Convergence in a Discrete Metric Space - What Does It Mean?

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The discussion centers on the concept of convergence in a discrete metric space, defined by the distance function d((x1,x2,...xn),(y1,y2,...yn)) which equals 0 if xi=yi and 1 otherwise. It establishes that a metric space is complete if every Cauchy sequence converges, with the condition that a Cauchy sequence must be eventually constant for convergence to occur. The conclusion emphasizes that in a discrete metric space, convergence is straightforward due to the nature of the distance function.

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how is discrete metric space given by d((x1,x2,...xn)(y1,y2,...yn))=0 if xi=yi else 1
disc
is complete
 
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Have you thought about what convergence of a sequence means in a discrete space?

A metric space is "complete" if and only if every Cauchy sequence converges. And, of course, a Cauchy sequence is one where \lim_{m,n\rightarrow \infty} d(a_n,a_m)= 0. Since d(x,y)= 1 for x\ne y, in order for that to happen the sequence must be "eventually constant", i.e. for some N, if n,m> N, a_n= a_m and it is easy to show that such a sequence converges.
 

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