# Open and Closed maps in topology atan example

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?

• I am puzzled that it considers "f: R-> R, x-> atan(x)" at all because that function is not continuous! – user247327 Nov 6 '18 at 20:21
• @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/… – zodiac Nov 6 '18 at 20:24

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).
• 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 $\ldots$." ? – irchans Nov 6 '18 at 21:44