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

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

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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].

 

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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