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Let $H$ be the open upper half-plane in $R^2$ and consider the metric space $(H\cup\{(0,0)\}, d)$ where $d$ is the usual euclidean metric. I want to show that $(H\cup\{(0,0)\}, d)$ is not locally compact.

Suppose that it is locally compact. Then $(0,0)$ has a compact neighborhood containing an open ball $B_r((0,0))$ that of course completely lies in $H\cup\{(0,0)\}$. But then $(0,-r/2)\in B_r((0,0)) \subset H\cup\{(0,0)\}$. Contradiction.

Is this correct?

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No, it is not correct. In this context, $B_r\bigl((0,0)\bigr)$ means the intersection of the usual $B_r\bigl((0,0)\bigr)$ with $H\cup\{(0,0)\}$.

If it was locally compact, then some closed ball centered at $(0,0)$ would be compact. Suppose that $r$ is the radius of this ball. The $\left(\frac r2,\frac1n\right)$ belongs to it if $n$ is large enought. But no subsequence converges to an element of the ball. Therefore, the ball is not compact.

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