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Math Help - metric spaces and infimums

  1. #1
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    Thumbs down metric spaces and infimums

    HI Ive got this question I dont know how to do;

    Let X be a metric space, and let Q,J be non-empty subsets of X. prove that

    inf{dist(x,J):x is a member of Q}= inf{dist(Q,y):y is a member of J}.


    I know that the dist(x,J):= inf{d(x,y)|y is a member of J}, but I am not sure how to tackle this question. Any help please?
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  2. #2
    Senior Member Tinyboss's Avatar
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    If d is a metric, then d(x,y)=d(y,x). You can get it from there--suppose the two infima were not equal, and derive a contradiction.
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    could i do this:

    inf{dist(x,J): x member of Q}
    = inf{inf{d(x,y)|y member of J}:x member of Q}

    Since we know inf{d(x,y)|x member of Q, y member of J}= d(Q,J)

    Then the above is = inf{d,(Q,J)},

    and then the same for the right hand side?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by choccookies View Post
    could i do this:

    inf{dist(x,J): x member of Q}
    = inf{inf{d(x,y)|y member of J}:x member of Q}

    Since we know inf{d(x,y)|x member of Q, y member of J}= d(Q,J)

    Then the above is = inf{d,(Q,J)},

    and then the same for the right hand side?
    \inf \{d(Q,J)\} makes very little sense.

    Try proving that each infimum is a lower bound of the other set.
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    So I shouldnt use any of what i wrote?
    Could you start me off on how to prove that?
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by choccookies View Post
    So I shouldnt use any of what i wrote?
    Could you start me off on how to prove that?
    So, if \alpha=\inf\left\{d(x,J):x\in Q\right\} and \beta=\inf \left\{d(Q,y):y\in J\right\} then you're really trying to prove that \alpha\leqslant \beta and \beta\leqslant \alpha. But, by definition it suffices to show that \alpha\leqslant \inf\left\{d(Q,y):y\in J\right\} and \beta\leqslant\left\{d(x,J):x\in Q\right\}. Can you work with that?
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  7. #7
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    Quote Originally Posted by Drexel28 View Post
    So, if \alpha=\inf\left\{d(x,J):x\in Q\right\} and \beta=\inf \left\{d(Q,y):y\in J\right\} then you're really trying to prove that \alpha\leqslant \beta and \beta\leqslant \alpha. But, by definition it suffices to show that \alpha\leqslant \inf\left\{d(Q,y):y\in J\right\} and \beta\leqslant\left\{d(x,J):x\in Q\right\}. Can you work with that?
    Thankyou so much for your help. I'm not just trying to get an answer off you!I really am trying to solve it, i just cant get my head around it!

    To prove alpha</beta, wouldnt we have to say that beta is also a lower bound of inf{d(Q,y):y member of J}? How can we just assume that?
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by choccookies View Post
    Thankyou so much for your help. I'm not just trying to get an answer off you!I really am trying to solve it, i just cant get my head around it!
    It's no problem! We all go through problems which just are too much info at first sight (which I can see how this would be one).

    To prove alpha</beta, wouldnt we have to say that beta is also a lower bound of inf{d(Q,y):y member of J}? How can we just assume that?
    We can't. That's exactly what I am saying. Recall that if \varnothing\subsetneq E\subseteq\mathbb{R} is bounded from below then x\leqslant \inf E if and only if x is a lower bound for E.

    So, to show that \alpha\leqslant \overbrace{\inf\left\{d(Q,y):y\in J\right\}}^{\beta} it suffices to show that \alpha is a lower bound for \left\{d(Q,y):y\in J\right\} and similarly to show that \beta\leqslant\overbrace{\inf\left\{d(x,J):x\in Q\right\}}^{\alpha} it suffices to show that \beta is a lower bound for \left\{d(x,J):x\in Q\right\}.

    But, to make things easier (we're back to the general case now) recall that to show that x is a lower bound of E it suffices to show that x\leqslant e for all e\in E.

    Can you apply this?
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  9. #9
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    Let’s simplify the notation.
    Let \alpha=\inf\{D(J,x):x\in Q\}~\&~\beta=\{D(Q,y):y\in J\}.
    Suppose that \alpha <\beta. So \left( {\exists q' \in Q} \right)\left[ {\alpha  \leqslant D(J,q') < \beta }\right].

    Suppose that \beta <\alpha. So \left( {\exists j' \in J} \right)\left[ {\beta  \leqslant D(Q,j') < \alpha }\right].

    But note that means \beta\leqslant d(j',q')~\&~\alpha\leqslant  d(j',q').

    Do you see a contradiction there?
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  10. #10
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    Quote Originally Posted by Drexel28 View Post
    It's no problem! We all go through problems which just are too much info at first sight (which I can see how this would be one).



    We can't. That's exactly what I am saying. Recall that if \varnothing\subsetneq E\subseteq\mathbb{R} is bounded from below then x\leqslant \inf E if and only if x is a lower bound for E.

    So, to show that \alpha\leqslant \overbrace{\inf\left\{d(Q,y):y\in J\right\}}^{\beta} it suffices to show that \alpha is a lower bound for \left\{d(Q,y):y\in J\right\} and similarly to show that \beta\leqslant\overbrace{\inf\left\{d(x,J):x\in Q\right\}}^{\alpha} it suffices to show that \beta is a lower bound for \left\{d(x,J):x\in Q\right\}.

    But, to make things easier (we're back to the general case now) recall that to show that x is a lower bound of E it suffices to show that x\leqslant e for all e\in E.

    Can you apply this?
    so i need to show that alpha is less than or equal to all elements in {d(Q,y):y member of J}?
    Also I just wanted to ask, have you just replaced 'dist' with the metric d throughout this question?
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  11. #11
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by choccookies View Post
    so i need to show that alpha is less than or equal to all elements in {d(Q,y):y member of J}?
    Also I just wanted to ask, have you just replaced 'dist' with the metric d throughout this question?
    Yes and yes in the sense that I call d(x,J)=\text{dist}(x,J)
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