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Math Help - Proving a function is continuous in a metric spaces

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    Proving a function is continuous in a metric spaces

    Let A be a non-empty set in a metric space (X,d). Define f : X \rightarrow \mathbb{R} by f(x) = inf \{d(a,x) : a \in A \}
    . Prove that f is continuous.

    I dont see how to bound f(x) and f(y)
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Proving a function is continuous in a metric spaces

    Hint The class of subsets of \mathbb{R} in the form G_{\alpha}=(\alpha,+\infty) or H_{\beta}=(-\infty,\beta) is a subbasis of (\mathbb{R},T_u) . Prove that f^{-1}(G_{\alpha}) and f^{-1}(H_{\beta}) are open sets.
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    Re: Proving a function is continuous in a metric spaces

    Quote Originally Posted by FGT12 View Post
    Let A be a non-empty set in a metric space (X,d). Define f : X \rightarrow \mathbb{R} by f(x) = \inf \{d(a,x) : a \in A \}. Prove that f is continuous.
    Suppose that a\in A.
    f(x)=\inf\{d(x,t):t\in A\}\le d(x,a)
    d(a,x)\le d(a,y)+d(y,x) or d(a,x)-d(y,x)\le d(a,y) for each a\in A

    Thus f(x)-d(x,y)\le f(y) or f(x)-f(y)\le d(x,y)

    Can you continue to prove that |f(x)-f(y)|\le d(x,y)~?
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    Super Member ILikeSerena's Avatar
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    Re: Proving a function is continuous in a metric spaces

    Quote Originally Posted by Plato View Post
    d(a,x)-d(y,x)\le d(a,y) for each a\in A

    Thus f(x)-d(x,y)\le f(y)
    How did you make this step?

    I understand that f(x)≤d(a,x), but how did you introduce f(y)?
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    Re: Proving a function is continuous in a metric spaces

    Quote Originally Posted by ILikeSerena View Post
    How did you make this step?
    I understand that f(x)≤d(a,x), but how did you introduce f(y)?
    Note that d(a,x)\le d(a,y)+d(y,x) or d(a,x)-d(y,x)\le d(a,y) for each a\in A
    implies that f(x)-d(x,y) is a lower bound for the set \{d(y,a):a\in A\}~.
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    Super Member ILikeSerena's Avatar
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    Re: Proving a function is continuous in a metric spaces

    Quote Originally Posted by Plato View Post
    Note that d(a,x)\le d(a,y)+d(y,x) or d(a,x)-d(y,x)\le d(a,y) for each a\in A
    implies that f(x)-d(x,y) is a lower bound for the set \{d(y,a):a\in A\}~.
    Thanks!
    I got it now.
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