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Could anyone please help me with the following problem, Let $f(z)={\rm Re}(z)$, then show that $f:\Bbb C\to \Bbb R$ is an open map but not a closed map whereas $f:\Bbb C\to \Bbb C$ is neither an open map nor a closed map. Thank you ....

Pedro
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Raven
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  • This has nothing to do with complex analysis. These are topological questions. Find counterexamples to show that the closed and open properties are not satisfied by your maps.

    In the future, write $\mathbb{R}$ and $\mathbb{C}$ instead of $R$ and $C$.

     – Jeremy Daniel Nov 22 '14 at 23:08
  • But I got this problem in Conway's book named "Functions of one complex variable." – Raven Nov 22 '14 at 23:15
  • Maybe, but there is no complex analysis involved. – Jeremy Daniel Nov 22 '14 at 23:17
  • @JeremyDaniel does that mean every question in the real analysis tag that has to do with metric spaces should be in Topology? – dustin Nov 22 '14 at 23:58

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$f:\Bbb C\to \Bbb R$ is a projection map, and each projection map is open. To show that it is not closed, let $G$ be the graph of $y=\frac1x$, then $G$ is closed but its image is the positive part (excluding $0$) of the $x$-axis, which is not closed. If you consider $f:\Bbb C\to \Bbb C$ then it is not closed for the same reason as above. It is not open since the image of $\Bbb C$ is the $x$-axis, which is not open in the plane.

Community
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Mirko
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  • But where did you use the fact that $f(z)={\rm Re}(z)$ – Raven Nov 23 '14 at 01:51
  • $Re(z)$ is a projection. If $z=(x,y)$ then $Re(z)=x$, this is by definition a projection (taking just the first coordinate, of course the standard representation of complex numbers as pairs of real numbers is used here). – Mirko Nov 23 '14 at 02:01