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I have found this theorem in a calculus book

We say a function $f: I \to \Bbb{R}$, $I$ interval of $\Bbb{R}$ is connected if $\forall J \subseteq I$, with $J$ interval, $f(J)$ is an interval. Prove that if $f$ is connected and if $\forall y \in \Bbb{R} \, f^{-1}(\{y\})$ is a closed set in the relative topology of $I$, then $f$ is continuous.

I have no idea where I can start, can you give me some hints?

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    $\begingroup$ To the OP: I would recommend adding the topology tag. This is a question that seems more suited toward a topological answer than a calculus-type of answer. (I have ideas from topology on how to approach this problem, but I personally am not sure how I’d be using strictly calculus topics to approach this problem. Hence my first question; you could be taking the two courses concurrently and mixed up the titles.) $\endgroup$
    – Clayton
    Nov 14, 2018 at 17:10
  • $\begingroup$ @Clayton if you can solve it using topology it's good too. Perhaps the problem is that I cannot distinguish between calculus and mathematical analysis since in my main language the latter is used to indicate both; but having seen some foreign books of both, I'm quite confident the one I use is a calculus book. I'm sorry for any misunderstandings due to this semantical problem... $\endgroup$
    – LuxGiammi
    Nov 14, 2018 at 17:10
  • $\begingroup$ Note: anyway, I'm still a student and I've not begun studying mathematical analysis yet... $\endgroup$
    – LuxGiammi
    Nov 14, 2018 at 17:12
  • $\begingroup$ Can you state a few of the topological definitions/theorems that you have available? I assume you know the “inverse image of an open set is open” characterization for continuous functions? $\endgroup$
    – Clayton
    Nov 14, 2018 at 17:16
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    $\begingroup$ @Clayton I know that $f: D \rightarrow \Bbb{R}$ is continuous $\iff$ $\forall X,\; \text{X open set of } \Bbb{R} \; f^{-1}(X) = f^\leftarrow(X)$ is opened in the relative topology of $D$ and that holds also if we replace "opened" with "closed". Anyway, I have some elementary general topology knowledge and the book states some of them (actually, three chapters of it are dedicated to topology...). $\endgroup$
    – LuxGiammi
    Nov 14, 2018 at 19:27

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For $x \in I$ let $I_n(x) = I \cap (x - \frac{1}{n},x + \frac{1}{n})$. This is an open neighborhood of $x$ in $I$. We have $$(*) \phantom{xx} \bigcap_{n=1}^\infty f(I_n(x)) = \{ f(x) \} .$$ "$\supset$" is trivial. To verify "$\subset$", let $y \in \bigcap_{n=1}^\infty f(I_n(x))$. Then there exist $x_n \in I_n(x)$ such that $f(x_n) = y$. Obviously $x_n \to x$. Since $f^{-1}(y)$ is closed in $I$ , we see that $x \in f^{-1}(y)$, i.e. $f(x) = y$.

Let $\varepsilon > 0$. Since $I_n(x)$ is an interval, also $J_n = f(I_n(x))$ is an interval. We have $J_n = \langle a_n,b_n \rangle$, where $\langle a_n,b_n \rangle$ stands for an open, half-open or closed interval such that $a_n \le f(x) \le b_n$. $a_n = -\infty$, $b_n = \infty$ is allowed. The sequence $(a_n)$ is increasing, the sequence $(b_n)$ decreasing. But $(*)$ shows that $a_n, b_n \to f(x)$, hence $f(I_n(x)) = J_n \subset (f(x)- \varepsilon, f(x)+ \varepsilon)$ for sufficiently large $n$. This means that $f$ is continuous.

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  • $\begingroup$ Very good proof. I've just skimmed through it and I've understood pretty much everything. I will look in more details into it in the next few days. Thanks! $\endgroup$
    – LuxGiammi
    Nov 16, 2018 at 15:56

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