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

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 $$$$z(\lambda)= \lambda x + (1-\lambda) y$$$$ points with $$\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.

• Maybe we should consider the linear homeomorphism $L_\epsilon (\zeta) = (1-\epsilon)\zeta + \epsilon y$, which maps $x$ to a point $x^{'}=(1-\epsilon)x + \epsilon y$ closer to $y$. For $\lambda\in(t,1]$, $\lambda x^{'} + (1-\lambda)y = \lambda(1-\epsilon)x+(1-\lambda(1-\epsilon)y) \in L_\epsilon(C)^\circ \subset C^\circ$. Dec 2, 2023 at 8:18

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

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 $$U$$.

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 $$C$$.

It seems there is a simple proof. Let $$L_t, t\in [0,1)$$ be a linear homeomorphism $$$$L_t(\xi) = (1-t)\xi + t y$$$$ Clear $$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 $$$$\cup_{t\in[0,1)} L_t(O_x)$$$$ We have $$\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$$.

• You need to explain why $L_t(C) \subset C$ if $y$ is in $\bar C\setminus C$.
– daw
Dec 3, 2023 at 16:37
• @daw Thank you for pointing it out. Indeed, not really a proof. Dec 4, 2023 at 13:42

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 $$C$$.