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

by dumbQuestion
Tags: boundary, limit, point
 P: 126 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?
 C. Spirit Sci Advisor Thanks P: 5,427 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 $U$.
 P: 126 thank you very much!
C. Spirit
Thanks
P: 5,427
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!
P: 1,622
 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.
C. Spirit
Thanks
P: 5,427
 Quote by jgens Unfortunately this does not work since open sets will not contain their boundary points.
Totally missed that detail haha. Maybe the OP didn't mean to say open but another class of subsets because he was trying to use connectedness explicitly. Anyways, I gotta go to class now so cheers!

 Related Discussions Calculus & Beyond Homework 4 Differential Equations 3 Classical Physics 0 Calculus 6 Classical Physics 0