Let f: (X1,d1) -> (X2, d2) be a continous function. Let C, a subset of X2 be compact.
Is f^-1(C) ("f inverse") compact in X2?
Having trouble finding a counterexample...
I'm assuming that you mean $\displaystyle X_1$
Consider the function
$\displaystyle f: \mathbb{R}\to \mathbb{R}$ with the usual metric
$\displaystyle f(x)=1$
The one point set $\displaystyle \{ 1 \}$ is compact in the range. (It is closed and bounded)
but $\displaystyle f^{-1}(1)=\mathbb{R}$ this set is unbouned and not compact.