A line segment in the closure of a convex subset of a topological vector space

Question

I am trying to solve the following problem. Let $C$ be a convex subset of a topological vector space $X$. Let x be in the interior $C^\circ$ and let y be in the closure $\bar{C}$. I am asked to prove that in the line segment \begin{equation} z(\lambda)= \lambda x + (1-\lambda) y \end{equation} points with $\lambda \in (0,1]$ all belong to the interior $C^\circ$.

If another point $x^{'}$ is in the interior, we have the open set $\lambda N_x + (1-\lambda) N_{x^{'}}$ to cover the points in between $\lambda x + (1-\lambda) x^{'}$ for $\lambda\in [0,1]$ ($N_x$ and $N_{x^{'}}$ are open neighborhoods of $x$ and $x^{'}$ in $C$). Therefore the preimage $z^{-1}(C^\circ)$ is an open and convex (connected) subset of $[0,1]$, which contains $1$ by definition. I want to rule out the possibility that for some $t>0$, $z([0,t))$ is contained in the closure $\bar{C}$ but not in $C$, presumably using the fact that $y\in \bar{C}$ and convexity of $C$. I don't know how to proceed though.

Answer 1

Here is my trial. I will try to explain my line of thought first. Preimages are a red herring here.

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Let $O$ be an open neighborhood of zero such that $x+O \subset C$. We want to prove that $z+ \lambda O \subset C$.

Since $y \in \bar C$, for every open neighborhood $U$ of zero we can find $x' \in ( y+ U) \cap C$ . The problem is to find $U$ that works. The idea is to ensure that $\lambda (x+O)+ (1-\lambda)x'$ contains $z$.

We still have to find $U$. Since $x' \in ( y+ U) \cap C$, there is $u\in U$ with $x'=y+u$. Then $$ \lambda (x+O)+ (1-\lambda)x' = z + \lambda O +(1-\lambda)u. $$ So we need that $0\in \lambda O +(1-\lambda)u$, or equivalently $ u \in -\frac{\lambda}{1-\lambda}O$. The latter set is our $U$.

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Lets start the proof.

Let $O$ be an open neighborhood of zero such that $x+O \subset C$. Since $y\in \bar C$, there is $x'\in ( y-\frac{\lambda}{1-\lambda}O) \cap C$. Let $u:=x'-y \in -\frac{\lambda}{1-\lambda}O$.

Then $$ \lambda (x+O)+ (1-\lambda)x' = z + \lambda O +(1-\lambda)u. $$ Since $(1-\lambda)u \in -\lambda O$, it follows that $0 \in \lambda O +(1-\lambda)u$, so that $z\in \lambda (x+O)+ (1-\lambda)x'$. Since $x'\in C$ and $x+O \subset C$, the open set $\lambda (x+O)+ (1-\lambda)x'$ is contained in $C$. And $z$ is an interior point of $C$.

Answer 2

It seems there is a simple proof. Let $L_t, t\in [0,1)$ be a linear homeomorphism \begin{equation} L_t(\xi) = (1-t)\xi + t y \end{equation} Clear $L_0$ is the identity map and it maps $x$ closer to $y$ in the line segment as $t$ increases.

By convexity of $C$, $L_t(C)\subset C$. Let $O_x$ be an open neighborhood of $x$ that is contained in $C$, $L_t(O_x)$ is an open set contained in $C$, so is the union \begin{equation} \cup_{t\in[0,1)} L_t(O_x) \end{equation} We have $\lambda x + (1-\lambda)y$ is contained in the open set above for $\lambda \in (0,1]$ (set $\xi=x,t=1-\lambda$), so they are all interior points of $C$.

Answer 3

Suppose that there is a $z$ in the relative interior of $[x,y]$ that is in the boundary of $C$. The boundary of $\overline{C}$ is the union of all the proper faces of $\overline{C}$. In particular, $z\in F$ for some proper face $F$ of $\overline{C}$. As the relative interior of $[x,y]$ meets $F$ and $F$ is a face of $\overline{C}$, we have that $[x,y]\subseteq F$. Since $F$ is disjoint from the interior of $\overline{C}$, which contains the interior of $C$, we conclude that $x$ is not in the interior of $C$. The resulting contradiction implies that all the points of $[x,y)$ are in the interior of $C$.