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Math Help - Strongly decreasing sequence of sets

  1. #1
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    Strongly decreasing sequence of sets

    This was a homework question few weeks ago, and the teacher didn't give
    out the solution to this problem (it doesn't directly concern complex analysis), so I was wondering weather someone knows the answer to this one.

    Let A\subset \mathbb{C}, and D be such that
    D(A) is the set A with all isolated points of A removed. So the question is, is there a set A\subset \mathbb{C}
    such that the sequence, (D^{n}(A))_{n\in\mathbb{N}} is strongly decreasing?
    That is D(A)\subset A, D^{2}(A)\subset D(A), but not equal?

    My first thoughts went something like this: Let M:=\{\frac{1}{n}:n\in\mathbb{N}\}\cup\{0\}. Then D(M)=\{0\} and
    D^{k}(M)=\emptyset if k\geq 2, so this wont do.

    Now for every x\in M x\ne0 we can make \varepsilon-balls such that \varepsilon_{x_1}\cap\varepsilon_{x_2}=\emptyset if x_1\ne x_2 and neither is 0. Then we can construct a sequence
    for every x\in M such that its limit is x and it is entirely contained in \varepsilon_x. Lets call the set of those sequences S_M.
    Now let M_1=M\cup S_M.
    M_1 is such that D(M_1)=M so now we got 4 strictly decreasing items in the sequence.

    For every given n\in\mathbb{N} we can continue this construction
    such that the sequence is strictly decreasing for the first n terms.
    But that is not the same as the infinite sequence is strictly decreasing.

    I am starting to think that this is impossible, but I am not sure.
    Does anyone have any ideas?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by hjortur View Post
    This was a homework question few weeks ago, and the teacher didn't give
    out the solution to this problem (it doesn't directly concern complex analysis), so I was wondering weather someone knows the answer to this one.

    Let A\subset \mathbb{C}, and D be such that
    D(A) is the set A with all isolated points of A removed. So the question is, is there a set A\subset \mathbb{C}
    such that the sequence, (D^{n}(A))_{n\in\mathbb{N}} is strongly decreasing?
    That is D(A)\subset A, D^{2}(A)\subset D(A), but not equal?

    My first thoughts went something like this: Let M:=\{\frac{1}{n}:n\in\mathbb{N}\}\cup\{0\}. Then D(M)=\{0\} and
    D^{k}(M)=\emptyset if k\geq 2, so this wont do.

    Now for every x\in M x\ne0 we can make \varepsilon-balls such that \varepsilon_{x_1}\cap\varepsilon_{x_2}=\emptyset if x_1\ne x_2 and neither is 0. Then we can construct a sequence
    for every x\in M such that its limit is x and it is entirely contained in \varepsilon_x. Lets call the set of those sequences S_M.
    Now let M_1=M\cup S_M.
    M_1 is such that D(M_1)=M so now we got 4 strictly decreasing items in the sequence.

    For every given n\in\mathbb{N} we can continue this construction
    such that the sequence is strictly decreasing for the first n terms.
    But that is not the same as the infinite sequence is strictly decreasing.

    I am starting to think that this is impossible, but I am not sure.
    Does anyone have any ideas?
    So, you're asking does there exist a set A\subseteq\mathbb{C} so that if D(A) is the set of limit points of A then

    A\supset D(A)\supset D\left(D\left(A\right)\right)\supset D\left(\left(\left(A\right)\right)\right)\supset\c  dots

    What makes sense to me is to go a route similar to what you did. So, let's think of, just for now, a subset of the real numbers which survives two applications of D. Something like D=[-5,-4]\cup\{0\}\cup\left\{\frac{1}{n}:n\in\mathbb{N}\rig  ht\}. What about a set which survives three? What if we took now the subset of the complex numbers which has the real line, appended to the vertical line \left\{\frac{i}{n}:n\in\mathbb{N}\} and then for each n_0\in\mathbb{N} we considered \frac{i}{n_0}+\frac{1}{m} for all m\in\mathbb{N}. Then, I feel that D\left(A\right)=\mathbb{R}\cup\left\{\frac{i}{n}:n  \in\mathbb{N}\right\} and D\left(D\left(A\right)\right)=\mathbb{R}. What if you added to each point of the \frac{i}{n_0}+\frac{1}{m} a sequence converging to it, and to each one of those points append a, etc.

    I don't feel this is correct though. Let me think about it.
    Last edited by Drexel28; November 14th 2010 at 05:14 PM.
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