Post
by **Nr** » Wed Nov 07, 2012 10:28 am

Hi,

Actually, you can solve question 1 and 3 together.

Take two alternating sequences. For example, 1, -1, 1, -1, ... and 2, 3, 2, 3, 2, 3, ... (there are of course lots of other examples.) Then both are bounded (they are always smaller than, let's say, 5.) and both have convergent sub-sequences, (just take the even sub-sequence which is just the constant sequence -1, -1,-1, ... or 3, 3, 3, ... )

However, these sequences are clearly neither Cauchy nor convergent.

It's basically obvious, but if you want to do it in a official way, just take epsilon to be 1/2. No difference is smaller than 1/2, let alone all differences from some point on. Now, it's clearly is not convergent (every convergent sequence in a metric space is Cauchy, right ?)

For question 2, for the proof we do something like this: Let {an} (n subscript) be a convergent sequence and {ank} (k a subscript of n) be some sub sequence and e>0. Then, because {an} is convergent there is a N such that if n>N then |an-a| < e. (where a is clearly just the limit of {an}. But now if nk >N then |ank-a| < e. So, we just find N1, such that whenever k>N1 then nk>N (the existence of N1 is clear because the nk are the index set of a sequence and so they go towards infinity). That's it (I'm really not sure if this is too technical or not technical enough for you. For examples, we just take two convergent sequences like {1/n}:1, 1/2, 1/3, ... and (n+1)/n:2, 3/2, 4/3, ... . Now for the first one a famous sub-sequence is 1/(n^2): 1, 1/4, 1/9 ... , which obviously converges to zero. For the other one you can take 1+1/2n: 3/2, 5/4, ... . And this converges to 1 (1/2n converges to 0 and 1 converges to 1 so their sum converges to 1.)

Is it clear or ... ???