Prove $f: \mathbb{R}^d \to \mathbb{R}$ is continuous iff $f^{-1}(O)$ is open for any open set $O \subset \mathbb{R}$.

Proof $(\implies)$. We need to show that any point $x\in f^{-1}(O)$ is an interior point.

Let $f(x) \in O$. Since $O$ is open there exists a ball $B_\epsilon\left(f(x)\right) \subset O$, for some $\epsilon > 0$. Since $f$ is continuous, we are guaranteed there exists a $\delta_\epsilon > 0$ such that

\begin{align} f\left(B_{\delta_\epsilon}(x)\right) \subset B_\epsilon\left(f(x)\right) \subset O \end{align}

Since $B_{\delta_\epsilon}(x)$ is in part of the domain of $f$ which maps to the open set $O$, it must be in $f^{-1}(O)$. And since $x$ is arbitrary, then for any $x\in f^{-1}(O)$, there exists a ball around $x$ that's completely contained in $f^{-1}(O)$. Therefore $f^{-1}(O)$ is open.

Proof $(\impliedby)$ Let $f^{-1}(O)$ be open for any open set $O$ in $\mathbb{R}$ and let $\epsilon >0$. We need to show we can find an open ball $B_{\delta_\epsilon}(x)$ in $f^{-1}(O)$, such that

\begin{align} f\left(B_{\delta_\epsilon}(x)\right) \subset B_\epsilon\left(f(x)\right) \end{align}

So, choose $\delta = \frac{|f(x) - \epsilon|}{2}$... [stuck here]

  • $\begingroup$ For the first part of your proof, how do you know that $O$ is not empty? $\endgroup$ – John Douma Oct 29 '18 at 1:50

Suppose $f$ is continuous and let $O$ be open in $\mathbb R$. If there are no $x$'s such that $f(x)\in O$ then $f^{-1}(O)$ is the empty set which is open. Suppose $f^{-1}(O)$ is not empty. Let $x\in f^{-1}(O)$. Since $O$ is open and $f(x)\in O$, there exists $\epsilon\gt 0$ such that $B_{\epsilon}(f(x))\subset O$. Since $f$ is continuous there exists $\delta\gt 0$ such that $f(B_{\delta}(x))\subset B_{\epsilon}(f(x))\subset O$. Since $f(B_{\delta}(x))\subset O$, $B_{\delta}(x)\subset f^{-1}(O)$. Since we can do this for any $x\in f^{-1}(O)$, $f^{-1}(O)$ is open.

Conversely, suppose $f^{-1}(O)$ is open for any open set $O$. Let $x\in\mathbb R^d$ and $\epsilon\gt 0$ be given. We know that $B_{\epsilon}(f(x))$ is open so by our hypothesis, $f^{-1}(B_{\epsilon}(f(x))$ is open. Moreover, $x\in f^{-1}(B_{\epsilon}(f(x))$ so there exists $B_{\delta}(x)\subset f^{-1}(B_{\epsilon}(f(x))\implies f(B_{\delta})\subset B_{\epsilon}(fx)$. Therefore, $f$ is continuous.

  • $\begingroup$ About reading the question: that last part, "...for any open set $O \subset \mathbb{R}$". Do we take that to apply to the left side of our iff. In other words, we proved $f$ is continuous for open sets, but it might not be continuous everywhere. Is that correct? $\endgroup$ – Zduff Oct 29 '18 at 2:40
  • 1
    $\begingroup$ @Zduff I am not sure I understand your question. We have proved that $f$ is continuous everywhere because we assumed nothing about $x$ except that it is a point of $\mathbb R^d$. Note this result holds for $f:X\to Y$ where $X$ and $Y$ are two metric spaces. $\endgroup$ – John Douma Oct 29 '18 at 2:50
  • 1
    $\begingroup$ @Zduff The real significance of this problem is that we can define continuity without using the concept of distance. This result is what makes Topology work. $\endgroup$ – John Douma Oct 29 '18 at 2:53
  • 1
    $\begingroup$ @Zduff Yes. I used open balls in the proof but consider the result. We get that a function is continuous if the inverse of every open set is open. No distance is needed. In practice, we may use distances if we are solving a problem with metric spaces but we can now talk about continuous functions on spaces that have no metric. $\endgroup$ – John Douma Oct 29 '18 at 3:14
  • 1
    $\begingroup$ @Zduff $f(B)=\{f(b): b\in B\}$. $\endgroup$ – John Douma Nov 1 '18 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.