- #1
JG89
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Is every sequence that converges a Cauchy sequence (in that for every e > 0, there is an integer N such that |a_n - a_m| < e whenever n,m > N)?
I think it is because if a sequence a_n converges to L, then you can mark off an open interval of any size about L such that this interval contains all of a_n except for at most a finite amount. So as the length of this open interval decreases then the distance between each point a_n gets closer and closer together. In fact maybe this open interval would be the epsilon neighborhood, because for two points a_n and a_m that are contained in this interval, we have |a_n - a_m| < e. Then within this epsilon neighborhood, we have points, a_j and a_p, where j,p > n, and j,p > m which are even closer to the limit, L, and again we have |a_j - a_p| < e, for an even smaller value of epsilon. We can obviously continue in this pattern, taking epsilon smaller and smaller as the integer n > N for the a_n gets larger and larger. And so in general, for large enough n and m, we have |a_n - a_m| < e.
Am I correct?
I think it is because if a sequence a_n converges to L, then you can mark off an open interval of any size about L such that this interval contains all of a_n except for at most a finite amount. So as the length of this open interval decreases then the distance between each point a_n gets closer and closer together. In fact maybe this open interval would be the epsilon neighborhood, because for two points a_n and a_m that are contained in this interval, we have |a_n - a_m| < e. Then within this epsilon neighborhood, we have points, a_j and a_p, where j,p > n, and j,p > m which are even closer to the limit, L, and again we have |a_j - a_p| < e, for an even smaller value of epsilon. We can obviously continue in this pattern, taking epsilon smaller and smaller as the integer n > N for the a_n gets larger and larger. And so in general, for large enough n and m, we have |a_n - a_m| < e.
Am I correct?
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