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Thread: irrationals, metrizable, Michael Line

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    irrationals, metrizable, Michael Line

    Prove that the irrationals have usual neighborhoods and the rationals are isolated is metrizable.


    This is the flip-flop of the Michael Line. I do not see how to prove this. I know we want to use Urysohn Metrization Theorem, but I do not see how to use it now. Thanks in advance.
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    Quote Originally Posted by xboxlive89128 View Post
    Prove that the irrationals have usual neighborhoods and the rationals are isolated is metrizable.


    This is the flip-flop of the Michael Line. I do not see how to prove this. I know we want to use Urysohn Metrization Theorem, but I do not see how to use it now. Thanks in advance.
    If I understand the question correctly, you only need to check two things in order to apply the Urysohn theorem to this space:
    1) The space is regular (it is easily seen to be Hausdorff);
    2) The space is second-countable.

    An open sets in the space is the union of a usual open set with a subset of the rationals. So a closed set is a usual closed set less a subset of the rationals. If F is such a set then F = F_0\cap Q^c, where F_0 is closed in the usual sense and Q\subseteq\mathbb{Q}. If x\notin F then either x\notin F_0 (in which case you can separate x from F by a usual open set), or x\in Q (in which case {x} is an open set separating x from F). That proves (1).

    For (2), you can take a countable base for the usual topology on \mathbb{R} and add to it all the singleton sets \{\{x\}:x\in\mathbb{Q}\} to get a countable base.
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