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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]

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ @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
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    $\begingroup$ @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
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    $\begingroup$ @Zduff $f(B)=\{f(b): b\in B\}$. $\endgroup$ – John Douma Nov 1 '18 at 17:06

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