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Show that if $U$ is an open connected subspace of $\mathbb{R}^2$, then $U$ is path connected. (Hint:Show that given $x_0 \in U$, the set of points can be joined to $x_0$ by a path in $U$ is both open and closed in $U$.)

This should not be too difficult, but I am stuck at some point.

Let $x_0 \in U$

Let $A = \{ x \in U | \text{there is a path connecting $x$ and $x_0$} \}$ $\subset U$.

I want to show that $A$ is open, so let $x \in A$. By openness of $U$, we can find a basic open set $\prod_{i=1}^2 (a_i,b_i)$ such that $\prod_{i=1}^2 (a_i,b_i) \subset U$. But I am stuck in showing that $\prod_{i=1}^2 (a_i,b_i)$ lies entirely in $A$.

Daniel
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    The keywords are: locally path-connected. – Julien Mar 03 '13 at 02:00
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    (Just a side comment, not really gonna help much.) If my memory serves me well, you can prove more than that: given any two points in U, you can connect them with a path which can be broken up in (finitely many) line segments that are parallel to the axes. Also, you can prove this result in general in $\mathbb{R}^{n}$ without additional difficulty. – Detached Laconian Mar 03 '13 at 02:41

2 Answers2

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Let $x\in A$. Then take an open ball $B$ containing $x$ small enough such that $B\cap U = B$. We know that open balls are path connected (you can easily construct a path), so $B\subseteq A$, and $A$ is open. To show that $A$ is closed, consider $x\in U\setminus A$ and show that there exists an open ball $B$ containing $x$ such that $B\cap A = \emptyset$ (hint: what would happen if $p\in B\cap A$?)

Stahl
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To show $A$ is closed:

$x\sim y$ iff $x$ and $y$ are path connected

is an equivalence relation on $U$.

$U-A$, as a union of open subsets, is therefore open.