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Math Help - Quesion on topology relating to produc spaces

  1. #1
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    Question on topology relating to product spaces

    Hello: I have what may be a silly question because did not find it in the books I looked for it and is this: Is the product of closed sets closed in the product topology?
    The answer seems to me to be "yes".
    Last edited by facenian; March 1st 2010 at 06:45 AM.
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  2. #2
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    Yes, the product of two closed sets is closed in the product topology, just as the product of two open sets is open.

    But be careful! just as not every set in X x Y is the product of sets in x and y, so closed (or open) sets in X x Y may not be the product of closed (or open) sets in X and Y.

    For example, the singleton set {(x, y)) in X x Y is the product of the two singleton sets {x} and {y} but {(x,y), (x, z)} in X x Y (so both y and z are in Y) is NOT the product of two sets in X and Y.
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  3. #3
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    Thank you but my question was more general in the sense that I asked whether the product of any number of closed sets y general product space not yust a finite product space
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by facenian View Post
    Thank you but my question was more general in the sense that I asked whether the product of any number of closed sets y general product space not yust a finite product space
    Theorem: let \left\{X_j\right\}_{j\in\mathcal{J}} be a non-empty class of topological spaces and let X=\prod_{j\in\mathcal{J}}X_j be under the product topology. Then, given any class of subsets \left\{E_j\right\}_{j\in\mathcal{J}} such that E_j\subseteq X_j,\text{ }\forall j\in\mathcal{J} we have that \prod_{j\in\mathcal{J}}\overline{E_j}=\overline{\p  rod_{j\in\mathcal{J}}E_j}

    Proof: Let \bold{x}\in\prod_{j\in\mathcal{J}}\overline{E_j}. Let N be any neighborhood of \bold{x}. Clearly we may find a basic open set B (in the defining open base) such that \bold{x}\in B\subseteq N. Since \pi_j(\bold{x}) is an adherent point of E_j we may pick some \alpha_j\in E_j\cap \pi_j(B). Clearly then \prod_{j\in\mathcal{J}}\{\alpha_j\}\in B\cap \prod_{j\in\mathcal{J}}E_j. Since N was arbitrary it follows that \bold{x} is an adherent point for \prod_{j\in\mathcal{J}}E_j and thus \bold{x}\in\overline{\prod_{j\in\mathcal{J}}E_j}

    Conversely, let \bold{x}\in\overline{\prod_{j\in\mathcal{J}}E_j}. Let j\in\mathcal{J} be arbitrary and consider any neighborhood N_j of \pi_j(\bold{x}). Since \pi_j^{-1}(N_j) is open in the product topology we see that it contains a point \bold{y}\in\prod_{j\in\mathcal{J}}E_j. It follows then that \pi_j(\bold{y})\in E_j\cap N_j. In other words, \pi_j(\bold{x}) is an adherent point for E_j and thus \pi_j(\bold{x})\in\overline{E_j}. It follows that \bold{x}\in\prod_{j\in\mathcal{J}}\overline{E_j}.

    The conclusion follows. \blacksquare

    There are two nice corollaries

    Corollary:Let \left\{E_j\right\}_{j\in\mathcal{J}} be a collection of closed sets such that E_j\subseteq X_j. Then \overline{\prod_{j\in\mathcal{J}}E_j}=\prod_{j\in\  mathcal{J}}\overline{E_j}=\prod_{j\in\mathcal{J}}E  _j from where it follows that \prod_{j\in\mathcal{J}}E_j is closed.

    Corollary: Let \left\{\mathfrak{D}_j\right\}_{j\in\mathcal{J}} be a collection of dense subsets of X_j. Then, \overline{\prod_{j\in\mathcal{J}}\mathfrak{D}_j}=\  prod_{j\in\mathcal{J}}\overline{\mathfrak{D}_j}=\p  rod_{j\in\mathcal{J}}X_j=X. From where it follows that \prod_{j\in\mathcal{J}}\mathfrak{D}_j is dense in X
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  5. #5
    Member mabruka's Avatar
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    but {(x,y), (x, z)} in X x Y (so both y and z are in Y) is NOT the product of two sets in X and Y.
    But drexel isnt

    {(x,y), (x, z)} = \{x\}\times \{y,z\} ?

    Sorry i did not understand your example
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mabruka View Post
    But drexel isnt

    {(x,y), (x, z)} = \{x\}\times \{y,z\} ?

    Sorry i did not understand your example
    I didn't say that.
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