# Proof of equivalence between open-set and $\varepsilon$-$\delta$ definition of continuity.

$\def\R{\mathbb{R}}$Here is my attempt.

Assume $f:\R \to \R$ is $\varepsilon-\delta$ continuous at $x_0$ . Let $N$ be a neighbourhood of $f(x_0)$. By definition, there exists an open interval $I$ such that $f(x_0)\in I\subseteq N$. Since $I$ is open, there exists $\varepsilon > 0$ such that $(f(x_0)-\varepsilon, f(x_0)+\varepsilon) \subseteq I \subseteq N$. $\varepsilon-\delta$ continuity at $x_0$ implies that there exists $\delta > 0$ such that $f((x_0-\delta,x_0+\delta)) \subseteq N$. Taking $f^{-1}$ of both sides implies that $(x_0-\delta,x_0+\delta) \subseteq f^{-1}(N)$, meaning $f^{-1}(N)$ is a neighbourhood of $x_0$. This is the open-set definition of continuity at $x_0$.

Assume $f:\R \to \R$ is open-set continuous at $x_0$. Let $N$ be a neighbourhood of $f(x_0)$. Then there exists $\varepsilon > 0$ and open interval $I'$ such that $f(x_0)\in (f(x_0)-\varepsilon, f(x_0)+\varepsilon)\subseteq I' \subseteq N$. By open-set continuity at $x_0$, there exists $\delta > 0$ and open interval $I''$ such that $x_0\in (x_0-\delta, x_0+\delta) \subseteq I'' \subseteq f^{-1}(N)$. This means, given any $\varepsilon > 0$, we can find a $\delta > 0$ such that $f((x_0-\delta, x_0+\delta)) \subseteq (f(x_0)-\varepsilon, f(x_0)+\varepsilon)$. This is the $\varepsilon-\delta$ definition of continuity.

I'm sure of the first implication but the end of the second is iffy. Is the conclusion correct?

In the first direction, it should be pointed out that, in general, $N \subseteq f^{-1}(f(N))$, and equality may fail. It's not clear that you assumed equality, but it's also not clear that you know of this potential difficulty. Of course, you can easily get around it by stating, $$(x_0 - \delta, x_0 + \delta) \subseteq f^{-1}f((x_0 - \delta, x_0 + \delta)) \subseteq f^{-1}(N).$$
Suppose $f : \mathbb{R} \to \mathbb{R}$ is continuous in the topological sense at $x_0$. Fix $\varepsilon > 0$. Since the interval $$I = (f(x_0) - \varepsilon, f(x_0) + \varepsilon)$$ is a neighbourhood of $f(x_0)$, it follows by the continuity of $f$ at $x_0$ that some neighbourhood $N$ of $x_0$ such that $N \subseteq f^{-1}(I)$. By the definition of $N$ being a neighbourhood of $x_0$, there must exist a symmetric open interval, centred at $x_0$, but contained in $N$. That is, there exists some $\delta > 0$ such that $$(x_0 - \delta, x_0 + \delta) \subseteq N.$$ Therefore, we have \begin{align*} 0 < |x - x_0| < \delta &\implies x \in (x_0 - \delta, x_0 + \delta) \subseteq N \subseteq f^{-1}(I) \\ &\implies f(x) \in I \\ &\implies |f(x) - f(x_0)| < \varepsilon. \end{align*} I haven't done anything different to you, but sometimes seeing it written slightly differently can mysteriously make a proof a little more convincing!
• I think you mean $N \subseteq f^{-1}(f(N))$? – Fimpellizieri Mar 4 '18 at 6:45
• To the proposer: In the second part you should not start with a nbhd of $f(x_0).$ Given $\epsilon >0,$ a random nbhd of $f(x_0)$ might not cover $(-\epsilon+f(x_0),\epsilon +f(x_0).$ Given $\epsilon >0,$ the only nbhd of $f(x_0)$ that you need to consider is $(-\epsilon+f(x_0),\epsilon+f(x_0))$. – DanielWainfleet Mar 5 '18 at 10:42