“A continuous map which is open but not closed”-can someone explain to me why this proof works

general-topologyopen-map

http://www.mathcounterexamples.net/continuous-maps-that-are-not-closed-or-not-open/

I need some explanation of a proof given in the above link. I'm looking at the proof that $f_1:(x,y)\longmapsto x$ is open but not closed. Why is it that they are able to use a different specific function $\varphi$ to show that the map is not closed? This might be a dumb question given that I'm in grad level topology, but what's the difference between a map and a function? Why is the different type of arrow ($\longmapsto$) used there? Any help is much appreciated! I'm in over my head.

Edit: is $\varphi$ just being used to show that the set $C=${ $(x, \frac{1}{x})|x>0 $} is closed, while its image $f_1(C)=(0,\infty)$ is open, implying that $f_1$ is not closed?

Best Answer

The first few words of the article are "an open map is function" followed by a condition. Then that article defines closed map. Although just "map" is not defined, a reasonable approximation is that a map is a function.

One convention for defining a map is to use "$\longrightarrow$" ($\TeX$'s \longrightarrow) between the domain and codomain and "$\longmapsto$" ($\TeX$'s \longmapsto) to indicate the image of a point in the domain. (Not everyone uses the long versions. You will see \rightarrow and \mapsto used sometimes.) The article you mention follows this convention. For instance, there is

$$ \begin{array}{r|lcl} f_1 : & \Bbb{R}^2 & \longrightarrow & \Bbb{R} \\ & (x,y) & \longmapsto & x \end{array} \text{.} $$

(I predict the writer has a macro that they use for this purpose. Perhaps something like \displayfunction{f_1}{\reals^2}{\reals}{(x,y)}{x} was used here.)

This says $f_1$ is a function from $\Bbb{R}^2$ to $\Bbb{R}$ defined by $f_1(x,y) = x$.

The function $\varphi$ is introduced only to show that a certain set is closed. That set, $C$, is the preimage of a closed set through a continuous map, so is closed. Then we let $f_1$ act on $C$ and discover that its image through $f_1$ is not closed, so $f_1$ failed to preserve its closedness, so must not be a closed function.

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