# 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

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May 17, 2022
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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
2. ### nycmathguy

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To get help with advanced mathematics, go to freemathhelp.com.

nycmathguy, May 19, 2022
3. ### 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) $d(p,q)\ge 0$ 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, $d(p, q)\le d(p, r)+ d(r, q)$.

HallsofIvy, Oct 11, 2023
4. ### RobertSmart

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A metric on a set X is essentially a rule that assigns non-negative numbers to pairs of elements from X, satisfying three key properties:

1. Non-negativity: The distance between any two elements p and q in X is always greater than or equal to zero. It's zero only when the elements are identical: p=q.

2. Symmetry: The distance between p and q is the same as the distance between q and p. This mirrors our intuition that the distance between two points should be independent of the order in which we consider them.

3. Triangle Inequality: For any three elements p, q, and r in X, the distance between p and q is always less than or equal to the sum of the distances between p and r and between r and q. This property captures the idea that the shortest path between two points is a straight line and ensures consistency in the notion of distance within the set.

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RobertSmart, Apr 12, 2024