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Math Help - Cardinality 2

  1. #1
    MHF Contributor Bruno J.'s Avatar
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    Cardinality 2

    Prove or disprove :

    The set of limit points of an uncountable set of real numbers is uncountable .
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Prove or disprove :

    The set of limit points of an uncountable set of real numbers is uncountable .
    This is easier said in more general terms.

    Lemma: If X is second countable and E\subseteq X is covered by \Omega then \Omega omits a countable subcover \Sigma.

    Proof:Call the countable open base \mathfrak{B}. For each e\in E there exists some set \omega\in\Omega such that e\in\omega. But, we have that there exists some B_e\in\mathfrak{B} such that e\in B_e\subseteq\omega. Clearly then the collection \left\{B_e\right\}_{e\in E} covers E and is countable. Taking an \omega which contains B_e for each e\in E finishes the argument. \blacksquare

    Now we can prove the problem at hand.

    Let X be second countable and E\subseteq X uncountable, then E\cap D(E) is uncountable.

    Assume not. Then, D(E) is countable and so E-D(E) is uncountable. But, since each point e\in E-D(E) is in particular not a limit point of E there exists a neighborhood N_e such that E\cap N_e=\{e\}. Clearly the class \left\{N_e\right\}_{e\in E-D(E)} is an open cover for E-D(E) and so by the lemma it must admit a finite subcover \left\{N_{e_n}\right\}_{n\in\mathbb{N}}. But, E-D(E)=\bigcup_{n\in\mathbb{N}}\left(E\cap N_{e_n}\right) but since each E\cap N_{e_n} is a singleton it follows that E-D(E) is countable Contradiction.


    Your problem is a corollary since \mathbb{Q} is a dense subset of \mathbb{R} and every separable metric space is second countable.
    Last edited by Drexel28; April 19th 2010 at 09:03 PM.
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  3. #3
    MHF Contributor Bruno J.'s Avatar
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    That doesn't work! (I thought you had it for a sec, I even wrote a post congratulating you )

    Removing the limit points from a set does not mean the resulting set has no limit points!
    Last edited by Bruno J.; April 19th 2010 at 08:51 PM.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    That doesn't work! (I thought you had it for a sec, I even wrote a post congratulating you )

    Removing the limit points from a set does not mean the resulting set has no limit points!
    That wasn't the screw up, but my head was somewhere else. See edited post.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    .
    Derived set of E. The set of limit points.
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    Derived set of E. The set of limit points.
    Figured it out just too late

    Thanks for the proof, I'll give it the attention which it deserves when I'm not dead tired! Exams begin tomorrow.

    (After all, what you had written before worked also. Why did you remove it? Every point of E-D(E) is an isolated point, and so E-D(E) has the cardinality of a collection of disjoint open sets, which is countable.)
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Figured it out just too late

    Thanks for the proof, I'll give it the attention which it deserves when I'm not dead tired! Exams begin tomorrow.

    (After all, what you had written before worked also. Why did you remove it? Every point of E-D(E) is an isolated point, and so E-D(E) has the cardinality of a collection of disjoint open sets, which is countable.)
    Because the needn't be disjoint. I was being stupid. Clearly they can intersect but they cannot contain each other. For example (0,2)(1,3) could be two of the sets if [math[\frac{1}{2},2\in E[/tex] since they're intersection with E could easily be a singleton but they aren't disjoint.
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    MHF Contributor Bruno J.'s Avatar
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    Right - but taking them small enough?

    Haha I really have to hit the bunk. 'Nite!
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Right - but taking them small enough?

    Haha I really have to hit the bunk. 'Nite!
    I'm not sure right this second but I'm really tired. I'll look again tomorrow. It is immediate how to finish the argument assuming that E-D(E) is closed.
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