I'm trying to show the Moore plane $M$ (or Niemytzki plane) is not locally compact (I'm told it isn't).

My guess is problems will arise somehow when considering compact neighbourhoods for a point $(p,0) \in M$, since the basic open sets around points in $\{(x,y)\in M \mid y>0\}$ behave just like open balls in Euclidean space.. But I'm not seeing it.

  • 3
    See Counterexamples in topology by Lynn Arthur Steen,J. Arthur Seebach page 102. – Jonas Teuwen Aug 16 '11 at 10:49
up vote 4 down vote accepted

Suppose $(0,0)$ (or any other point on the x-axis) has a compact neighbourhood $\mathcal{N}$. Then $\mathcal{N}$ contains a basic open neighbourhood $\mathcal{U}$, say $\mathcal{U}$ is $(0,0)$ along with the open disc of radius $r$ centred at $(0,r)$. Since the Moore plane is Hausdorff and $\mathcal{N}$ is compact, $\mathcal{N}$ must be closed. Thus $\mathcal{N}$ contains the boundary $\operatorname{bd}(\mathcal{U})$ of $\mathcal{U}$. Thus $\operatorname{bd}(\mathcal{U})$ is also compact. But $\operatorname{bd}(\mathcal{U})$ is the circle radius $r$ centred at $(0,r)$ with the point $(0,0)$ deleted. In particular, it's contained in the complement of the x-axis, which is homeomorphic to the upper half-plane with the usual topology. But of course a circle with a point deleted is not compact in the usual topology.

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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