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

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

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

    For every given $\displaystyle n\in\mathbb{N}$ we can continue this construction
    such that the sequence is strictly decreasing for the first $\displaystyle n$ 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 $\displaystyle A\subset \mathbb{C}$, and $\displaystyle D$ be such that
    $\displaystyle D(A)$ is the set $\displaystyle A$ with all isolated points of $\displaystyle A$ removed. So the question is, is there a set $\displaystyle A\subset \mathbb{C}$
    such that the sequence, $\displaystyle (D^{n}(A))_{n\in\mathbb{N}}$ is strongly decreasing?
    That is $\displaystyle D(A)\subset A$, $\displaystyle D^{2}(A)\subset D(A)$, but not equal?

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

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

    For every given $\displaystyle n\in\mathbb{N}$ we can continue this construction
    such that the sequence is strictly decreasing for the first $\displaystyle n$ 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 $\displaystyle A\subseteq\mathbb{C}$ so that if $\displaystyle D(A)$ is the set of limit points of $\displaystyle A$ then

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

    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 $\displaystyle D$. Something like $\displaystyle D=[-5,-4]\cup\{0\}\cup\left\{\frac{1}{n}:n\in\mathbb{N}\rig ht\}$. 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 $\displaystyle \left\{\frac{i}{n}:n\in\mathbb{N}\}$ and then for each $\displaystyle n_0\in\mathbb{N}$ we considered $\displaystyle \frac{i}{n_0}+\frac{1}{m}$ for all $\displaystyle m\in\mathbb{N}$. Then, I feel that $\displaystyle D\left(A\right)=\mathbb{R}\cup\left\{\frac{i}{n}:n \in\mathbb{N}\right\}$ and $\displaystyle D\left(D\left(A\right)\right)=\mathbb{R}$. What if you added to each point of the $\displaystyle \frac{i}{n_0}+\frac{1}{m}$ 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; Nov 14th 2010 at 05:14 PM.
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