Interactive Real Analysis3.3. Subsequences
Proposition 3.3.3: Subsequences from Convergent Sequence
is a convergent
sequence, then every subsequence of that sequence converges to the same
limit
If is a sequence
such that every possible subsequence extracted from that sequences
converge to the same limit, then the original sequence also converges to
that limit.
ContextProof:
The first statement is easy to prove: Suppose the original sequence {aj} converges to some limit L. Take any sequence nj of the natural numbers and consider the corresponding subsequence of the original sequence. For any
> 0 there
exists an integer N such that
| an - L | <as long as n > N. But then we also have the same inequality for the subsequence as long as nj > N. Therefore any subsequence must converge to the same limit L.
The second statement is just as easy. Suppose {aj}
is a sequence such that every subsequence extracted from it converges to
the same limit L. Now take any
> 0. Extract from the
original sequence every other element, starting with the first. The
resulting subsequence converges to L by assumption, i.e. there
exists an integer N such that
| aj - L | <where j is odd and j > N. Now extract every other element, starting with the second. The resulting subsequence again converges to L, so that
| aj - L | <where j is even and j > N. But now we take any j, even or odd, and assume that j > N
- if j is odd, then
| aj - L | <
because aj is part of the first subsequence
- if j is even, then
| aj - L | <
because aj is part of the second subsequence
Note that we can see from the proof that if the "even" and "odd" subsequence of a sequence converge to the same limit L, then the full sequence must also converge to L. It is not enough to just say that the "even" and "odd" subsequence simply converge, they must converge to the same limit.
