For continuous functions, preimage of open set is open.

Let $f$ be a continuous function from a metric space $X$ into $Y$. If $V\subset Y$ and $V$ is open, then show that $f^{-1}(V)$ is open.

The proofs I've seen of the fact that open sets have open preimages either use the fact that continuous functions map limit points to limit points, or they use a completely topological proof.

Is there a more basic metric feeling proof? Something that just uses the basic definition of open sets, and the basic definition of continuity? Or are these sequential/topological arguments the only arguments to make?

Solution 1:

Let $f$ be a continuous function from a metric space $X$ into $Y$. If $V\subset Y$ and $V$ is open, then we shall prove that $f^{-1}(V)$ is open.

Suppose that $p\in X$ and $f(p)\in V$. Since $V$ is open, there exists $\varepsilon>0$ such that $y=f(x) \in V$ if $d_{Y}(f(x),f(p))<\varepsilon$, and since $f$ is continuous at $p$, there exists $\delta>0$ such that $d_{Y}(f(x),f(p))<\varepsilon$ if $d_{X}(x,p)<\delta$. Thus $x\in f^{-1}{(V)}$ as soon as $d_X(x,p)<\delta$.