Prove $\epsilon$-$\delta$ definition of continuity implies the open set definition for real function

Solution 1:

Since the OP's work was reviewed already in the comments, I collect together the entire argument in case future visitors find it useful.

If $f$ is $\varepsilon$-$\delta$-continuous, then it is open-set-continuous. Suppose $f : \mathbb R \to \mathbb R$ is continuous by the $\varepsilon$-$\delta$ definition; we want to prove that it is continuous by the open sets definition.

Take an arbitrary open set $V \subseteq \mathbb R$; we want to prove $f^{-1}(V)$ is open. This is true if $f^{-1}(V)$ is empty, so assume $x \in f^{-1}(V)$. Since $f(x) \in V$ and $V$ is open, there exists some $\varepsilon > 0$ such that $(f(x) - \varepsilon, f(x) + \varepsilon) \subseteq V$. By continuity at $x$, there exists some $\delta > 0$ such that $(x - \delta, x+ \delta) \subseteq f^{-1}(V)$. That is, $x$ is an interior point of $f^{-1}(V)$. Since this is true for arbitrary $x \in f^{-1}(V)$, it follows that $f^{-1}(V)$ is open.

If $f$ is open-set-continuous, then it is $\varepsilon$-$\delta$-continuous. Suppose $f : \mathbb R \to \mathbb R$ is continuous by the open sets definition; we want to prove that it is continuous by the $\varepsilon$-$\delta$ definition.

Fix $x \in \mathbb R$ and $\varepsilon > 0$. Then $(f(x) - \varepsilon, f(x) + \varepsilon)$ is an open set in $\mathbb R$ (containing $f(x)$). By continuity, $U = f^{-1}((f(x) - \varepsilon, f(x) + \varepsilon))$ is an open set in $\mathbb R$. It's easy to see that $U$ contains $x$; then $x$ is an interior point of $U$ by openness of $U$. That is, there exists $\delta >0$ such that $(x - \delta, x+\delta) \subseteq U = f^{-1}((f(x) - \varepsilon, f(x) + \varepsilon))$. Then it follows that $f((x - \delta, x+\delta)) \subseteq (f(x) - \varepsilon, f(x) + \varepsilon)$.