Prove that every convergent sequence is a Cauchy sequence?

Answer 1

Assume you have a convergent sequence in a metric space with metric d. Let the sequence be determined by {x_n} and the limit by x. Then we have

d( x_n, x ) --> 0, since the sequence converges to x.

But d( x_n, x_m ) <= d( x_n, x ) + d( x_m, x ), where the right hand side tends to zero. Therefore the sequence is Cauchy.

This is not completely rigorous, but hopefully is enough that you can work out the details on your own.

Steve

Answer 2

The triangle inequality is your friend.

If two points are each within epsilon/2 of the limit, then they are also within epsilon of each other.