How to prove that if the preimage of an open set is open the function is continuous?

Question

The definition I use for continuity is as follows: $f:X\to Y$ (X and Y are metric spaces) is continous means that: $\forall y\in X$ held constant,$\forall \varepsilon >0$ $\exists \delta$, s.t for all x satisfying $d_X(x,y)<\delta$: $d_Y(f(x),f(y))<\varepsilon$.

Let $V\subset Y$ be an open set.

I mananged to show that if f is continous then $f^-1(V)$ is an open set. However I can't manange to prove the other way around, because $d_Y(f(x),f(y))$ can be smaller than epsilon, yet $d(x,y)$ can still be greater than delta. Can someone help me prove that if $f^-1(V)$ is an open set then f is continous?

Answer 1

ou shloud not look at a fixed open set: $f$ is continuous if, for every open set $V$,$f^{-1}(V)$ is an open set. Thus, (for fixed $y)$, you look at the open set $V=B(f(y),\epsilon)$; as $f^{-1}(V)$ is open and as $y\in f^{-1}(V)$, there is a ball $B(y,\delta)$ contained in $f^{-1}(V)$.

Of course, there may be points $x$ such that $d(x,y)\geq \delta$ and $x\in f^{-1}(V)$, but you are interested only in points in the ball $B(y,\delta)$.