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

I need to prove that the $$\epsilon$$-$$\delta$$ definition of continuity implies the open set definition continuity for a real function. Here's my attempt.

For any basis $$V: (a, b)$$ in the range, for each $$f(x) \in V$$,
let $$\epsilon = \min(f(x) - a, b - f(x))$$, then for any $$x$$ that $$f(x) \in V$$ according the $$\epsilon-\delta$$ definition of continuity there must exists a $$\delta$$ that the open set $$U_x : (x - \delta, x + \delta) \subset f^{-1}((f(x) - \epsilon, f(x) + \epsilon)) \subset f^{-1}(V)$$
In conclusion, $$f^{-1}(V) = \bigcup_{x \in f^{-1}(V)} U_x .$$ $$f^{-1}(V)$$ is an open set. Then for any open set $$W$$, $$f^{-1}(W) = \bigcup_{V \subset W} f^{-1}(V)$$ $$f^{-1}(W)$$ is an open set. So for any open $$W$$, $$f^{-1}(W)$$ is also an open set. This is exactly the open set definition of continuity. QED.

Is my answer correct? Thanks.

• Not every open set of the real line is of the form $(a,b)$; though it suffices to consider such sets, you need to argue why this is the case. In addition, a single element of $V$ need not be the image of a single $x$ in the domain; but you are considering only a single $x$. What if $f(x)=f(y)$ but $x\neq y$? You seem to only select a single $U_x$ for each element of $V$, so one of the two might be "left out". The main idea is right, but the devil is in the details, as usual. Commented Sep 19, 2011 at 1:52
• Let me explain again: you pick a single $f(x)$ in $V$, and you look at the corresponding $x$. But if $f(x)=f(y)$, you never address what happens to $y$, if $y\neq x$; one can read what you write as saying "for every $x$ such that $f(x)\in V$, we do the following..." in which case what you say is complete. But one can also read it as saying: "for each point in $V$ which is the image of someone, let $x$ in the domain such that $f(x)$ is that point; then..." in which case your argument is not complete. So, why leave it up to the reader? Be clear and unambiguous instead. Commented Sep 19, 2011 at 2:29
• (+1) for your continued involvement (as evidenced by comments above) Commented Sep 19, 2011 at 2:39
• The union describing $f^{-1}(V)$ should be over all $x\in f^{-1}(V)$, not over all $f(x)\in V$; and after the second displayed equation, you should say that the $V$ range over all basic open sets $(a,b)$ that are contained in $W$. Otherwise, it looks fine. Commented Sep 19, 2011 at 2:49
• @Jichao Can you please post a brief answer summarizing your understanding so that this post appears as answered in the future? Thank you! Commented Nov 2, 2011 at 2:15

## 2 Answers

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 $f((x - \delta, x+\delta)) \subseteq (f(x) - \varepsilon, f(x) + \varepsilon)$.

To prove the $$ϵ-δ$$ definition implies the open set definition, we proceed as follows:

Let $$U⊆R$$ be open in the range $$R$$. By the definition of openness in metric spaces, there exists for each $$y∈U$$ some $$ϵ_y>0$$ such that $$B(ϵ_y,y)⊆U$$.

It is clear that
$$U=⋃_{y∈U}B(ϵ_y,y) . \; \; \; \; \; (*)$$ We claim that $$f^{-1} (U)$$ is open in the domain $$R$$.

Suppose that $$x_0∈f^{-1} (U)$$. Then $$f(x_0 )∈U$$, so $$f(x_0 )∈B(ϵ_{y_0},y_0 )$$ for some $$y_0∈U$$ by (*). Then, $$|f(x_0 )-y_0 |=d(f(x_0 ),y_0 )<ϵ_{y_0}$$.

Define $$ξ≡ϵ_{y_0}-|f(x_0 )-y_0 | \; \; \; \; \; (**)$$ We have by the ϵ-δ definition of continuity, for the positive number $$ξ$$, there exists some $$δ>0$$ such that if $$x∈R$$ (domain) and $$|x-x_0 |=d(x,x_0 )<δ$$, then $$|f(x)-f(x_0 )|=d(f(x),f(x_0 ))<ξ$$.

Let $$x∈B(δ,x_0 )$$, that is, $$|x-x_0 |=d(x,x_0 )<δ$$, so that $$|f(x)-f(x_0 )|=d(f(x),f(x_0 ))<ξ$$.

The triangle inequality and (**) imply that $$|f(x)-y_0 |≤|f(x)-f(x_0 )|+|f(x_0 )-y_0 |≤ϵ_{y_0}.$$ Hence, $$f(x)∈B(ϵ_{y_0},y_0 )$$ so that $$x∈f^{-1} (U)$$. Then, $$B(δ,x_0 )⊆f^{-1} (U)$$.

It follows that $$f^{-1} (U)$$ is open.