Definition
A subsequence of a sequence \((x_n)_n\) is a sequence of the form \((x_{n_k})_k\) with \(1 < n_1 < n_2 < n_3 \cdots \).

Example: Suppose we have a sequence \((x_n)_n = 1,2,3,4,5,6,7,8,9,\ldots\). Then a subsequence of this could be \(2,4,6,8,\ldots\). Here we pick \(n_1 = 2, n_2 = 4, n_3 = 6\). That is \(n_k\) refers to the indices we picked from the original sequence. The outside \(k\) is just the indices into the new subsequence. So for this example, we can write \((x_{2k})_k\) which gets us \(x_{2}, x_{3}, x_{4},\cdots\).

Another subsequence is \((x_{k^2})_k\). So now we get \(x_1, x_4, x_9, x_{16},\ldots\). Basically the inside index, refers to what indices we’re picking from the original sequence for each \(k = 1,2,3,...\)etc. So now we have this theorem next:

Theorem
\((x_n)_n\) converges to \(x\) if and only if every subsequence of \((x_n)_n\) converges to \(x\)

Proof

\(\Rightarrow:\) Suppose \((x_n)_n\) converges to \(x\). This means that for every \(\epsilon > 0\), there exists an \(N \in \mathbb{N}\) such that if \(n \geq N\), then

$$ \begin{align*} |x_n - x| < \epsilon. \end{align*} $$

Now, let \((x_{n_k})_k\) be any subsequence of \((x_n)_n\). We can show by induction that since \(n_1 < n_2 < n_3 < \ldots\), then we must have that \(n_k \geq k\) for all \(k \in \mathbb{N}\). (This is easy to see since we can never go backwards. So at any point in the original sequence, the index in the original sequence must be \(\geq\) the index of the element in the subsequence). So now in the subsequence, choose \(k\) such that \(k \geq N\). This means that \(n_k \geq k \geq N\) so \(n_k \geq N\). But this implies that

$$ \begin{align*} |x_{n_k} - x| < \epsilon. \end{align*} $$

But this just says that \(x_{n_k} \rightarrow x\) as \(k \rightarrow \infty\).

\(\Leftarrow:\) This direction is trivial since we can just take the original sequence as a subsequence of itself. So since every subsequence converges, then we must have that \((x_n)_n\) converges as well. \(\ \blacksquare\)

References