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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 $U$.)

as i found the answer here Show that if $U$ is an open connected subspace of $\mathbb{R}^2$, then $U$ is path connected but i did n't the red line

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Pliz help me,,,,,,,,

jasmine
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    What don't you understand? Do you not see why the underlined portion shows that A is closed? Do you not know why we want to show that A is closed? Do you not understand why the open ball B has to exist? Your question is very unclear. – G Tony Jacobs Apr 07 '18 at 20:14

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When you want to show that $A$ is closed, one way to do so is to show that the complement is open. So, you take some $x\in A^c = U\setminus A$ (the complement in U) and show that there exists some $\varepsilon>0$ small enough so that $B(x,\varepsilon)\subseteq U\setminus A$ (the ball centered at $x$ with radius $\varepsilon$ is contained in $U\setminus A$).

Yu Zhao
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