# Interactive Real Analysis

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## 3.3. Subsequences

### Proposition 3.3.3: Subsequences from Convergent Sequence

If 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.

### Proof:

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
Hence, the original sequence must also converge to L.

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. Next | Previous | Glossary | Map