20 at 20:52
  • Hi Felix, actually I think this result may not be true. There is also this answer which has 8 upvotes https://math.stackexchange.com/q/20473 which I think is not true too. It says If $C$ is a convex subset of a topological space, and if $x\in \mathrm{int} C$ and $y\in \mathrm{cl} C$, then $[x,y)\subset \mathrm{int} C$ (where the half-open segment $[x,y)$ is a right-open convex combination). – Namch96 Mar 19 '20 at 14:51