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Math Help - Theorem from Baby Rudin (Thm. 2.36)

  1. #1
    Member Mollier's Avatar
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    Theorem from Baby Rudin (Thm. 2.36)

    Theorem. If \{K_{\alpha}\} is a collection of compact subsets of a metric space X such that the intersection of every finite subcollection of \{K_{\alpha}\} is nonempty, then
    \bigcup_{\alpha}K_{\alpha} is nonempty.

    I try and prove theorems my self before reading them (seldom looks pretty), and the one I came up with for this theorem is the following.

    Theorem 2.34 says that compact subsets of metric spaces are closed. From this I know that every set in \{K_{\alpha}\} is closed.

    Theorem 2.24(b) says that for any collection \{K_{\alpha}\} of closed sets, \bigcap_{\alpha}K_{\alpha} is closed.

    Now comes my question. Since the empty set is both closed and open, can I just say that \bigcap_{\alpha}K_{\alpha} is nonempty since it is closed? I am pretty sure the answer is no, but it would be nice if someone could explain this a bit.

    Thanks.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Mollier View Post
    Theorem. If \{K_{\alpha}\} is a collection of compact subsets of a metric space X such that the intersection of every finite subcollection of \{K_{\alpha}\} is nonempty, then
    \bigcup_{\alpha}K_{\alpha} is nonempty.

    I try and prove theorems my self before reading them (seldom looks pretty), and the one I came up with for this theorem is the following.

    Theorem 2.34 says that compact subsets of metric spaces are closed. From this I know that every set in \{K_{\alpha}\} is closed.

    Theorem 2.24(b) says that for any collection \{K_{\alpha}\} of closed sets, \bigcap_{\alpha}K_{\alpha} is closed.

    Now comes my question. Since the empty set is both closed and open, can I just say that \bigcap_{\alpha}K_{\alpha} is nonempty since it is closed? I am pretty sure the answer is no, but it would be nice if someone could explain this a bit.

    Thanks.
    Ok, let me see if I can get a handle on this. One knows that compact subspaces of metric spaces are closed-true. Next, one knows that the arbitrary intersection of closed sets is closed-true. So your argument is that by these two theorems one would have that \displaystyle \bigcap_{\alpha\in\mathcal{A}}K_\alpha is a closed subspace of X and thus can't be empty since \varnothing is both open and closed? Unfortunately this is not the case :|. It's often true that a subspace of a metric space is both open and closed (clopen) for it not to be in general the space would have to be connected. It's often impossible (in full generality) to complete arguments such as this by looking at openess/closedness (although it works sometimes, like proving things like the fundamental lemma of path connectedness in alg. top.).


    For the argument I'll try to give you two well placed hints:


    Hint 1: If \displaystyle \bigcap_{\alpha\in\mathcal{A}}K_\alpha=\varnothing then \displaystyle \bigcup_{\alpha\in\mathcal{A}}\left(X-K_\alpha\right)=X.


    Hint 2: If for every finite \mathcal{B}\subseteq\mathcal{A} one has that \displaystyle \bigcap_{\beta\in\mathcal{B}}K_\beta\ne\varnothing then for every finite \mathcal{B}\subseteq\mathcal{A} one has \displaystyle \bigcup_{\beta\in\mathcal{B}}\left(X-K_\beta\right)\subsetneq X.


    So, since the K_\alpha are closed we know that the X-K_\alpha are open and so the above shows we've found a ____________ with no _____________.
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  3. #3
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    In fact, it's easy to see from Drexel28's argument that your property (which I've heard being called "finite intersection property") is equivalent to compactness (even in arbitrary topological spaces).
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    Member Mollier's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Ok, let me see if I can get a handle on this. One knows that compact subspaces of metric spaces are closed-true. Next, one knows that the arbitrary intersection of closed sets is closed-true. So your argument is that by these two theorems one would have that \displaystyle \bigcap_{\alpha\in\mathcal{A}}K_\alpha is a closed subspace of X and thus can't be empty since \varnothing is both open and closed? Unfortunately this is not the case :|. It's often true that a subspace of a metric space is both open and closed (clopen) for it not to be in general the space would have to be connected. It's often impossible (in full generality) to complete arguments such as this by looking at openess/closedness (although it works sometimes, like proving things like the fundamental lemma of path connectedness in alg. top.).


    For the argument I'll try to give you two well placed hints:


    Hint 1: If \displaystyle \bigcap_{\alpha\in\mathcal{A}}K_\alpha=\varnothing then \displaystyle \bigcup_{\alpha\in\mathcal{A}}\left(X-K_\alpha\right)=X.


    Hint 2: If for every finite \mathcal{B}\subseteq\mathcal{A} one has that \displaystyle \bigcap_{\beta\in\mathcal{B}}K_\beta\ne\varnothing then for every finite \mathcal{B}\subseteq\mathcal{A} one has \displaystyle \bigcup_{\beta\in\mathcal{B}}\left(X-K_\beta\right)\subsetneq X.


    So, since the K_\alpha are closed we know that the X-K_\alpha are open and so the above shows we've found a __member__ with no __brain__.
    Sorry for the late reply.
    I understand the proof in the book to some extent, but I fail to see how your argument works.
    If you have the time, I would appreciate it if you explain some more.
    Thanks.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Mollier View Post
    Sorry for the late reply.
    I understand the proof in the book to some extent, but I fail to see how your argument works.
    If you have the time, I would appreciate it if you explain some more.
    Thanks.
    Suppose that \displaystyle \bigcap_{\alpha\in\mathcal{A}}K_\alpha=\varnothing then by DeMorgan's law we get that \displaystyle \bigcup_{\alpha\in\mathcal{A}}\left(X-K_\alpha\right)=X but since the K_\alpha are closed we see that X-K_\alpha are open and so by compactness there exists some finite subcover X-K_{\alpha_1},\cdots,X-K_{\alpha_n} but \displaystyle \bigcup_{j=1}^{n}\left(X-K_{\alpha_j}\right)=X implies, once again by DeMorgan's law, that \displaystyle \bigcap_{j=1}^{n}K_{\alpha_j}=\varnothing contradicting the FIP of \left\{K_{\alpha}\right\}_{\alpha\in\mathcal{A}}.
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  6. #6
    Member Mollier's Avatar
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    Thanks Drexel, much appreciated.
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