Quotient of a locally compact space need not be locally compact - the non Hausdorff case

Question

In my search for a simple example that proves that a quotient space of a locally compact space need not be locally compact, I stumbled on this previous entry:

Closed image of locally compact space

Although the example is perfectly clear to me, the definition for local compactness that is used is not the same as the one I yield. Clearly, the poster uses as definition that for any neighbourhood U, its closure should be compact. I however use the definition that any point should have a compact neighbourhood, or, equivalently, a fundamental set of compact neighbourhoods. I believe that the definition mentioned in the aforementioned link is called 'strong local compactness' in some textbook (Counterexamples in Topology?). And I thought this strong local compactness only implies local compactness in the Hausdorff case (correct me if I'm wrong).

I was looking for a more simple example based on this one, refined to suit my definition, so I came up with something along this: Put $q:\mathbb{R}\longrightarrow Y:=\mathbb{R}|R$ the canonical quotient map where $R$ identifies all natural numbers.

I want to prove that no neighbourhood $U$ can be compact. Without loss of generality, $q^{-1}(U)$ should contain a set $\displaystyle\bigcup_{n\in\mathbb{N}}]n-\varepsilon_n,n+\varepsilon_n[$ with any $\varepsilon_n \in \left]0,\dfrac{1}{2}\right[$ (to avoid overlap). I rewrite this set as $\displaystyle\bigcup_{n\in\mathbb{N}}\Big(]n-\varepsilon_n,n+\varepsilon_n[\setminus\{n\}\Big) \bigcup \mathbb{N}$, and then consider the sets $$V_n := \displaystyle\bigcup_{n\in\mathbb{N},n\neq m}\Big(]n-\varepsilon_n,n+\varepsilon_n[\setminus\{n\}\Big) \bigcup \mathbb{N}$$

Putting $(q(V_m))_{m\in\mathbb{N}}$, I suppose this should be an open cover of $\overline{0}$ in the quotient space without finite subcover. I'm not sure if this does the trick, though.

score 1 · Accepted Answer · edited May 20 '18 at 07:31

1

The reals with the rationals identified as a point.

Let $U$ be an open set with $\mathbb{Q} \subseteq U$.
Let $K$ be a compact set with $[U] \subseteq K$.
As $K$ is unbounded, let $(x_k)_k$ be an increasing
sequence of irrationals without limit into $K$.

For all $j \in \mathbb{N}$, let $U_j = R - \{x_k : j \leq k \}$.
As $U_j$ is open and contains $\mathbb{Q}$, $[U_j]$ is open.

$\{[Uj] : j \in \mathbb{N} \}$ covers $K$. If there is a finite subcover,
then, as the $U_j$'s are increasing, there's some $k$ with $[U_j]$
covering K which cannot be.

edited May 20 '18 at 07:31

answered Mar 30 '17 at 08:38

William Elliot

18,025

Yes of course! Much more simple! I was a bit overblown by the aforementioned example! – Werner Mar 30 '17 at 08:42
Can you provide a quick argument why this quotient is not locally compact? – May 17 '18 at 21:33
@Math_QED. Consider the point Q in the quotient space. An open nhood U of Q misses a countable number of irrationals. How is there a compact set containing U? – William Elliot May 18 '18 at 01:44
I don't see how this argument shows the desired result sorry. – May 19 '18 at 08:58
@Math_QED. Is the edit convincing? – William Elliot May 20 '18 at 06:48
Before I look at the content of your answer, is my edit of your post correct? It makes things easier to read :) – May 20 '18 at 07:34
@Math_QED. Correct. – William Elliot May 20 '18 at 11:39
@WilliamElliot how and why can we say that $K$ is unbounded? It lives in $\mathbb{R} / \mathbb{Q}$, so how is there an order on this space? – Geigercounter May 22 '22 at 09:29

Quotient of a locally compact space need not be locally compact - the non Hausdorff case

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