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.


$\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 $.

  • 1
    $\begingroup$ For the converse, you should start with arbitrary $x\in\Bbb R$ and $\varepsilon>0$ given, and find $\delta$ that witnesses continuity at $x$. $\endgroup$
    – Berci
    Sep 26 at 12:17
  • $\begingroup$ @Berci, I edited my answer, based on your suggestion. I am assuming that $O$ is the image set of $g$. $\endgroup$
    – Quasar
    Sep 26 at 12:51
  • 3
    $\begingroup$ 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 $\mathbb R$. $\endgroup$
    – Lee Mosher
    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.


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