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I am reading a basic introductory book on topology. It is written that a continuous map f from one topological space X to a second topological space Y is open ( closed ) if it maps open ( closed ) sets from X to open ( closed ) sets in Y. Next the following examples are shown:

$$ f:\ \mathbb{R} \rightarrow \mathbb{R}, x \mapsto x, $$ is closed and open. This is clear, since it maps an interval to the same interval. So open sets will stay open and closed once will be closed.

$$ f:\ \mathbb{R} \rightarrow \mathbb{R}, x \mapsto 0, $$ This is closed but not open. Since everything is mapped to 0 what is always a closed set.If I understand correctly.

$$ f:\ \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \arctan x $$ This is closed but not open.

$$ f:\ \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \left | \arctan x \right | $$ is neither open nor closed.

The last two examples I am not able to understand because to me it would seem that the atan is open and closed, what is definitely wrong since it is written in the book. But why?

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  • $\begingroup$ I am puzzled that it considers "f: R-> R, x-> atan(x)" at all because that function is not continuous! $\endgroup$ – user247327 Nov 6 '18 at 20:21
  • $\begingroup$ @user247327 As far as I know the atan(x) or arctan(x) is a continuous function like is also shown here math.stackexchange.com/questions/294683/… $\endgroup$ – zodiac Nov 6 '18 at 20:24
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Observe that $f(x)=\arctan(x)$ is a homeomorphism (actually a diffeomorphism) between $\mathbb{R}$ and $(-\pi/2, \pi/2)$, because it is continuous (smooth) and has a continuous (smooth) inverse, $f^{-1}(x)=\tan(x)$.
But if you think of it as a map $\mathbb{R}\rightarrow \mathbb{R}$ is not closed since, as observed before it sends $\mathbb{R}$ (closed) to $(-\pi/2, \pi/2)$ which is not closed, otherwise it would disconect $\mathbb{R}$. Now, it is open since as a homeomorphism $\mathbb{R}\rightarrow (-\pi/2, \pi/2)$ it sends open subsets of $\mathbb{R}$ to open subsets of $(-\pi/2, \pi/2)$, which, in turn, are open in $\mathbb{R}$, as $(-\pi/2, \pi/2)$ has the subspace topology. So, $f(x)=\arctan(x)$ is actually open but not closed.
As of $f(x)=|\arctan(x)|$, it is not closed because it sends $\mathbb{R}$ (closed) to $[0,\pi/2)$ (not closed). It is not open by the same reason, it sends $\mathbb{R}$ (open) to $[0,\pi/2)$ (not open).

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  • $\begingroup$ You wrote "But if you think of it as a map $\mathbb R \rightarrow \mathbb R$ is not open since, as observed before it $\ldots$." Did you mean to write "But if you think of it as a map $\mathbb R \rightarrow \mathbb R$, then it is not closed since, as observed before it $\ldots$." ? $\endgroup$ – irchans Nov 6 '18 at 21:44
  • $\begingroup$ Yes, thanks @irchans, I edited my answer. $\endgroup$ – Laz Nov 6 '18 at 21:45

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