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Let $S$ be a convex set. If $x\in$ int$S$ and $y\in$ cl$S$, show that relint[x,y] $\subset$ int$S$.

I easily proved this for a case where y is in the interior of S, but am stuck if y is in the boundary. S need not be closed by assumption, so I either need to contradict that or need to show that y is in S, making S closed.

user182475
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Here are some broad strokes. I think they'll lead to a complete proof rather directly. But please let me know if I'm wrong.

Since $S$ is a convex set, its closure is a convex set.

Since $cl(S)$ is a convex set, the line segment containing any two points in the set lies in $cl(S)$.

Since $x\in \text{int}(S)$, at least part of the line segment connecting x and y lies in $\text{int}(S)$.

If you suppose that a point on $\text{relint}[x,y]$ other than $y$ lies on the boundary of S, this leads to a contradiction.

Since $y$ lies on the boundary of $S$ and $x$ lies in the interior of $S$, all points on $\text{relint}[x,y]$ lie in the interior of $S$.

NicNic8
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  • I still haven't been able to produce a contradiction. – user182475 Oct 14 '14 at 04:58
  • Ok; I'll try to take a look at this today. – NicNic8 Oct 14 '14 at 15:46
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    I had another idea too, if we take a ball around $x$ of radius $\delta$ contained in $S$, move $y$ to the origin, and let $u\in$ relint[$x,y$] be $\lambda x$, we can prove that the ball around u of radius $\lambda \delta$ is contained in $S$. We need to take $p$ in $B(y, \epsilon) \cap S \cap$ relint$[x,y]$. Then we can draw a line through $z\in B(u, \lambda \delta)$ and prove that it intersects $B(x, \delta)$. Then Convexity implies that $z\in S$. – user182475 Oct 14 '14 at 17:06