I'm trying to prove that the radial plane is not locally compact.
I assume on the contrary that is locally compact and take any arbitrary point in it (say 0). Now 0 has a compact neighborhood $K$. $\Big[$Then $K$ contains a circle (or part of a circle) $C$ intersecting 0$\Big]$. Since $C$ is closed and $K$ is compact, $K \cap C=C$ is compact. But the subspace topology of $C$ is discrete in the radial plane. Thus $\{\{x\}:x\in C\}$ is an open cover of $C$ that has not a finite subcover.
I could not prove nor disprove what in brackets, can anyone help?
If my claim is not right, then can we prove the theorem using a similar idea?
Best Answer
Since for all $(x,y) \in \mathbb{R}^2$ and $r > 0$ the circle $$C_r (x,y) = \{ (u,v) \in \mathbb{R}^2 : (x-u)^2 + (y-v)^2 = r^2 \}$$ of radius $r$ centered at $(x,y)$ is a closed discrete subset of the radial plane, it suffices to show that for each open set $U$ and each $(x,y) \in U$ there is an $r > 0$ such that $C_r$ has infinite intersection with $U$. (Then if $K$ is a compact neighborhood of $(x,y)$ there is an $r>0$ such that $C_r(x,y) \cap \operatorname{Int} (K) \subseteq C_r(x,y) \cap K$ is infinite. This latter intersection cannot be compact, because it is infinite and discrete. But it must be compact, being a closed subset of the compact $K$. Because of this contradiction, $(x,y)$ cannot have a compact neighborhood.)
By definition of the topology, for each $\theta \in [0,2\pi)$ there is an $r_\theta > 0$ such that $(x+y) + t ( \cos \theta , \sin \theta ) \in U$ for all $0 \leq t < r_\theta$. Then there must be an $n > 0$ such that $\{ \theta \in [0,2\pi) : \frac{1}{n} < r_\theta \}$ is infinite (even uncountable). As $(x,y) + \frac{1}{n} ( \cos \theta , \sin \theta ) \in U$ for all of these $\theta$ it is clear that the circle $C_{1/n} (x,y)$ has infinite (even uncountable) intersection with $U$.