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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.