|
Assume we have a sequence S that is a counter example to the claimed statement. We start by picking two different sequence members a and b. Next we pick t with a<t<b such that t is not in S. Because t is not a subsequential limit of S we can choose an ε>0 so that the interval [t-ε,t+ε] does not contain any other sequence members. We define M0:={xi | xi<t-ε} and M1:={xi | xi>t+ε}. The xi's in M0 form a subsequence of S, that fulfills the same condition as S itself, i.e. its subsequential limits are exactly its sequence elements. The same holds for M1.
Therefore we can continue the process and split M0 into two subsets M00 and M01 which are separated by some ε'. The same can be done for M1 and we get M10 and M11, separated by some ε''. Likewise, in a third step we arrive at sets M000,M001,..,M111, in step four we have M0000,... and so forth. All M's are infinite subsets of the members of S. This process can be continued infinitely, because otherwise we would end up with a M-set that contains only one member. But since this member x must be a subsequential limit and because the M-set is separated by some ε from other members and because the series x,x,x,... is not a subsequence of S, there must be infinitely many other sequence members in our M-set.
Now comes the trick. For any infinite binary sequence B=b1b2b3... (e.g. B=00101000...) we construct a subsequence SB of S as follows: For the first sequence member y1 of SB we pick any element in Mb1, for the second member y2 we pick an element in Mb1b2, next comes a y3 in Mb1b2b3 and so on.
This sequence SB contains a converging subsequence (except maybe for B=000.. and B=111.., which may be unbounded) and therefore has a subsequential limit that we call xB. Thanks to our choice of M's, which are all separated by some ε, all the xB's are different. Being subsequential limits of SB and therefore of S, all xB's are members of S. But how many such xB's do we have? Obviously there are as many as there are real numbers in the interval [0,1] (written in binary notation), which is known to be an uncountable set. Because our sequence is countable by definition, we have a contradiction.
|
