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Let $f:\mathbb{R} \to \mathbb{R}$ assume that for an arbitrary interval $(a,b) \subseteq \mathbb{R}$ inverse image $f^{-1}((a,b))$ is union (not necessarily finite) of open intervals. Show that function $f$ is continuous.

I'd be grateful for any hints, since I have problem with tackling this one

Martin
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2 Answers2

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Hint: let $x \in \mathbb{R}$, $\varepsilon > 0$, and define $y=f(x)$. Then $f^{-1}((y-\varepsilon,y+\varepsilon))$ is an open set which contains $x$. How can you use this to obtain a $\delta$ as in the $\varepsilon,\delta$ definition of continuity? Once you've done that, you conclude that $f$ is continuous at $x$, and since $x$ was arbitrary you get that $f$ is continuous on $\mathbb{R}$.

If you get stuck, here's a followup hint: if $O \subset \mathbb{R}$ is open and $x \in O$ then there is an interval $I \subset O$ centered at $x$.

Ian
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Let $U\subset\mathbb R$ be open, then for each $x\in U$, there is an open interval $I_x$ centered at $x$ contained in $U$. So we can write $U=\bigcup_{x\in U}I_x$. Hence $$f^{-1}(U) = f^{-1}\left(\bigcup_{x\in U}I_x\right) = \bigcup_{x\in U}f^{-1}(I_x). $$ By assumption, each set $f^{-1}(I_x)$ is a union of open intervals, and therefore is open. It follows that $f^{-1}(U)$ is open, which implies that $f$ is continuous.

Math1000
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  • why $f^{-1}(U)$ is open implies that $f$ is continuous ?? – Martin Jan 14 '15 at 01:25
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    Because the topological definition of continuity is that if for all open $U \subset \mathbb R$, $f^{-1}(U)$ is open then $f$ is continuous. If you haven't yet proven that this definition is equivalent to the analysis definition then that is likely the goal of this exercise. – dalastboss Jan 14 '15 at 02:25