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Math Help - [SOLVED] Cardinality (easy)

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    MHF Contributor Bruno J.'s Avatar
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    [SOLVED] Cardinality (easy)

    It's trivial that we can have a subset S \subset \mathbb{R} such that any bounded neighbourhood of 0 contains \aleph_0 elements of S, and such that the complement of any bounded neighbourhood of 0 contains finitely many elements of S. Now is it possible to replace " \aleph_0" by " \aleph_1" and "finitely many" by " \aleph_0"?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    It's trivial that we can have a subset S \subset \mathbb{R} such that any bounded neighbourhood of 0 contains \aleph_0 elements of S, and such that the complement of any bounded neighbourhood of 0 contains finitely many elements of S. Now is it possible to replace " \aleph_0" by " \aleph_1" and "finitely many" by " \aleph_0"?
    I don't think so.

    Let E be the set in question. We may assume WLOG that 0\notin E (since it won't affect the cardinality). U_n=E\cap\left(\mathbb{R}-B_{\frac{1}{n}}(0)\right) (where B_{\frac{1}{n}}(0) is the open ball of radius \frac{1}{n} around 0). Then, I claim that E=\bigcup_{n=1}^{\infty} U_n. To see this let x\in E then since x\ne 0 we see d(x,0)>0 and by the Archimedean principle there exists some m\in\mathbb{N} such that 0<\frac{1}{m}<d(x,0). Thus, x\notin B_{\frac{1}{m}}\implies x\in U_m\subseteq\bigcup_{n=1}^{\infty}U_n and since it's trivial that \bigcup_{n=1}^{\infty}U_n\subseteq E the conclusion follows. But, each U_n is the intersection with E of the complement of a bounded neighborhood of 0 and thus countable. So, E=\bigcup_{n=1}^{\infty}U_n is the countable union of countable sets and thus countable. But, this contradicts the assumption that E\cap [-1,1] is uncountable.
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    MHF Contributor Bruno J.'s Avatar
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    Perfect!

    My own problem, hope you liked it.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    Perfect!

    My own problem, hope you liked it.
    Nice problem!

    P.S. I think the result (with a slight tweaking) can be extended to any first-countable Hausdorff space.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Nice problem!

    P.S. I think the result (with a slight tweaking) can be extended to any first-countable Hausdorff space.
    Just an update, it does.

    Theorem: Let X be a T_1 topological space and x\in X be such that \{x\} has a countable neighborhood base. Then, if E is a set such that E\cap N' is countable for each neighborhood N of x then E is countable.

    Proof: We may assume WLOG again that x\notin E. Then, if \mathfrak{N} is the countable neighborhood base at x then

    E=E\cap X=E\cap\left(\varnothing\right)'=E\left(E\cap\bigc  ap_{N\in\mathfrak{N}}N\right)'=E\cap\left(E'\cup \bigcup_{N\in\mathfrak{N}}N'\right) =E\cap\bigcup_{N\in\mathfrak{N}}N'=\bigcup_{N\in\m  athfrak{N}}\left(E\cap N'\right)

    And since each E\cap N' is countable and \mathfrak{N} countable it follows that E is the countable union of countable sets and thus countable. The conclusion follows.
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