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Thread: Triangle inequality applied to sets and a point

  1. #1
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    Triangle inequality applied to sets and a point

    Let A and B be non-empty subsets of Y, and let $\displaystyle z \in Y$. Then

    $\displaystyle d(A,B) \leq d(A,z)+d(z,B)$.

    Obviously, this is a restatement of the triangle inequality, but I'm having a hard time putting the proof into words. Should I begin by assuming $\displaystyle d(A,B)=t_1=|a-b|$, where (a,b) is in AxB? Because then I run into a problem when I assume that
    $\displaystyle d(A,z)=t_2=|a-z|$ because I'm using the same element $\displaystyle a \in A$, which I don't think I can assume.

    Can anyone give me some help with this?
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    Senior Member vincisonfire's Avatar
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    Re: Triangle inequality applied to sets and a point

    Assuming that $\displaystyle d$ is a metric, the above is true.
    Isn't the problem more like :
    Here is a definition of $\displaystyle d$, prove that it is a distance, and hence that it satisfies the triangle inequality.
    An intuitive definition of $\displaystyle d$ would be $\displaystyle min\{d(a,b) | a\in A b\in B\}$.
    In this case you can proceed as you said from a pain $\displaystyle (a,b)$ such that d(a,b) = d(A,B).
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