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Consider $\mathbb{R}_\ell$ be the the 'Sorgenfrey line':

Real line with the topology constructed from the intervals $\{[a,b):a<b\}$.

Prove that $\mathbb{R}_\ell$ is not locally compact.

Don Fanucci
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3 Answers3

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I show here that every compact subset of $\mathbb{R}_\ell$ is countable, so has empty interior in particular. So there are no non-empty open sets inside of some compact set, so whatever your definition of local compactness, $\mathbb{R}_\ell$ fails it.

Henno Brandsma
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  • Does non-existence of compact sets with compact closure imply that all notions of local compactness fail? I mean, this space is Hausdorff, so they are all equivalent, but in general, there are spaces with bases of compact neighbourhoods, but such that the closure of an open set is never compact if it's not empty, no? – tomasz Aug 31 '20 at 08:51
  • @tomasz well, I showed there is no non-empty open subset that is even a subset of a compact set. And that kills all versions of local compactness that I know about. – Henno Brandsma Aug 31 '20 at 16:34
  • Ah, yes, this statement does it, I guess. :) – tomasz Sep 01 '20 at 10:51
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Suppose it is. Then every $x$ has a compact nbd $K$ such that $x \in [a,b) \subseteq K \subseteq \mathbb R$. Since $\mathbb R_l$ is $T_2$ so $K$ is a compact, $T_2$ space. Since $[a,b)$ is closed in $\mathbb R_l$, so $[a,b) \cap K=[a,b)$ is closed in $K$. But in a compact $T_2$ space, a set is closed iff compact. So $[a,b)$ is compact in $K$ and thus in $\mathbb R_l$, which is absurd.

ZSMJ
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Suppose it is. Then every $x$ has a compact nbd $K$ such that $x \in [a,b) \subseteq K \subseteq \mathbb R$. But for a suitable $n$, $$(-\infty,a) \cup\big(\bigcup_{i=n}^{\infty}[a,b-\frac{1}{i})\big)\cup[b,\infty)$$ is an open cover of $K$ with no finite subcover, which is absurd.

ZSMJ
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