## Boundary of an open set in R2 is a limit point?

I have kind of a simple point set topology question. If I am in ℝ2 and I have a connected open set, call it O, then is it true that all points on the boundary ∂O are limit points of O? I guess I'm stuck envisioning as O as, at least homeomorphic, to an open disk of radius epsilon. So it seems obvious that any points on the boundary would be limit points. But is that true in general?
 Recognitions: Gold Member Science Advisor Let $U\subseteq \mathbb{R}^{n}$ be connected and open and non empty. $p\in \partial U$ if and only if every neighborhood of $p$ contains both a point in $U$ (and in $\mathbb{R}^{n}\setminus U$ but we don't care about that here). Let $p\in \partial U$ and assume there exists a neighborhood $V$ of $p$ in $\mathbb{R}^{n}$ such that $V\cap U = \left \{ p \right \}$ (we know of course that $U\supset \left \{ p \right \}$). This implies $\left \{ p \right \}$ is a non - empty proper clopen subset of $U$ which is a contradiction because $U$ is connected. Thus, $p$ is a limit point of $U$.
 thank you very much!

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## Boundary of an open set in R2 is a limit point?

 Quote by dumbQuestion thank you very much!
Should work for any Hausdorff connected space and not just euclidean space as far as I can see. Was there a particular reason for this question or did it just pop into your head for fun or something= D? Cheers!

 Quote by WannabeNewton Let $U\subseteq \mathbb{R}^{n}$ be connected and open and non empty. $p\in \partial U$ if and only if every neighborhood of $p$ contains both a point in $U$ (and in $\mathbb{R}^{n}\setminus U$ but we don't care about that here). Let $p\in \partial U$ and assume there exists a neighborhood $V$ of $p$ in $\mathbb{R}^{n}$ such that $V\cap U = \left \{ p \right \}$ (we know of course that $U\supset \left \{ p \right \}$). This implies $\left \{ p \right \}$ is a non - empty proper clopen subset of $U$ which is a contradiction because $U$ is connected. Thus, $p$ is a limit point of $U$.
Unfortunately this does not work since open sets will not contain their boundary points. Luckily that observation gives us a way to fix the proof. All we now have to note is that the boundary point condition implies that every nbhd of p non-trivially intersects U - {p} = U.

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