Question

Prove that a function f from a metric space X to a metric space Y is continuous if and only if the inverse image of any open subset of Y is open in X

Answer 1

1. Suppose that $f:X\to Y$ is a continuous function, $V\subset Y$ is an arbitrary open set, $U=f^{-1}(V)$ is inverse image of $V$. We are to show that $U$ is an open subset of $X$.

Let $x\in U$, thus $f(x)\in V$. Since V is an open set, there exists $\varepsilon>0$ such that the open ball $B_{\varepsilon}(f(x))\subset V$.

By the definition of continuity, there exists $\delta>0$ such that $f(B_{\delta}(x))\subset B_{\varepsilon}(f(x))\subset V$.

This means that $B_{\delta}(x)\subset f^{-1}(V)=U$. We have proved that for every $x\in U$ there exists $\delta>0$ such that $B_{\delta}(x)\subset U$. Therefore, $U$ is an open subset of $X$.

2. Conversely, suppose that $f:X\to Y$ is a such function that for every open subset $V\subset Y$ its inverse image $U=f^{-1}(V)$ is also open. Let $x\in X$ be an arbitrary point, we are to show that $f$ is continuous at $x$.

For every $\varepsilon>0$ the open ball $B_{\varepsilon}(f(x))$ is an open subset of $Y$. Hence, its inverse image, $f^{-1}(B_{\varepsilon}(f(x)))$ , must be open. This means there exists $\delta>0$ such that $B_{\delta}(x)\subset f^{-1}(B_{\varepsilon}(f(x)))$ .

Therefore $f(B_{\delta}(x))\subset B_{\varepsilon}(f(x))$ and $f$ is continuous at $x$. The assertion is proved.