# Closed image of locally compact space

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?

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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

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. $\Box$
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.) –  Arthur Fischer Mar 8 '12 at 21:04