Let $X$ and $Y$ be topological spaces, $X$ locally compact and $f:X\to Y$ a function. Prove or disprove:
a) If $f$ is continuous, then $f(X)$ is locally compact.
b) If $f$ is continuous and open, then $f(X)$ is locally compact.
I proved b), therefore a) must be false.
So a counterexample is needed to show a) is false.
What could be the counterexample?
Please any help or suggestion.
Best Answer
Hint: Try to think of a discrete space and a non-locally compact space. Note that any function from a discrete space is continuous (why?), hence you can construct a function from the discrete space to the non-locally compact space and it will be a counterexample.
Edit: A simple, well-known discrete space is $\Bbb N\subseteq \Bbb R$ with the usual topology. Hence, if we let $\{q_n\}_{n=1}^\infty$ be an enumeration of $\Bbb Q$, we can define $f:\Bbb N\to\Bbb Q$ by $f(n)=q_n$, in which case we have a continuous function from a locally compact space to a non-locally compact space.