Open and Connectedness on the Topological Vector Space over $\mathbb{R}$ Implies Path-Connectedness

Question

I want to prove the following proposition (which is also the title).

Open and connectedness of a subset G of the topological vector space X over $\mathbb{R}$ implies path-connectedness.

I haven't proved this kind of proposition earlier, and I don't even know where to start. If $G$ is convex, then the proposition is trivial. But if $G$ is not convex, I have no idea how I can find the path between two arbitrary points. I suppose being a vector space over $\mathbb{R}$ is important, but I don't know how to use this fact. Should I instead prove that this is a locally path-connected space?

Please give me any hints. Thanks in advance.

Edit. Here is my another try. Fix $x\in G$ and let $U, V$ be $$ U:=\{y\in G|y,x\textrm{ are path-connected.}\}\\ V:=\{y\in G|y,x\textrm{ are not path-connected.}\} $$ I claim that $U$ is open. If $G$ has a open convex neighborhood $W$ of $0$ (I am not sure about the existence of $W$ here), $y+\epsilon W \subset U$ when $y\in U$(again here I am not sure about the existence of $\epsilon$). Hence $U$ is open. Similarly, if $y\in U$, $y+\epsilon W\subset V$. $y\in U$ and $U\cap V=\emptyset$, but $U\cup V=G$ and since $G$ is connected, $U=G$. Thus $G$ is path-connected.

There are some missing parts in the proof, and I don't know how I can complete it.

This is true for normed spaces; I'm not sure about arbitrary topological vector spaces though. — Math1000, May 09 '18 at 19:09
This is from the exercise of Conway so I think this is true. — J1U, May 09 '18 at 19:36
You are not told that he space is locally convex, so existence of $W$ is not guaranteed. — Kavi Rama Murthy, May 10 '18 at 00:22
That's exactly the reason why I am not sure about the existence of $W$. — J1U, May 10 '18 at 01:49
In a space like $l^{p}$ with $0<p<1$ convex neighborhoods don't exist. I think Conway should have given a hint at least since standard arguments don't work. — Kavi Rama Murthy, May 10 '18 at 08:02

score 3 · Accepted Answer · answered Jan 16 '20 at 20:48

As you mention, it is enough to prove that $X$ is locally path-connected. For this, you don't need local convexity, just local star-convexity. In other words, it is enough to know that if $U$ is any open neighborhood of $0$, then there is a neighborhood $V$ of $0$ contained in $U$ such that $cV\subseteq V$ for all $c\in [0,1]$. That way, any element of $V$ can be connected by a path to $0$ (just multiply by scalars ranging from $1$ to $0$) so $V$ is path-connected.

To prove the existence of such a $V$, you the use continuity of scalar multiplication. Since $U$ is open and contains $0$, the set $\{(t,v)\in\mathbb{R}\times X:tv\in U\}$ is open and contains $(0,0)$, and so contains an open rectangle $(-\epsilon,\epsilon)\times W$ where $W\subseteq X$ is an open neighborhood of $0$. Now let $V=\bigcup_{t\in(0,\epsilon)}tW$. Then $V$ is an open neighborhood of $0$ contained in $U$, and $cV\subseteq V$ for all $c\in[0,1]$ since $cV=\bigcup_{t\in(0,\epsilon)}ctW=\bigcup_{t\in(0,c\epsilon)}tW$.

Open and Connectedness on the Topological Vector Space over $\mathbb{R}$ Implies Path-Connectedness

1 Answers1

Linked