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Math Help - Product Topology

  1. #1
    MHF Contributor Drexel28's Avatar
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    Product Topology

    Hello everyone. I have the following problem.

    Let X_j=[0,1] with the usual topology for every j\in[0,1] and let X=\prod_{j\in[0,1]}X_j. Prove that X is not second countable.
    The quote says to consider that if X were second countable that every open base must have a countable subset which is also an open base.

    But, I don't understand why this doesn't work.

    Clearly we have that G_j=\prod_{k\in[0,1]}E_k where E_k=\begin{cases} (0,1) & \mbox{if} \quad k=j \\ X_k & \mbox{if} \quad k\ne j\end{cases} is open in X (in fact it's a subbasic open set). So, we have that O=\bigcup_{j\in[0,1]}G_j is an open set in X. If X were second countable we would have, by Lindelof's theorem, that O must have a countable subcover. But, that isn't the case here. Right?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Drexel28 View Post
    Hello everyone. I have the following problem.



    The quote says to consider that if X were second countable that every open base must have a countable subset which is also an open base.

    But, I don't understand why this doesn't work.

    Clearly we have that G_j=\prod_{k\in[0,1]}E_k where E_k=\begin{cases} (0,1) & \mbox{if} \quad k=j \\ X_k & \mbox{if} \quad k\ne j\end{cases} is open in X (in fact it's a subbasic open set). So, we have that O=\bigcup_{j\in[0,1]}G_j is an open set in X. If X were second countable we would have, by Lindelof's theorem, that O must have a countable subcover. But, that isn't the case here. Right?
    Just as a follow-up. I posted this on the AOPS website and got an affirmative response.

    Clearly O is an open set engineered to have an uncountable open cover with no proper subcover. The fact that it cannot satisfy Lindelof's theorem, and thus cannot be second countable follows.
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  3. #3
    Member mabruka's Avatar
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    Right, i agree it can not have a countable subcover.

    Suppose there is, O \subset \bigcup_{j=i_1,i_2,...}G_j for some numbers \{i_1,i_2,...\} \subset [0,1]

    Now let x \in X be such an element that \pi_j (x) = 1 for every j \in \{i_1,i_2,...\} where \pi_j is the projection on the jth coordenate.

    Clearly x \not \in\bigcup_{j=i_1,i_2,...}G_j but x \in \bigcup_{j\in[0,1]}G_j = O




    What do you think?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by mabruka View Post
    Right, i agree it can not have a countable subcover.

    Suppose there is, O \subset \bigcup_{j=i_1,i_2,...}G_j for some numbers \{i_1,i_2,...\} \subset [0,1]

    Now let x \in X be such an element that \pi_j (x) = 1 for every j \in \{i_1,i_2,...\} where \pi_j is the projection on the jth coordenate.

    Clearly x \not \in\bigcup_{j=i_1,i_2,...}G_j but x \in \bigcup_{j\in[0,1]}G_j = O




    What do you think?
    Yes. I agree. Here is what I said. Let \bold{x}_\ell be such that \pi_\ell\left(\bold{x}_\ell\right)=\frac{1}{2} and \pi_{k}\left(\bold{x}_j\right)=1,\text{ }k\ne \ell. Clearly we have that \bold{x}_\ell is only in G_\ell and so removing it from \left\{G_j\right\}_{j\in[0,1]} will result in a cover that no longer covers O. Since \ell was arbitrary it follows that the removal of any set in \left\{G_j\right\}_{j\in[0,1]} will result in a non-cover. Thus, \left\{G_j\right\}_{j\in[0,1]} has no proper subcover and thus clearly no countable subcover.
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