## Proposition 3.3.3: Subsequences from Convergent Sequence

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*{a*converges to some limit

_{j}}*L*. Take any sequence

*n*of the natural numbers and consider the corresponding subsequence of the original sequence. For any

_{j}*> 0*there exists an integer

*N*such that

as long as| a_{n}- L | <

*n > N*. But then we also have the same inequality for the subsequence as long as

*n*. Therefore any subsequence must converge to the same limit

_{j}> N*L*.

The second statement is just as easy. Suppose *{a _{j}}*
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

where| a_{j}- L | <

*j*is odd and

*j > N*. Now extract every other element, starting with the second. The resulting subsequence again converges to

*L*, so that

where| a_{j}- L | <

*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*| a*because_{j}- L | <*a*is part of the first subsequence_{j} - if
*j*is even, then*| a*because_{j}- L | <*a*is part of the second subsequence_{j}

*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.