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Given $z\in\mathbb(C)$ and the sequence $ \\{ p_n(z) \\} $ $\in\mathbb{C}, n\in\mathbb{N}$ defined by recursion : $p_0(z) = z$ and $p_{n+1}(z)=(p_n(z))^2+z$, I can show that if $ \\{ p_n(z) \\}$ is bounded then $ |p_n(z)| \le 2$ for every $n\in\mathbb{N}$. I can also prove that $p_n(z)$ is continuous for every $n\in\mathbb{N}$. But I can't show that if $M$ is the set such that for every $z\in M$ the sequence $\\{ p_n(z) \\}$ is bounded then $M$ is closed. Can anybody please help me? Thank you. RG

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What do you know about the intersection of closed sets? – Erick Wong Mar 18 '13 at 1:00
I can't figure out how $M$ could be seen as an intersection of closed sets. – rhogamma Mar 18 '13 at 1:03
Can you show that $\{ z\in \mathbb C: |p_n(z)| \le 2\}$ is closed? – Erick Wong Mar 18 '13 at 1:08
The point is that I don't know how to use the definition of closed set in this case. – rhogamma Mar 18 '13 at 1:47
Do you know any facts relating continuous functions and closed sets? – Erick Wong Mar 18 '13 at 1:56

Hint distilled from the comments: for each $n$, show that the set $S_n = \{ z \in \mathbb C: |p_n(z)|\le 2\}$ is the preimage of a closed set under a continuous function. How is the Mandelbrot set related to the intersection of all $S_n$?

More hints: Show that $S_n$ is the preimage of the closed disk $D = \{z \in \mathbb C: |z|\le 2\}$ under $p_n$ (what is the definition of preimage?). One usually, at this stage, takes for granted that a closed disk is actually closed, but this is very easy to prove by showing that the complement $D^c = \{z \in \mathbb C: |z| > 2\}$ is open (what is the definition of open?).

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Thank you for your hints. Now I've got that M is the intersection of all the $S_n$. The problem is to show that, if $f:z\mapsto p_n(z)$, $f(S_n)$ is closed. I tried this:let $z_* \in \mathbb{C}\setminus f(S_n)$ then I must show that for every $\delta >0$ it is $|z_* -z| \geq \delta$ for a $z\in f(S_n)$ . Now since $z\in f(S_n)$ there exists a $\bar{z} \in S_n$ such that $z=p_n(\bar{z})$ and so $|z|=|p_n(\bar{z})| \leq 2$. Hence $|z_* -z| \geq |z_*| - |z| \geq |z_*|-2$ . And now I can't end.... – rhogamma Mar 18 '13 at 20:20
@rhogamma There's no need to refer to $f(S_n)$ so abstractly. $S_n$ can be identified as the preimage (under $f$) of a very simple geometric subset of $\mathbb C$, regardless of what $f(S_n)$ looks like (I guess the two sets are equal but that's besides the point). – Erick Wong Mar 18 '13 at 22:43
I'm sorry, I can't get this. What do you mean with "a very simple geometric subset of C"? I do not know any other method to prove that $f(S_n)$ is a closed set. I am at the very beginning of my Topology class. – rhogamma Mar 18 '13 at 23:01
@rhogamma You need to prove that $S_n$ is the preimage of a closed set, which is not equivalent to proving $f(S_n)$ is closed. $S_n$ is already the preimage of a circular disk, so looking at $f(S_n)$ is both beside the point and somewhat incorrect. – Erick Wong Mar 19 '13 at 0:20
@rhogamma For example, say $f(z) = z^2$ and $S$ is the half-open interval $[-2,1)$. Then $f(S)$ is the interval $[0,4]$, which is closed. But $S$ itself is certainly not closed, so proving $f(S)$ is closed cannot help you establish that $S$ is closed. – Erick Wong Mar 19 '13 at 0:24

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