Shouldn’t $\tan^{-1}(\mathbb{R})$ be closed?

Question

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:

$\mathbb{R}$ is closed (and is also opened, but that’s not what confuses me)
$\tan$ 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?

Oh, right. That’s the same reason which explains why $\mathbb{R}$ is closed. Thanks! — Jujustum, Jun 28 '21 at 21:38
@TimurBakiev That's a fine answer to a natural question. Why don't you post it as an answer? — José Carlos Santos, Jun 28 '21 at 21:55
@TimurBakiev There's nothing wrong with a short answer. Quite the opposite, a compact answer is read more and has greater impact. — Joonas Ilmavirta, Jun 28 '21 at 22:17

score 9 · Accepted Answer · answered Jun 28 '21 at 22:02

9

$\tan$ 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).

answered Jun 28 '21 at 22:02

Timur Bakiev

2,215

Shouldn’t $\tan^{-1}(\mathbb{R})$ be closed?

1 Answers1