# Topological characterization of Continuity

I am self-learning Real Analysis, and I would like to prove the following result in the exercise set 4.4 in Stephen Abbott's, Understanding Analysis. I am not completely confident about the reverse implication. I'd like someone to check my proof, if its rigorous.

[Abbott, 4.4.11] (Topological Characterization of Continuity). Let $$\displaystyle g$$ be >defined on all of $$\displaystyle \mathbf{R}$$. If $$\displaystyle B$$ is a subset of $$\displaystyle \mathbf{R}$$, define the set $$\displaystyle g^{-1}( B)$$ by

$$\begin{equation*} g^{-1}( B) =\{x\in \mathbf{R} :g( x) \in B\} \end{equation*}$$

Show that $$\displaystyle g$$ is continuous if and only if $$\displaystyle g^{-1}( O)$$ is open whenever $$\displaystyle O\subseteq \mathbf{R}$$ is an open set.

Proof.

$$\displaystyle \Longrightarrow$$ direction.

Let $$\displaystyle c$$ be an arbitrary point in $$\displaystyle g^{-1}( O)$$. Therefore, $$\displaystyle g( c) \in O$$.

Assume that $$\displaystyle g$$ is continuous and that the image set $$\displaystyle O$$ is open. By definition, there exists an $$\displaystyle \epsilon >0$$, such that the $$\displaystyle \epsilon$$-neighbourhood, $$\displaystyle V_{\epsilon }( g( c)) =( g( c) -\epsilon ,g( c) +\epsilon )$$ is contained in $$\displaystyle O$$.

We assumed that the function $$\displaystyle g$$ is continuous. So, for all $$\displaystyle \xi >0$$, there exists $$\displaystyle \delta >0$$, such that $$\displaystyle | x-c| < \delta$$ $$\displaystyle \Longrightarrow$$ $$\displaystyle | g( x) -g( c)| < \xi$$.

If we set $$\displaystyle \xi =\epsilon$$ above, we are guaranteed that, if $$\displaystyle x$$ belongs to some $$\displaystyle \delta$$-neighbourhood of $$\displaystyle c$$, $$\displaystyle V_{\delta }( c) =( c-\delta ,c+\delta )$$, then $$\displaystyle g( x) \in V_{\epsilon }( g( c))$$.

Thus, $$\displaystyle g( V_{\delta }( c)) \subseteq V_{\epsilon }( g( c))$$.

Since $$\displaystyle O$$ is open, it must be the union of open sets. $$\displaystyle O=\bigcup _{y=g( c)} V_{\epsilon }( y)$$, that is, the union of all $$\displaystyle \epsilon$$-neighbourhoods of $$\displaystyle y$$, such that $$\displaystyle y=g( c)$$ fills up $$\displaystyle O$$. Therefore, we must have that $$\displaystyle \bigcup V_{\delta }( c)$$ is the pre-image of $$\displaystyle O$$. But, that implies the pre-image $$\displaystyle g^{-1}( O)$$ is an open set.

$$\Longleftarrow$$ direction. (Updated)

Assume that whenever $$\displaystyle O$$ is an open set, $$\displaystyle g^{-1}( O)$$ is open. Pick an arbitrary $$\displaystyle \epsilon >0$$.

Let $$\displaystyle y\in O$$ be an arbitrary point, such that $$\displaystyle y=g( x)$$. $$\displaystyle O$$ is open $$\displaystyle \Longrightarrow$$ $$\displaystyle \exists \xi \leq \epsilon$$, such that $$\displaystyle V_{\xi }( y) \subseteq O$$. Consider the open set $$\displaystyle V_{\xi }( y)$$. Let $$\displaystyle U$$ be the pre-image of this $$\displaystyle \xi$$-neighbourhood. From the above assumption, $$\displaystyle U$$ is open. Since, $$\displaystyle x\in U$$, we can construct a $$\displaystyle \delta$$-neighbourhood around $$\displaystyle x$$, such that $$\displaystyle V_{\delta }( x) \subseteq U$$. Consequently, $$\displaystyle g( V_{\delta }( x)) \subseteq V_{\xi }( y) \subseteq V_{\epsilon }( y)$$.

Let $$\displaystyle c$$ be a fixed point in the pre-image of $$\displaystyle O$$, so $$\displaystyle g( c) \in O$$. Since, $$\displaystyle c$$ is in the pre-image of $$\displaystyle O$$, which is an open set, it must belong to the $$\displaystyle \delta$$-ball of some $$\displaystyle x$$, which means that $$\displaystyle g( c)$$ must belong to the corresponding $$\displaystyle \xi$$-ball of $$\displaystyle g( x)$$. Thus, if $$\displaystyle | x-c| < \delta$$, it follows that $$\displaystyle | g( x) -g( c)| < \epsilon$$.

• For the converse, you should start with arbitrary $x\in\Bbb R$ and $\varepsilon>0$ given, and find $\delta$ that witnesses continuity at $x$. Sep 26 at 12:17
• @Berci, I edited my answer, based on your suggestion. I am assuming that $O$ is the image set of $g$. Sep 26 at 12:51
• It is not valid to assume that $O$ is the image set of $g$. You must take $O \subset \mathbb R$ to be an arbitrary open subset of $\mathbb R$. Sep 26 at 13:12

In $$\Rightarrow$$ we're done after we've shown $$g[V_\delta(c)] \subseteq V_\varepsilon(g(c)) \subseteq O$$, because this implies $$V_\delta (c) \subseteq g^{-1}[O]$$ so that $$c$$ is by definition an interior point of $$g^{-1}[O]$$, and as $$c$$ was arbitrary, we know $$g^{-1}[O]$$ is open (every point is an interior point); the stuff about unions you added is not needed.
For $$\Leftarrow$$, take $$x \in X$$ and you show $$g$$ is continuous at $$x$$. So let $$\xi>0$$ and note that $$O:=V_\xi(g(x))$$ is open in $$\Bbb R$$. So $$g^{-1}[O]$$ is open and contains $$x$$ so that gives us the required $$\delta$$ for continuity at $$x$$, as you can check... So pick this specific $$O$$ to do the proof.