An example of a not locally compact space in $\mathbb R^2$

Question

Are the two subspaces $X$ and $\operatorname{cl}(X)$ of Euclidean space $\mathbb R^2$ locally compact?

$$X = \{(x,\sin 1/x) \mid 0 < x \le 4\}\cup\{(x,\sin 1/x) \mid -4 \le x \lt 0\} \cup \{(0,0)\}$$

Clearly since $\operatorname{cl}(X)$ is closed and bounded so $\operatorname{cl}(X)$ is compact and thus locally compact.

However, I'm having trouble proving that $X$ is not locally compact. My conjecture is that I need to show there is no open set $U$ and compact set $C$ such that $(0,0) \in U \subset C$. But I can't come up with a rigorous proof for this part. How can I construct a written proof of this?

Answer 1

Hint. In a compact set in the plane every sequence has a convergent subsequence. For convenience, instead of an open ball, consider a rectangular neighborhood of $(0,0)$, $U=[-\epsilon,\epsilon]^2$ where $\epsilon>0$. Take $0<\delta<\epsilon$ and try to come up with a sequence $x_n\to 0$ such that $\delta=\sin1/{x_n}$ for all $n$. What does the sequence of points in the plane $(x_n,\sin1/{x_n})$ converge to? Is the limit of this sequence going to belong to $X$?

Answer 2

Every open nbhd of the origin contains a set of the form

$$U_\epsilon=X\cap\Big((-\epsilon,\epsilon)\times(-\epsilon,\epsilon)\Big)$$

for some $\epsilon>0$. Show that the closure of this set in $X$ is

$$\operatorname{cl}_XU_\epsilon=X\cap\Big([-\epsilon,\epsilon]\times[-\epsilon,\epsilon]\Big)\;.$$

Then let $a=\dfrac{\epsilon}2$ and find a sequence in

$$\big(\Bbb R\times\{a\}\big)\cap\operatorname{cl}_XU_\epsilon$$

that has no convergent subsequence in $X$ (because in $\Bbb R^2$ it converges to a point not in $X$).