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My notes claim the following theorem:

A function $f: \mathbb{R}^n \to \mathbb{R}^m$ is continuous if and only if whenever $U \subseteq R^m$ is an open set, then $f^{-1}(U) \subseteq \mathbb{R}^n$ is an open set.

Why do we require both sets to be open? What if they were closed?

TimelordViktorious
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    Can you show that these definitions are equivalent? – Fimpellizzeri Nov 16 '16 at 00:05
  • If you prove the theorem, it will be apparent that open sets are crucial to demonstrating the equivalence; the open set criterion is typically the definition of continuity in general topological spaces, and in a space like $\mathbb{R}^n$ it reduces down to the familiar one involving $\epsilon>0$.... Perhaps you want to ask if there exists a function which takes the inverse image of closed sets to closed sets but is not continuous. – Nap D. Lover Nov 16 '16 at 00:09
  • Yes. Perhaps I was not understanding the proof entirely. I am going through the proof now and the open sets play an important role to establish the epsilon-neighbourhood, but otherwise I still feel a bit uncertain about it all.. – TimelordViktorious Nov 16 '16 at 00:14
  • The definition says essentially that given any neighborhood $N$ of a point $f(x)$ in the codomain, one can find a neighborhood $M$ of $x$ such that if $z \in M$ then $f(z) \in N$. In other words, given an arbitrarily small tolerance for change in the codomain, there is a sufficiently small change in the domain to ensure that the change in the codomain is smaller than your tolerance. It might be useful to consider preimages of open sets for discontinuous functions; for example, if $f(x)=\begin{cases} 0 & x \leq 0 \ 1 & x>1 \end{cases}$, what is $f^{-1}((-1/2,1/2))$? – Ian Nov 16 '16 at 00:20
  • We get [0,1], which is not open, right? Which means if a function is continuous we are guaranteed that f^{-1}(U) is open. – TimelordViktorious Nov 16 '16 at 00:28
  • You get $(-\infty,0]$ which is still not open. – Ian Nov 16 '16 at 01:04
  • Oh, I'm sorry, I had a typo, I meant to write $f(x)=\begin{cases} 0 & x \leq 0 \ 1 & x>0 \end{cases}$. What I wrote before is not fully defined... – Ian Nov 16 '16 at 01:25
  • Ah gotcha! What is the "motivation" behind your example? From this, I see that even though a function is discontinuous, it is possible to take the preimage of an open set and get back a set that is not open (possibly closed). Does this mean that for all discontinuous functions, there exists an open set such that the preimage is not open? – TimelordViktorious Nov 16 '16 at 01:35
  • If you have a discontinuous function, some open sets will have open preimages. The whole space and the empty set, for example. But in my example we "zoom in" on the values in the codomain near the discontinuity, and the topology tells us about this discontinuity by giving us a closed preimage. Similarly, $f^{-1}({ 1 })$ would be closed if $f$ were continuous but it isn't in my example. – Ian Nov 16 '16 at 02:14
  • Oh I see. I think I understand it. Thank you! – TimelordViktorious Nov 16 '16 at 02:21

1 Answers1

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If $f^{-1}(C)$ is closed for any closed set $C$ then $f$ is continuous. Why? If $C$ is closed then $\mathbb{R}^m\setminus C$ is open. Since $f^{-1}(C) $ is closed we have that

$$f^{-1}(\mathbb{R}^m\setminus C) =\mathbb{R}^n\setminus f^{-1}(C)$$ is open. So, both conditions are equivalent.

mfl
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