## metric space

Forum for the GRE subject test in mathematics.
johnny_01
Posts: 7
Joined: Wed Oct 31, 2012 9:17 am

### metric space

Problem
1#every bounded sequence in a metric space (X,d) may not Cauchy sequence.
give two examples and explain
2#If a sequence in a metric space (X,d) converges to a point x of X
then every subsequence of it also converges that point of X.
give two examples and explain
3#If a sequence has a convergent subsequence in a metric space then it is not necessary that
whole sequence should also converge.
give two examples and explain

thank you

Nr
Posts: 96
Joined: Tue Jan 24, 2012 6:09 am

### Re: metric space

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 ... ???

johnny_01
Posts: 7
Joined: Wed Oct 31, 2012 9:17 am

thank you Nr