I am having a bit of trouble seeing why the following $f_\epsilon$ is continuous.
Be $M \subset \mathbb{R}^n $ bounded and closed and $f : M \to \mathbb{R}^k $ continuous, then for a $\epsilon > 0$ is: $$f_\epsilon = sup\{\delta > 0 | f(B(x,\delta))\cap M \subset B(f(x),\epsilon) \}$$ also continuous, but why is that?
P.S. the definition for continuity we're using is $\forall x \in M: \forall \epsilon > 0: \exists \delta > 0: f(B(x,\delta)) \subset B(f(x),\epsilon)$