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I am reading Stromberg's Analysis and the definition of the locally compact space is given as:

A locally compact space is a topological space $X$ such that each $x\in X$ has some neighborhood $V$ such that closure of $V$, $V^{-}$ is compact.

I tried to use this definition on right ray topology $\mathcal{T}$ on $\mathbb{R}$ where

$$ \mathcal{T}=\{ (a,\infty):a\in\mathbb{R} \}\cup \{ \emptyset, \mathbb{R} \}. $$

If we take $x\in \mathbb{R}$ and the neighborhood $V=(a,\infty)$ then $V^{-}=\mathbb{R}$ as any $t\in \mathbb{R}$ is a limit point of $(a,\infty)$. However, this means I am trying to prove $\mathbb{R}$ is compact under right ray topology.

I would appreciate if you point out my mistake in here.

EDIT: I believe $\mathbb{R}$ is not compact under right ray topology unless every open covering contains $\mathbb{R}$. But this contradicts with the arbitrariness of a cover.

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What you say is correct and the space you are considering is not an example satisfying the definition.

The fact is that there are, in mathematical literature, at least three inequivalent definitions of locally compact topological space:

  1. Every point has a compact neighbourhood.
  2. Every point is contained in some relatively compact open set (i.e. every point has a relatively compact neighbourhood).
  3. Every point has a fundamental system of compact neighbourhoods.

There's also Bourbaki's, which is $(1)\lor (2)\lor(3)$. Stromberg is using the second one. The scheme of implications in general is $\text{compact}\Rightarrow (2)\Rightarrow (1)$ and $(3)\Rightarrow (1)$. The reason for this ambiguity is the fact that $T2+(1)\Rightarrow (2)\land (3)$, so that in the Hausdorff case the three conditions are equivalent (and also compactness implies them all).

What you have at hand is in fact an instance of some space which satisfies $(3)\land \neg (2)$ (and, of course, $(1)$ too): can you guess what could be a compact neighbourhood of $x$ in this topology?

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I don't see any particular mistake here, other than hesitation in following through with your convictions. You've correctly identified that the closure of any neighbourhood has to be the entirity of $\mathbb{R}$. Therefore, as you say, showing that the right-ray topology is locally compact is exactly the same as showing that it is compact.

Is there any particular reason why you were expecting a result on the compactness of the right-ray topology different than the one you actually got?

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