00A05-Subsequence.tex

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\pmtitle{subsequence}
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\newcommand{\Nats}{\mathbb{N}}
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\begin{document}
Given a sequence $\{x_n\}_{n\in \Nats}$, any infinite subset of the sequence forms a subsequence. We formalize this as follows:

\begin{defn}
If $X$ is a set and $\{a_n\}_{n \in \mathbb{N}}$ is a sequence in $X$, then a \emph{subsequence} of $\{a_n\}$ is a sequence of the form $\{a_{n_r}\}_{r \in \mathbb{N}}$ where $\{n_r\}_{r \in \mathbb{N}}$ is a strictly increasing sequence of natural numbers.
\end{defn}

Equivalently, $\{y_n\}_{n\in \Nats}$ is a subsequence of $\{x_n\}_{n\in \Nats}$ if
\begin{enumerate}
\item $\{y_n\}_{n\in\Nats}$ is a sequence of elements of $X$, and

\item there is a strictly increasing function $a:\Nats \to \Nats$ such that $$y_n = x_{a(n)} \quad \text{ for all } n\in\Nats.$$
\end{enumerate}

\begin{exa}
Let $X=\Reals$ and let $\{x_n\}$ be the sequence 
$$\left\{\frac{1}{n}\right\}_{n\in\Nats}=\left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\right\}.$$ Then, the sequence $$\{y_n\}_{n\in\Nats}=\left\{\frac{1}{n^2}\right\}_{n\in\Nats}=\left\{1,\frac{1}{4},\frac{1}{9},\frac{1}{16},\ldots\right\}$$
is a subsequence of $\{x_n\}$. The subsequence of natural numbers mentioned in the definition is $\{n^2\}_{n\in\Nats}$ and the function $a:\Nats\to\Nats$ mentioned above is $a(n)=n^2$.
\end{exa}
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