d((x_n),(y_n)) := 2 ^{−min{n∈N: x_n≠y_n}} defines a metric on S={0,1}^ℕ

Discussion in 'Analysis and Topology' started by Polleei, May 18, 2022.

  1. Polleei

    Polleei

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    im currently struggling to prove the statement in the title. I'm aware of how a metric is defined but i have no clue how to prove this. Here is the whole problem, but better readable:

    Let S := {0,1}^ℕ the set of 0-1 sequences. Prove that d((x_n),(y_n)) := 2 ^{−min{n∈N: x_n≠y_n}} defines a metric on S.

    Found nothing that could help to solve this problem on the internet. I'd be very thankfull for help on this topic :)

    Best regards
     
    Polleei, May 18, 2022
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  2. Polleei

    nycmathguy

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    To get help with advanced mathematics, go to freemathhelp.com.
     
    nycmathguy, May 19, 2022
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  3. Polleei

    HallsofIvy

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    A metric on a set, X, is a function, d, from pairs (p, q) in X x Y to non-negative numbers, satisfying
    1) [math]d(p,q)\ge 0[/math] and d(p,q)= 0 if and only if p= q.
    2) d(p, q) is "symmetric". That is, d(p, q)= d(q, p).
    3) The "triangle inequality": for any p, q, r in X, [math]d(p, q)\le d(p, r)+ d(r, q)[/math].
     
    HallsofIvy, Oct 11, 2023
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