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?

  • $\begingroup$ If $X$ were locally compact, then for every sufficiently small $r > 0$, the set $\overline{B_r(0)}\cap X$ would be compact, in particular, a closed subset of $\mathbb{R}^2$. $\endgroup$ – Daniel Fischer Nov 28 '14 at 22:38
  • $\begingroup$ Note: \bigcup is used for things like $\displaystyle\bigcup_{n=1}^\infty A_n$, and \cup for things like $A\cup B$ and $A_1\cup\cdots\cup A_n$. Also note that "a\sin b" yields $a\sin b$ and "a sin b" yields $a sin b$ without proper spacing and with "$\sin$" improperly italicized. And in $\{(x,\sin1/x)\mid 0<x\le 4\}$, the $\{\text{curly braces}\}$ are INSIDE the math environment. I did these and a variety of other edits to the question. ${}\qquad{}$ $\endgroup$ – Michael Hardy Nov 28 '14 at 22:44
  • $\begingroup$ @DanielFischer I see the existence of such closed compact balls if X were locally compact but how does the fact that such closed subsets exist lead to a contradiction? $\endgroup$ – nomadicmathematician Nov 28 '14 at 22:46
  • $\begingroup$ The point is that such sets are not closed in $\mathbb{R}^2$. $\endgroup$ – Daniel Fischer Nov 28 '14 at 22:51

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 $X$?


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


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


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 $X$).


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