## Example 3.4.5:

If is the sequence
of all rational numbers in the interval [0, 1], enumerated in any way,
find the

Since the numbers 1 and 0 are itself rational numbers, it is clear that
*lim sup*and*lim inf*of that sequence.sup{ aand_{n}} = 1inf{ aTherefore, we already know that_{n}} = 0To find the0 lim inf{ a_{n}} lim sup{ a_{n}} 1lim sup, we will construct a subsequence that converges to 1:These numbers exist because the rational numbers in the interval [0, 1] are arbitrarily close to any real number in that interval, according to the Density principle.

- there exists
0 < awith_{j1}< 11 - a_{j1}< 1- there exists
0 < awith_{j2}< 11 - aand_{j2}< 1/2a_{j1}# a_{j2}- there exists
0 < awith_{j3}< 11 - aand_{j3}< 1/3adifferent from the previous ones_{j3}- and so on ...
The subsequence

{ aconstructed in the above way converges to 1. We already know that any limit of any convergent subsequence must be less than or equal to 1. Therefore, since the_{jk}}lim supis the greatest limit of any convergent subsequence, we haveSimilarly, we can extract a subsequencelim sup { a_{n}} = 1{ athat converges to 0. We also know that every limit of any convergent subsequence must be greater or equal to zero. Therefore, since the_{jk}}lim infis the smallest possible limit of all convergent subsequence, we have:lim inf { a_{n}} = 0