Is the following subset of $\mathbb{R}^2$ locally compact?

Question

Consider the set $\mathcal{S}$ defined as: $$ \mathcal{S} = \{(x,y)\in\mathbb{R}^2\lvert x^2-y<0, y< a \}\cup\{(0,0)\} $$ where $a > 0$. This set is not open nor closed. However, I'm confused regarding whether it is locally compact or not. The point $(0,0)$ obviously is the issue. It's not hard to show that it belongs to a relatively compact neighborhood in $\mathcal{S}$. In particular, a relatively open neighborhood containing $(0,0)$ is relatively compact (right?). Hence it satisfies the wiki definition here: https://en.wikipedia.org/wiki/Locally_compact_space#Formal_definition

So, is it locally compact?

Answer 1

Assume that $\mathcal S$ is locally compact. Then $(0,0)$ must have a compact neighborhood in $\mathcal S$. That is, there exists an open $V \subset \mathcal S$ and a compact $K \subset \mathcal S$ such that $(0,0) \in V \subset K$. We can write $V = W \cap \mathcal S$ with an open $W \subset \mathbb R^2$. Choose $\epsilon$ with $\min(4a,1) > \epsilon > 0$ such that $(x,y) \in W$ for $\lVert (x,y) \rVert < \epsilon$. Then $\xi_n = (\frac{\epsilon n}{2(n+1)},\frac{\epsilon^2}{4})$ is contained in $\mathcal S$ because $\frac{\epsilon^2}{4} < \frac{\epsilon}{4} < a$ and $\frac{\epsilon^2 n^2}{4(n+1)^2} - \frac{\epsilon^2}{4} < \frac{\epsilon^2}{4} - \frac{\epsilon^2}{4} = 0$. Moreover $\lVert \xi_n \rVert^2 = \frac{\epsilon^2 n^2}{4(n+1)^2} + \frac{\epsilon^4}{16} < \frac{\epsilon^2}{4} +\frac{\epsilon^2}{16} < \epsilon^2$, thus $\xi_n \in W \cap \mathcal S = V \subset K$. But $\xi_n \to \xi = (\frac{\epsilon}{2},\frac{\epsilon^2}{4})$. Since $K$ is compact, we conclude $\xi \in K \subset \mathcal S$. But clearly $\xi \notin \mathcal S$ because $(\frac{\epsilon}{2})^2 - \frac{\epsilon^2}{4} = 0$.

This is a contradiction which proves that $\mathcal S$ is not locally compact.