# Check proof that every point in open set $E \subset\mathbb R^2$ is a limit point of $E$

Would appreciate if somebody can comment if my writing is clear and accurate.

Prove that every point in open set $E \subset\mathbb R^2$ is a limit point of $E$.

Suppose $x \in E.$ Then, there exists a neighborhood of $x$ with radius $r, N_r(x)$ such that $N_r(x) \subset E.$ Now consider any neighborhood of $x, N_s(x).$ If $s \leq r$ then $N_s(x) \subset N_r(x)$ and contains a point of $E$ other than $x.$ If $s \geq r$ clearly it contains some other point of $E$ other than $x$ as $N_r(x)$ contains a point of $E$ other than $x.$ Hence, $x$ is a limit point.

• What is $E$? What if $E$ is a finite set - say, $E = \{(2, 3)\}$? Commented Aug 21, 2017 at 6:15
• I think $E$ is meant to be an open set, in which case yeah Commented Aug 21, 2017 at 6:29
• Oh sorry!! $E$ is an open set! Commented Aug 21, 2017 at 6:31
• It might be good to start with mentioning that $N_r(x)$ contains an element $y\in N_r(x)-\{x\}$ for every $r>0$. Commented Aug 21, 2017 at 7:23
• Great idea. Thank you. Commented Aug 21, 2017 at 7:50

Fact 1. If $x$ is a limit point of a subset of $E$, then it is also a limit point of $E$ (easy proof).
Fact 2. If $s>0$, then $N_r(x)\cap N_s(x)=N_{\min\{r,s\}}(x)$ is infinite, so every neighborhood of $x$ intersects $N_r(x)$ in points different from $x$.