# In proving every open connected subset of $\mathbb{R}^n$ is path connected

I would like to prove that every open connected subset of $\mathbb{R}^n$ is path connected.

Let us choose $E$ to be a such open connected subset, then given any point $p \in E$, we will define $F$ be the set of all points in E that can be joined to $p$ by a path in $E$.

The idea is to show that $F = E$ by showing $F$ is clopen as well as $F$ is path connected.

We choose $q \in F \subseteq E$, then since $E$ is open, we can find an open ball $D_x(q,\epsilon)$ in $E$.

Here are my questions.

1. My lecturer said $D_x(q,\epsilon)$ is path connected as any point of distance smaller than $\epsilon$ from $q$ can be connected by a polygonal (= $D_x(q,\epsilon)$ can be contained by $F$) but I am not sure how every point in $D_x(q,\epsilon)$ can be contained by $F$.

2. How does this ensure that $F$ is open?

Any help will be greatly appreciated.

1. Because if there is a path $\gamma$ from $p$ to $q$ and if $r\in D_x(q,\varepsilon)$, you can sider the path $\gamma$ followed by a line segment from $q$ to $r$.
2. Because for each $q\in F$, there is an open ball centeres at $q$ and contained in $F$.
• Thank you very much, but for 2, $F$ could be closed and $q$ might be on its boundary, in such a case we cannot find an open ball centred at $q$. – James May 31 '18 at 13:20
• @James That cannot be, because when we get to point 2., point 1. has already been proved. So, now we know that $D_x(q,\varepsilon)\subset F$. – José Carlos Santos May 31 '18 at 13:23
• @James the problem with your intuition is that you are assuming there is a "boundary" for $F$. It doesn't have to have one - in fact you are deliberately using the fact that only clopen sets are the emptyset or the whole set $E$, which do not have a boundary – user160738 May 31 '18 at 13:26
Note that $D_x(q, \epsilon)$ is convex: every pair of points in the open ball can be connected by a straight line.
In particular, if $p$ can be path-connected to $q$ ($\iff q \in F$), and $q$ can be path-connected to every point in $D_x(q, \epsilon)$, then every point of $D_x(q, \epsilon)$ is contained in $F$.