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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.

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See Counterexamples in topology by Lynn Arthur Steen,J. Arthur Seebach page 102. –  Jonas Teuwen Aug 16 '11 at 10:49

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up vote 3 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.

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