Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If $X$ is locally compact and $f : X \rightarrow Y$ is continuous closed and surjective, must $Y$ be locally compact? This seems like it should be relatively simply to answer, but I am unable to find either a proof or a counterexample. Any ideas?

share|cite|improve this question
Is $X$ assumed to be Hausdorff? – Jonas Meyer Mar 8 '12 at 20:01
Am I correct to assume that $f$ is also assumed continuous? Otherwise a counterexample (inspired by the recently deleted answer) would be given by $f:\mathbb R\to\mathbb Q$, $f(x)=x$ for $x$ rational, $f(x)=0$ for $x$ irrational. (A trivial counterexample if Hausdorff isn't assumed comes from letting $X$ be an indiscrete space.) – Jonas Meyer Mar 8 '12 at 20:12
Yes, lets add continuity. That rules out indiscrete spaces since the only way $f $ can be continuous if $X$ has the trivial topology is if $Y$ also has the trivial topology. – user15464 Mar 8 '12 at 20:14
If you leave out Hausdorffness for $X$ you need to define what you mean by "locally compact", as the various definitions are not equivalent in this case. – Najib Idrissi Mar 8 '12 at 20:31
$X$ is locally compact if for each $x\in X$ there is a compact $K\subset X$ such that $x$ is in the interior of $K$. – user15464 Mar 8 '12 at 20:54
up vote 4 down vote accepted

Here is a counterexample.

First note that $\mathbb{R}$ is locally compact.

Consider the quotient space $Y = \mathbb{R} / \mathbb{N}$ (i.e., identifying all natural numbers to a point $*$). Note that the quotient mapping $f : \mathbb{R} \to Y$ is closed (and continuous). (This essentially follows because we are identifying a discrete subset of $\mathbb{R}$.)

Claim: $Y$ is not locally compact.

proof: If $U$ is a neighbourhood of $*$, we may without loss of generality assume that it is of the form $$U = \bigcup_{n\in \mathbb{N}} ( (n-\varepsilon_n , n + \varepsilon_n ) \setminus \{ n \} ) \cup \{ * \},$$ where $\varepsilon_n < \frac{1}{4}$ for all $n$. It follows that $$\overline{U} = \bigcup_{n\in \mathbb{N}} ( [n-\varepsilon_n , n + \varepsilon_n ] \setminus \{ n \} ) \cup \{ * \}.$$ For each $m \in \mathbb{N}$ define the open set $V_m$ to be $$\left( \bigcup_{n < m} ( ( n - \varepsilon_n - \frac{1}{4} , n + \varepsilon_n + \frac{1}{4} ) \setminus \{ n \} ) \right) \cup \left( \bigcup_{n > M} ( (n-\varepsilon_n , n + \varepsilon_n ) \setminus \{ n \} ) \right) \cup \{ * \}.$$ It is clear that $\{ V_m : m \in \mathbb{N} \}$ is an open cover of $\overline{U}$, however it has no finite subcover, and so $\overline{U}$ is not compact. $\Box$

share|cite|improve this answer
Note that $\bigcup_{n=1}^\infty ( (n-\frac{1}{n} , n+\frac{1}{n} ) \setminus \{ n \} ) \cup \{ * \}$ is a neighbourhood of $*$ that is disjoint from $\{ n + \frac{1}{n} : n \geq 1 \}$. (Or a simple modification depending on what your $\mathbb{N}$ is.) – arjafi Mar 8 '12 at 21:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.