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Math Help - Compactness question

  1. #1
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    Compactness question

    I'm trying to figure out a problem here and I don't quite get it.

    Let n \leftrightarrow q_{n} be a bijection between N and Q\cap[0, 1]. For \epsilon\in(0,  1), define G_{n} to be the open set B_{\frac{\epsilon}{2^n}}-ball around the point q_{n}. Define F=G^{c}\cap[0, 1]. Show that F is compact.

    Now I don't want the answer to this question; I want to figure it out. It's just that I could swear that the set F is empty, which I know is compact. However, the next part of the question asks me to show that 0<inf(F) and that sup(F)<1.

    Basically, it looks to me like G_{n} completely engulfs [0, 1] and then some, so G^{c}\cap[0, 1] wouldn't contain any values. Thus the set F is empty.
    Please Help! Thank you very much.
    Last edited by miatchguy; April 20th 2010 at 08:45 AM.
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by miatchguy View Post
    I'm trying to figure out a problem here and I don't quite get it.

    Let n \leftrightarrow G_{n} be a bijection between N and Q\cap[0, 1]. For \epsilon\in(0, 1), define G_{n} to be the open set B_{\frac{\epsilon}{2^n}}-ball around the point q_{n}. Define F=G^{c}\cap[0, 1]. Show that F is compact.

    Now I don't want the answer to this question; I want to figure it out. It's just that I could swear that the set F is empty, which I know is compact. However, the next part of the question asks me to show that 0<inf(F) and that sup(F)<1.

    Basically, it looks to me like G_{n} completely engulfs [0, 1] and then some, so G^{c}\cap[0, 1] wouldn't contain any values. Thus the set F is empty.
    No, that's not necessarily true. You can, for example, sum all the diameters of these balls (geometric series), and that sum just might turn out to be strictly smaller than the length of the [0;1].
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  3. #3
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    Quote Originally Posted by miatchguy View Post
    Let n \leftrightarrow G_{n} be a bijection between N and Q\cap[0, 1]. For \epsilon\in(0,  1), define G_{n} to be the open set B_{\frac{\epsilon}{2^n}}-ball around the point q_{n}. Define F=G^{c}\cap[0, 1]. Show that F is compact.
    What do you know about a closed subset of a compact set?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by miatchguy View Post
    I'm trying to figure out a problem here and I don't quite get it.

    Let n \leftrightarrow G_{n} be a bijection between N and Q\cap[0, 1]. For \epsilon\in(0,  1), define G_{n} to be the open set B_{\frac{\epsilon}{2^n}}-ball around the point q_{n}. Define F=G^{c}\cap[0, 1]. Show that F is compact.

    Now I don't want the answer to this question; I want to figure it out. It's just that I could swear that the set F is empty, which I know is compact. However, the next part of the question asks me to show that 0<inf(F) and that sup(F)<1.

    Basically, it looks to me like G_{n} completely engulfs [0, 1] and then some, so G^{c}\cap[0, 1] wouldn't contain any values. Thus the set F is empty.
    Please Help! Thank you very much.
    A) Apparently Plato and Failure get it, but I don't understand what G is? You've never said.

    B) What topology is this? Is this the usual real analysis kind of spaces?
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    Quote Originally Posted by Drexel28 View Post
    A) Apparently Plato and Failure get it, but I don't understand what G is? You've never said.

    B) What topology is this? Is this the usual real analysis kind of spaces?
    So sorry. Made a horrible typo. At the beginning I meant n \leftrightarrow q_{n} rather than n \leftrightarrow G_{n}
    G_{n} is the collection of balls around the points in the sequence q_{n}
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by miatchguy View Post
    So sorry. Made a horrible typo. At the beginning I meant n \leftrightarrow q_{n} rather than n \leftrightarrow G_{n}
    G_{n} is the collection of balls around the points in the sequence q_{n}
    Ok, so I get what G_n is. What's G???
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    Quote Originally Posted by Drexel28 View Post
    Ok, so I get what G_n is. What's G???
    G is the set of all balls around q_{n} defined by the sequence G_{n}. Of course G^{c} is the complement of that set.
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  8. #8
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    Because I know this problem well, I just assumed things that were not explicitly stated.
    Actually G = \bigcup\limits_n {B\left( {q_n ;\frac{\varepsilon }{{2^n }}} \right)} so this is an open set covering the rationals in [0,1].
    But \sum\limits_{n = 1}^\infty  {\frac{\varepsilon }{{2^n }}}  = \frac{\varepsilon }{{2 - \varepsilon }} < 1, so G^c\cap [0,1] is not empty.
    Moreover G^c is closed and [0,1] is compact.
    So?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Plato View Post
    Because I know this problem well..
    Good, because I had no idea what the hell was going on. So \varepsilon was fixed?
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    Quote Originally Posted by Plato View Post
    Because I know this problem well, I just assumed things that were not explicitly stated.
    Actually G = \bigcup\limits_n {B\left( {q_n ;\frac{\varepsilon }{{2^n }}} \right)} so this is an open set covering the rationals in [0,1].
    But \sum\limits_{n = 1}^\infty  {\frac{\varepsilon }{{2^n }}}  = \frac{\varepsilon }{{2 - \varepsilon }} < 1, so G^c\cap [0,1] is not empty.
    Moreover G^c is closed and [0,1] is compact.
    So?
    Thanks. I got that part now. I still don't quite understand why 0 and 1 are not elements of G^{c}, but I'll try and figure it out.
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  11. #11
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    Quote Originally Posted by miatchguy View Post
    why 0 and 1 are not elements of G^{c}, but I'll try and figure it out.
    Well both 0 & 1 are rational.
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    Quote Originally Posted by Plato View Post
    Well both 0 & 1 are rational.
    Well yeah, but the can't be included in G^{c} because inf(G^{c})>0 and sup(G^{c})<1 and since G is open over [0, 1] it can't include 0 or 1 either. So that's what I'm having trouble figuring out.
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    Quote Originally Posted by miatchguy View Post
    Well yeah, but the can't be included in G^{c} because inf(G^{c})>0 and sup(G^{c})<1 and since G is open over [0, 1] it can't include 0 or 1 either. So that's what I'm having trouble figuring out.
    1) G is not a cover of [0,1]. Just the rationals in [0,1]

    2) 0=q_j~\&~1=q_k for some j~\&~k.
    So 0\in B(0;\epsilon/2^j) and 1\in B(1;\epsilon/2^k).
    Thus \inf(F)\ge\epsilon/2^j and \sup(F)\le1-\epsilon/2^k.
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  14. #14
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    Quote Originally Posted by Plato View Post
    1) G is not a cover of [0,1]. Just the rationals in [0,1]

    2) 0=q_j~\&~1=q_k for some j~\&~k.
    So 0\in B(0;\epsilon/2^j) and 1\in B(1;\epsilon/2^k).
    Thus \inf(F)\ge\epsilon/2^j and \sup(F)\le1-\epsilon/2^k.
    Cool, this is exactly what I was looking for.
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