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Math Help - uncountable set: limit point

  1. #1
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    uncountable set: limit point

    Prove that an uncountable set of  \bold{R} must have a limit point.

    If a set  S \subset \bold{R} and is uncountable, then  \aleph_{0} < \text{card}(S) . Now a point  x_0 is a limit point of  S \subset \bold{R} if for every  \varepsilon > 0 ,  (x_{0}- \varepsilon, \ x_{0}+ \varepsilon) \cap S \ \backslash \{x_0 \} \neq \emptyset .

    Suppose for contradiction that  S did not have a limit point. Would this then create a gap, contradicting the fact that the set is uncountable?
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    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by particlejohn View Post
    Prove that an uncountable set of  \bold{R} must have a limit point.

    If a set  S \subset \bold{R} and is uncountable, then  \aleph_{0} < \text{card}(S) . Now a point  x_0 is a limit point of  S \subset \bold{R} if for every  \varepsilon > 0 ,  (x_{0}- \varepsilon, \ x_{0}+ \varepsilon) \cap S \ \backslash \{x_0 \} \neq \emptyset .

    Suppose for contradiction that  S did not have a limit point. Would this then create a gap, contradicting the fact that the set is uncountable?
    i'll try it this way..

    if  S does not have a limit point, then every s\in S, there is an epsilon neighborhood V_{\varepsilon,s} of s contains finitely many points in S.

    Then, S \subseteq \bigcup_{s \in S} V_{\varepsilon,s} is still finite.

    EDIT: my mistake, sorry john..
    Last edited by kalagota; July 10th 2008 at 03:28 AM.
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    Quote Originally Posted by kalagota View Post
    i'll try it this way..

    if  S does not have a limit point, then every s\in S, there is an epsilon neighborhood V_{\varepsilon,s} of s contains finitely many points in S.

    Then, S \subseteq \bigcup_{s \in S} V_{\varepsilon,s} is still finite.
    which is the contradiction I was aiming for?
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    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by kalagota View Post
    if  S does not have a limit point, then every s\in S, there is an epsilon neighborhood V_{\varepsilon,s} of s contains finitely many points in S.

    Then, S \subseteq \bigcup_{s \in S} V_{\varepsilon,s} is still finite.
    I don't know how to answer the original question but this solution doesn't seem right to me : take S=[0,1], for s\in S,\, \{s\} is a finite set but \bigcup_{s\in S}\{s\}=S is uncountable. (a union of countable sets is not necessarily countable)
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    MHF Contributor kalagota's Avatar
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    oh yeah! it is the intersection of an arbitrary collection of finite sets is finite..

    thanks for that and sorry john for the mistakes..

    EDIT: I also jumbled the concepts of countability and finity..
    Last edited by kalagota; July 10th 2008 at 05:46 AM.
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    I did not try an answer before because I was unsure of what you are working with,
    But here are some observations.
    If S has no limits points, then every point of S is not a limit point of S.
    If t \in S there is an open interval such that V_t  = \left( {t - \delta ,t + \delta } \right)\,\& \,S \cap V_t  = \left\{ t \right\}.
    Because between any two numbers there is a rational number, \left( {\exists x_t  \in Q} \right)\left[ {t \in \left( {x_t  - \varepsilon ,x_t  + \varepsilon } \right) \subseteq \left( {t - \delta ,t + \delta } \right)} \right].
    Name \left( {x_t  - \varepsilon ,x_t  + \varepsilon } \right) = U_{x_t } . Note that U_{x_t } contains exactly one point of S.

    How many U_{x_t } could there be? (This is a modification of a proof by John Kelly.)
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    Quote Originally Posted by Plato View Post
    I did not try an answer before because I was unsure of what you are working with,
    But here are some observations.
    If S has no limits points, then every point of S is not a limit point of S.
    If t \in S there is an open interval such that V_t  = \left( {t - \delta ,t + \delta } \right)\,\& \,S \cap V_t  = \left\{ t \right\}.
    Because between any two numbers there is a rational number, \left( {\exists x_t  \in Q} \right)\left[ {t \in \left( {x_t  - \varepsilon ,x_t  + \varepsilon } \right) \subseteq \left( {t - \delta ,t + \delta } \right)} \right].
    Name \left( {x_t  - \varepsilon ,x_t  + \varepsilon } \right) = U_{x_t } . Note that U_{x_t } contains exactly one point of S.

    How many U_{x_t } could there be? (This is a modification of a proof by John Kelly.)
    There can be an infinite number of  U_{x_t} .
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    Quote Originally Posted by particlejohn View Post
    There can be an infinite number of  U_{x_t} .
    Yes, but only countably many! The rationals are countable.
    But you are given that S is uncountable.
    Is that a contradiction?
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    That is a contradiction. Thanks. Just a quick question: \left( {\exists x_t  \in Q} \right)\left[ {t \in \left( {x_t  - \varepsilon ,x_t  + \varepsilon } \right) \subseteq \left( {t - \delta ,t + \delta } \right)} \right].

    Is this basically saying that there is a rational number x_t , such that the neighborhood of  x_t is a subset of the neighborhood of the real number?
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    Quote Originally Posted by particlejohn View Post
    Is this basically saying that there is a rational number x_t , such that the neighborhood of  x_t is a subset of the neighborhood of the real number?
    What in the world does that mean: “subset of a number”?

    It says that given any point t \in S there is a neighborhood, V, of t that contains no other point of S.
    Then there is neighborhood, U, with a rational center that contains t and is a subset of neighborhood, V.
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    Quote Originally Posted by Plato View Post
    What in the world does that mean: “subset of a number”?

    It says that given any point t \in S there is a neighborhood, V, of t that contains no other point of S.
    Then there is neighborhood, U, with a rational center that contains t and is a subset of neighborhood, V.
    I said subset of neighborhood.
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    Quote Originally Posted by particlejohn View Post
    Prove that an uncountable set of  \bold{R} must have a limit point.

    If a set  S \subset \bold{R} and is uncountable, then  \aleph_{0} < \text{card}(S) . Now a point  x_0 is a limit point of  S \subset \bold{R} if for every  \varepsilon > 0 ,  (x_{0}- \varepsilon, \ x_{0}+ \varepsilon) \cap S \ \backslash \{x_0 \} \neq \emptyset .

    Suppose for contradiction that  S did not have a limit point. Would this then create a gap, contradicting the fact that the set is uncountable?
    Break the real line into countably many disjoint, bounded intervals ..., [-2,1), [-1,0), [0, 1), [1, 2), ...

    At least one of these intervals must contain infinitely many members of S-- otherwise S would be the union of countably many finite sets and would be countable. So...
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    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by awkward View Post
    Break the real line into countably many disjoint, bounded intervals ..., [-2,1), [-1,0), [0, 1), [1, 2), ...

    At least one of these intervals must contain infinitely many members of S-- otherwise S would be the union of countably many finite sets and would be countable. So...
    i think, we have to distinguish the definition of limit points in topology.
    the points in any interval in R are limit points. so in fact, all of those intervals are set of limit points, even the end points.
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  14. #14
    MHF Contributor kalagota's Avatar
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    I then list some candidates for S.

    1. a set that contains an interval.
    2. a set that contains the irrationals.
    3. (both of course)
    4. (and of course, R)

    what are the other sets that are uncountable
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