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

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]

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

## 1 Answer

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.

• 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? – Zduff Oct 29 '18 at 2:40
• @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. – John Douma Oct 29 '18 at 2:50
• @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. – John Douma Oct 29 '18 at 2:53
• @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. – John Douma Oct 29 '18 at 3:14
• @Zduff $f(B)=\{f(b): b\in B\}$. – John Douma Nov 1 '18 at 17:06