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.

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    $\begingroup$ 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. $\endgroup$ Commented Sep 19, 2011 at 1:52
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    $\begingroup$ 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. $\endgroup$ Commented Sep 19, 2011 at 2:29
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    $\begingroup$ (+1) for your continued involvement (as evidenced by comments above) $\endgroup$ Commented Sep 19, 2011 at 2:39
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    $\begingroup$ 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. $\endgroup$ Commented Sep 19, 2011 at 2:49
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    $\begingroup$ @Jichao Can you please post a brief answer summarizing your understanding so that this post appears as answered in the future? Thank you! $\endgroup$
    – Srivatsan
    Commented Nov 2, 2011 at 2:15

2 Answers 2


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.


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