Continuous functions and open covers

Question

Let $\{V_{i}\}$ be an open cover of $f(X)$.

Why is it that if $f$ is continuous then $f^{-1}(V_{i})$ is open?

What does the continuity of $f$ have to do with $f^{-1}(V_{i})$ been open?

Answer 1

The topological definition of continuity is that $f:X\to Y$ is continuous iff $f^{-1}V$ is open in $X$ whenever $V$ is open in $Y,$ where $f^{-1}V=\{p\in X: f(x)\in V\}.$ Continuity is an important and useful concept and has many equivalent def'ns.

Answer 2

Let $x$ be any point in $f^{-1}(V_i)$. So $f(x) \in V_i$.

As $V_i$ is open, for there is an $\epsilon > 0$ such that $B_{\epsilon} \subset V_i$.

As $f$ is continuous, there exists a $\delta$ such that $d(x,y) < \delta$ implies $d(f(x),f(y)) < \epsilon$ so $f(y) \in B_{\epsilon} \subset V_i$.

Let $B_{\delta}(x)$ be a neighborhood around $x$. Let $y \in B_{\delta}(x) \implies d(x,y) < \delta \implies d(f(x), f(y)) < \epsilon$

$ \implies f(y) \in V_i \implies y \in f^{-1}(V_i) $

$\implies B_{\delta}(x) \subset f^{-1}(V_i) \implies f^{-1}(V_i)$ is open.