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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?

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  • $\begingroup$ You did use the fact that $U$ is open, and yes the proof works for $\mathbb R^n.$ $\endgroup$
    – Matematleta
    Nov 16, 2019 at 2:59
  • $\begingroup$ 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? $\endgroup$
    – Math1000
    Nov 16, 2019 at 3:09
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    $\begingroup$ 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. $\endgroup$
    – Matematleta
    Nov 16, 2019 at 4:22
  • $\begingroup$ 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. $\endgroup$
    – Math1000
    Nov 16, 2019 at 4:33
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    $\begingroup$ 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. $\endgroup$
    – Matematleta
    Nov 16, 2019 at 4:56

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As advised by @Matematleta, I am revising my proof and writing it as a solution.

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.

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    $\begingroup$ 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! :) $\endgroup$
    – Fabrizio Gambelín
    May 28, 2022 at 20:30
  • $\begingroup$ Neighborhoods are in general not path connected. The point is that there exist arbitrary small path connected neighborhoods. $\endgroup$
    – Paul Frost
    May 29, 2022 at 7:07

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