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...

Printable View

- Mar 15th 2009, 01:20 PMmath_helpContinous function
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... - Mar 15th 2009, 02:50 PMTheEmptySet
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. :) - Mar 16th 2009, 11:26 AMmath_help
What do you mean by saying I assume you mean X1?

- Mar 16th 2009, 11:31 AMTheEmptySet