A Hausdorff space which is not locally compact

Question

It is said thet the set $A=\{(x,y): x>0, y\in\mathbb{R} \}\cup \{(0,0)\}\subset\mathbb{R}^2$ is Hausdorff but not locally compact. Can anyone tell how?

Answer 1

Suppose that $U$ is a neighbourhood of $(0,0)$ in $A$ such that $\overline{U}$ (closure in $A$) is compact.

Then it's easy to see that $U$ must contain a sequence of the form $(\frac 1n, p)$ for some $p >0$, for large enough $n$, say. And this sequence does not have a convergent subsequence in $\overline{U}$ (it converges to $(0,p) \notin A$ in $\Bbb R^2$ etc.). This would contradict compactness of $\overline{U}$.

Answer 2

I think the issue here is the origin. You cannot find a compact neighborhood of $(0,0)$. To see that, imagine a ball in $A$ containing $(0,0)$. Then this ball is basically the right half of the ball in $\mathbb{R}^2$ but missing the part of the boundary that lies on the y-axis (excluding the origin of course), which is not a compact set.