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Thread: Trouble with a product topology problem

  1. #1
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    Trouble with a product topology problem

    Hello, This is my problem: let $\displaystyle d:X\times X\rightarrow R$ be a distance on X. We know d is continous with the prodcut topology when the topology given to X is the metric topology. Show tha if d is continous with the product topology when the topology given to X is $\displaystyle \mathcal{F}$ then this topology is finner than the metric toplogy.
    Attempts: we know that there is a coarsest topology for which d is continous is $\displaystyle \{d^{-1}(A):A$ open set in R$\displaystyle \}$ I tried to prove that this topology is the product topology of the metric but this doesn't seem to be the case.
    I also tried to prove directly that if U is open in the metric topology then U is in $\displaystyle \mathcal{F}$
    Any comments will be appreciated.
    Last edited by facenian; Aug 24th 2010 at 03:50 AM. Reason: gramma correction
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  2. #2
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    The open balls $\displaystyle B(x_{0},\epsilon) := \{ y \in X : d(x_{0},y)<\epsilon \}$ are a basis for the topology on X, which is induced by the metric d.
    It is therefor sufficient to show that every open ball is also in $\displaystyle \mathcal{F}$.

    For every $\displaystyle x_{0} \in X$ consider the map

    $\displaystyle h: X \rightarrow \mathbb{R}$
    $\displaystyle h(x):=d(x,x_0)$
    , which is the composition of
    $\displaystyle f:X \rightarrow X \times X$
    ,$\displaystyle f(x):=(x,x_{0})$
    and
    $\displaystyle d:X\times X \rightarrow \mathbb{R}
    $

    both are continious in the topology $\displaystyle \mathcal{F}$, so the composition is also continous and we have
    $\displaystyle B(x_{0},\epsilon)=h^{-1}((-\infty , \epsilon)) \in \mathcal{F}$
    Last edited by Iondor; Aug 25th 2010 at 02:57 PM.
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  3. #3
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    I think you solved my problem, thanks
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