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A topological space is said to be locally compact if each point $x\in X$ has at least one neighbourhood which is compact. Prove continuous image of a locally compact space is not necessarily locally compact.

Attempted solution:

I was able to find a counter example I believe:

$g:(X',\tau')\to(Y',\tau'')$ where $\tau$ is the discrete topology and $\tau_2$ is the indiscrete topology where $Y'$ is infinite, hence it is not compact(there is no finite open set but $\emptyset$).

However I am striving to see why the image is not necessarily locally compact.

Admitting $f:(X,\tau)\to (Y,\tau_1)$ is a continuous function. Then if $y\in Y$ then $f^{-1}(y)=x$ since $(X,\tau)$ is locally compact then $x\in U$ such that $U$ is compact. Them image of a compact set is a compact set so $y\in f(U)$ so that $f(U)$ is compact. So the image is locally compact.

Question:

1) Since the author states not "necessarily compact". What am I missing?

2) Is my counterexample right?

Thanks in advance!

José Carlos Santos
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Pedro Gomes
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  • What is your definition of compact and locally compact? The indiscrete space is compact (although not Hausdorff compact) just like any space with finite number of open subsets. Note that in the indiscrete space also whole $X$ is open. – freakish Dec 26 '18 at 12:15
  • You may want to read this as well: https://math.stackexchange.com/questions/1287344/continuous-image-of-a-locally-compact-space-is-locally-compact – freakish Dec 26 '18 at 12:17
  • @freakish But I have assumed $Y'$ to be infinite in order to avoid the possibility there would be an open covering that is $Y'$ itself. – Pedro Gomes Dec 26 '18 at 12:23
  • @freakish The link you provided the function is assumed to be surjective. Here there is no assumption whatsoever regarding the function but it is continuous. – Pedro Gomes Dec 26 '18 at 12:25
  • @freakish I am using the cover definition for compactness and locally compactness is defined in the beginning of the post. – Pedro Gomes Dec 26 '18 at 12:26
  • every space that has finite number of open subsets is compact. Simply because there are no infinite coverings. Also in every space $Y$ itself is open. As for the surjective function: every function is surjective onto its own image. – freakish Dec 26 '18 at 13:48

2 Answers2

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Your counterexample is wrong because the indiscrete topology is locally compact in your definition, but it can be fixed:

If $\tau_d$ is the discrete topology on a set $X$ then $f(x)=x$ is continuous as a map between $(X,\tau_d)$ to $(X,\tau)$ where $\tau$ is any topology on $X$ we like. This observation does hold. Also, $(X,\tau_d)$ is locally compact as every point $x$ has the compact neighbourhood $\{x\}$ in $\tau_d$.

So let $(X,\tau)$ be any non-locally compact topology. (Like $\mathbb{Q}$ in the standard topology inherited from $\mathbb{R}$ (or its order)) and use that special case. The same "example template" can also be used for metrisable spaces, locally connected spaces etc.

So you then have shown that the continuous image of a locally compact space is not necessarily locally compact (you have at least one concrete example where this is not the case). E.g. the continuous image of a compact space is "necessarily compact", because there is a theorem that says so.

Henno Brandsma
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  • How is $\mathbb{Q}$ not locally compact with the Euclidian distance? It has been hard for me to understand that. Thanks for your answer! – Pedro Gomes Dec 26 '18 at 12:56
  • @PedroGomes did you try searching this site? There are loads of answers that address this question. Using sequences: in any open rational interval there are rational sequences converging (in the reals) to an irrational number. – Henno Brandsma Dec 26 '18 at 13:01
  • I have understood the example. However there is still a problem regarding the my attempted proof. It contradicts the example and I do not understand why. I do not know how to reconcile them. What is failing? How should I correct it? thanks in advance! – Pedro Gomes Dec 26 '18 at 18:54
  • @PedroGomes there is no contradiction. You happen to give an example where the image is locally compact and I give one where it is not. The question was to show that it was not necessarily (always) locally compact, so my example was the one you need. Yours shows that the image could be locally compact as well. – Henno Brandsma Dec 26 '18 at 18:59
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Your example is wrong because every subset is compact in indiscrete topology. Also in the proof that follows $f(U)$ is compact but it need not be a neighborhood of $y$.

Kavi Rama Murthy
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