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I’m quite confused about some topological results. I know there must be something wrong in my reasoning, but I cannot find out what is wrong here. We know that:

  1. $\mathbb{R}$ is closed (and is also opened, but that’s not what confuses me)
  2. $\tan$ is a continuous function on $]-\pi/2, \pi/2[$

My question is quite simple: since the inverse image under a continuous function of a closed set is closed, why do we have $\tan^{-1}(\mathbb{R})=]-\pi/2,\pi/2[$, which is not closed?

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    $\begingroup$ The interval is closed in the restriction topology $\endgroup$
    – Timur Bakiev
    Commented Jun 28, 2021 at 21:34
  • $\begingroup$ Oh, right. That’s the same reason which explains why $\mathbb{R}$ is closed. Thanks! $\endgroup$
    – Jujustum
    Commented Jun 28, 2021 at 21:38
  • $\begingroup$ @TimurBakiev That's a fine answer to a natural question. Why don't you post it as an answer? $\endgroup$
    – José Carlos Santos
    Commented Jun 28, 2021 at 21:55
  • $\begingroup$ @JoséCarlosSantos don’t know, it is so short :) $\endgroup$
    – Timur Bakiev
    Commented Jun 28, 2021 at 21:58
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    $\begingroup$ @TimurBakiev There's nothing wrong with a short answer. Quite the opposite, a compact answer is read more and has greater impact. $\endgroup$
    – Joonas Ilmavirta
    Commented Jun 28, 2021 at 22:17

1 Answer 1

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$\tan$ is continuous as a function $(-\frac \pi 2, \frac \pi 2) \to \mathbb R$, and the interval is closed (and open) in the restriction topology (which is a standard topology if we consider it as a topological space).

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