# Proof verification: An open connected subset of $\mathbb R^2$ is path connected.

Let $$U$$ be an open connected subset of $$\mathbb R^2$$, $$x_0\in U$$, and $$S=\{x\in U: \text{there exists a path from x_0 to x}\}.$$ We show that $$S$$ is clopen in $$U$$, and because $$U$$ is connected, is therefore equal to $$U$$. First let $$x\in S$$, then because $$U$$ is open, there exists a neighborhood (i.e. open ball) $$V$$ of $$x$$ with $$V\subset U$$. Since neighborhoods are path connected, there exists a path from $$x$$ to $$y$$, so by the pasting lemma there exists a path from $$x_0$$ to $$y$$. Hence $$V\subset S$$, so that $$S$$ is open in $$U$$.

Now let $$x$$ be a limit point of $$S$$. Then each neighborhood $$V$$ of $$S$$ intersects $$S$$ in a point $$y\ne x$$. Since $$y\in S$$, there exists a path from $$x_0$$ to $$y$$, and since $$V$$ is path connected, there exists a path from $$y$$ to $$x$$. So again by the pasting lemma, there exists a path from $$x_0$$ to $$x$$. It follows that $$x\in S$$, so that $$S$$ is closed in $$U$$.

I don't see any error in my proof, but I did not use the fact that $$U$$ is open in showing that $$S$$ is closed, so I am unsure of its validity.

As an aside, this same argument (provided it is correct) can be used for $$\mathbb R^n$$, right?

• You did use the fact that $U$ is open, and yes the proof works for $\mathbb R^n.$ Nov 16, 2019 at 2:59
• I used the fact that $U$ is open in showing that $S$ is open, but not in showing that $S$ is closed. Or are you saying that $S$ is closed even if $U$ is not open? Nov 16, 2019 at 3:09
• No, I am saying that you need the fact that $U$ is open in the first part of your proof. In fact, once you know path components of $U$ are open, which is what you proved, then you are done because then since $U$ the disjoint union of its path connected components, the complement of any component $C$ is open so $C$ is closed. Nov 16, 2019 at 4:22
• The complement of any path component is open because it is the union of the other path components, correct? Write this as an answer and I will accept it, thanks. Nov 16, 2019 at 4:33
• Yes, indeed. Why don't you answer your own question? After all, you did most of the work. I will be happy to upvote it. Nov 16, 2019 at 4:56

We show that $$S$$ is clopen in $$U$$, and because $$U$$ is connected, is therefore equal to $$U$$. First let $$x\in S$$, then because $$U$$ is open, there exists a neighborhood (i.e. open ball) $$V$$ of $$x$$ with $$V\subset U$$. Since neighborhoods are path connected, there exists a path from $$x$$ to $$y$$, so by the pasting lemma there exists a path from $$x_0$$ to $$y$$. Hence $$V\subset S$$, so that $$S$$ is open in $$U$$.
Now, $$U$$ can be partitioned into its path components, and $$S$$ is a path component of $$U$$. It follows that $$U\setminus S$$ is open, as the union of the other path components of $$U$$ which also are open. Therefore $$S$$ is closed, and since $$U$$ is connected, we conclude that $$S=U$$; in other words, $$U$$ is path connected.
• When you said "Therefore $U$ is closed" in the second part, it should be $S$ is closed, in order to be $S$ clopen in $U$ and therefore, $S=U$. Amazing proof by the way! :) May 28, 2022 at 20:30