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Math Help - continuity of a function; uniform topology on R^\omega

  1. #1
    Senior Member abhishekkgp's Avatar
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    continuity of a function; uniform topology on R^\omega

    Consider \mathbb{R} in the standard topology and \mathbb{R}^{\omega} in the uniform topology.
    Consider the function g:\mathbb{R} \Rightarrow \mathbb{R}^{\omega} given by g(t)=(t,t,t, \ldots).
    QUESTION: Is g continuous??

    MY ATTEMPT:

    Let  \bold{x} \in \mathbb{R}^{\omega} where \bold{x}=(x_1,x_2,\ldots).

    Let 1> \epsilon >0.

    To see whether g^{-1}(B_{\overline{\rho}}(\bold{x},\epsilon)) is open in \mathbb{R} or not.

    I have worked out that B_{\overline{\rho}}(\bold{x},\epsilon)=\bigcup_{ \delta < \epsilon}U(\bold{x},\delta), where

    U(\bold{x},\delta)=(x_1-\delta ,x_1+\delta )\times (x_2-\delta ,x_2+\delta )\times \ldots

    Using this i proved that g^{-1}(B_{\overline{\rho}}(\bold{x},\epsilon))=\bigcup  _{ \delta < \epsilon}\left(\bigcap_{i \in \mathbb{Z}^+}(x_i- \delta, x_i + \delta) \right)

    I am not able to see whether \bigcup_{ \delta < \epsilon}\left(\bigcap_{i \in \mathbb{Z}^+}(x_i- \delta, x_i + \delta) \right) is open in \mathbb{R} or not. Can someone please help me on this??
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Re: continuity of a function; uniform topology on R^\omega

    Okay i think i've got it. 'g' is continuous.
    Anyways.. thanks to anyone who cared to have a look.
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  3. #3
    MHF Contributor Amer's Avatar
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    Re: continuity of a function; uniform topology on R^\omega

    why you did not try to prove that g is continuous at every point x in R

    g is continuous at x ( x in the domain sure ) if for every open set "v" containing g(x) we can find an open set "u" in
    R such x \in u and g(u) \subset v

    how is the open sets in the uniform topology ?
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