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For case a), we want to show that for any given a, and every open set containing a, that all the an are in the open set as long as n is greater than some M which we get to choose. Since the open set in only missing a finite number of elements, we can put them in order {b1,b2,…,bp}. Then in the seriesan, since b1 does not appear infinitely many times, there is a last occurrence, call it b1f. Similarly there is a last occurrence of b2, call it b2f. Then if n>M=max(b1f,b2f,…bpf), an will be in our open set. So the sequence converges to a
The arguments for b) and c) are similar.
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answered Nov 6 '11 at 19:48
Ross Millikan
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