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

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?

• No, S is not locally compact because of (0,0). – William Elliot Oct 27 '19 at 1:52
• – Mirko Oct 27 '19 at 5:03

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.