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Section 5.5 Subsequences

Consider the sequence \(x_n=(-1)^{n}\text{,}\) the simplest example of a bounded divergent sequence. Note that, while this sequence doesn't approach one value as \(n\rightarrow\infty\text{,}\) it concentrates on two values, and if you throw out all the odd numbered terms of the sequence, you get the new sequence \(x_{2n}=(-1)^{2n}=1\text{,}\) which does converge. This was quite a simple sequence, but it turns out this is more than just coincidence: for any bounded sequence, we can throw out some terms so that the sequence becomes convergent. Below we make more precise what we mean by “throwing out” terms.

Definition 5.5.1.

Given a sequence \((x_{n})\text{,}\) a subsequence is a sequence of the form \((x_{n_{k}})\) where \(n_{1}\lt n_{2}\lt \cdots\) are positive integers. A real number \(x\) is a limit point of a sequence \((x_{n})\) if there is a subsequence \((x_{n_{k}})\) so that \(x_{n_{k}}\rightarrow x\text{.}\)

Show that if \(n_{1}\lt n_{2}\lt \cdots\) is a strictly increasing sequence of positive integers, then \(n_k\geq k\) for all \(k\in\NN\text{.}\)

The following gives a criterion for detecting limit points:

If \(L\) is a limit point, we will show that for all \(\epsilon>0\) and for all integers \(N\text{,}\) we can find \(n> N\) so that \(|x_n-L|\lt \epsilon\text{.}\) Let \(\epsilon>0\) and \(N\in\NN\text{.}\) Since \(L\) is a limit point, there is a subsequence \((x_{n_{k}})\) converging to \(L\) with \(n_{1}\lt n_{2}\lt \cdots\text{,}\) so there is \(K\in\NN\) so that \(k> K\) implies \(|x_{n_{k}}-L|\lt \epsilon\text{.}\) By the previous exercise, \(n_{k}\geq k\text{,}\) so if \(k> \max\{N,K\}\text{,}\) then \(n_k\) is an integer bigger than \(N\) so that \(|x_{n_{k}}-L|\lt \epsilon\text{.}\) This proves the forward implication of the proposition.

Now we prove the converse. Assume \(L\) is a number so that for all \(\epsilon>0\) and for all integers \(N\text{,}\) we can find \(n> N\) so that \(|x_n-L|\lt \epsilon\text{.}\) We will show \(L\) is a limit point.

By our assumption (with \(\epsilon=1\) and \(N=1\)), we can find \(n_1\in\NN\) so that

\begin{equation*} |x_{n_{1}}-L|\lt 1\text{.} \end{equation*}

By our assumption again (now with \(\epsilon=\frac{1}{2}\) and \(N=n_{1}+1\)) we can find \(n_2>n_1\) so that

\begin{equation*} |x_{n_{2}}-L|\lt \frac{1}{2}\text{.} \end{equation*}

Now suppose we have picked \(n_{k-1}\) for some integer \(k\geq 2\text{.}\) Using the same logic, we can find \(n_{k}>n_{k-1}\) so that

\begin{equation} |x_{n_{k}}-L|\lt \frac{1}{k}\text{.}\label{e_xn-L_1_k}\tag{5.5.1} \end{equation}

If we continue in this way, we get a sequence of integers \(n_1\lt n_2\lt \cdots\) so that (5.5.1) holds for all \(k\text{.}\) This means \(x_{n_k}\rightarrow L\text{:}\) For \(\epsilon>0\text{,}\) if \(N=\frac{1}{\epsilon}\text{,}\) then for \(k> N\text{,}\)

\begin{equation*} |x_{n_{k}}-L|\lt \frac{1}{k}\lt \frac{1}{N}=\epsilon\text{,} \end{equation*}

which implies \(x_{n_k}\rightarrow L\text{.}\)

We have already seen that \(x_n=(-1)^{n}\) has a subsequence \(x_{2k}=1\) that converges to \(1\text{.}\) Also, \(x_{2k+1}=(-1)^{2k+1}=-1\) is a subsequence that converges to \(-1\text{.}\) We claim that \(\{-1,1\}\) are the only limit points:

Suppose \(L\) is a limit point not equal to \(-1\) or \(1\text{.}\) Then for all \(n\geq 1\text{,}\) \(|x_{n}-L|=|(-1)^{n}-L|\) is either \(|1-L|\) or \(|-1-L|=|1+L|\text{,}\) hence

\begin{equation*} |x_{n}-L|\geq \min\{|1-L|,|1+L|\}\text{.} \end{equation*}

Let \(\epsilon=\min\{|1-L|,|1+L|\}\) and \(N=1\text{.}\)Note \(\epsilon>0\) since \(L\neq \pm 1\text{.}\) Then \(\epsilon>0\) and \(N\) so that for all \(n\geq N\text{,}\) \(|x_{n}-L|\geq \epsilon\text{.}\) This is the negation of Proposition 5.5.3, and so \(L\) is not a limit point. Thus, the only limit points are \(\pm 1\text{.}\)

Our next convergence theorem says that, while a sequence does not need to converge, if it is bounded then we can always find a convergent subsequence, no matter how wild the original sequence is.

Let

\begin{equation*} S=\{x : \text{ there are infinitely many \(n\in\NN\) so that } x_n\geq x\} \end{equation*}

and set \(L=\sup (S)\text{.}\) Since \((x_n)\) is bounded, \(\sup(S)\) exists. By Proposition 5.5.3, \(L\) is a limit point if we can verify the following:

Claim: For all \(\epsilon>0\) and \(N\in\NN\) there is \(n>N\) so that \(|x_{n}-L|\lt \epsilon\text{.}\)

We prove by contradiction. Suppose instead that there exists \(\epsilon>0\) and \(N\in\NN\) so that \(|x_{n}-L|\geq \epsilon\) for all \(n> N\text{.}\) In other words, there are at most \(N\) many integers \(n\) (hence only finitely many) for which \(|x_{n}-L|\lt \epsilon\text{,}\) or equivalently

\begin{equation} L-\epsilon\lt x_n\lt L+\epsilon\text{.}\label{e_finitely-many-betweenL-eL_e}\tag{5.5.2} \end{equation}

Since \(L+\epsilon>L=\sup(S)\) and \(\sup(S)\) is an upper bound for \(S\text{,}\) we can't have \(L+\epsilon\in S\text{,}\) so by the definition of \(S\text{,}\) there are only finitely many \(n\in\NN\) so that

\begin{equation} x_n\geq L+\epsilon\label{e_finitely-many-aboveL_e}\tag{5.5.3} \end{equation}

Thus, there are only finitely many \(n\) satisfying (5.5.2) or (5.5.3), so there can only be finitely many \(n\) so that \(x_n>L-\epsilon\text{.}\) But then \(L-\epsilon\geq \sup(S)\text{,}\) since if \(L-\epsilon\lt \sup(S)\text{,}\) by definition of the sup, \(L-\epsilon\) is not an upper bound for \(S\text{,}\) so there is \(t>L-\epsilon\) so that there are infinitely many \(x_n\geq t\text{,}\) which is impossible since there are only finitely many \(x_n>\epsilon\text{.}\) Thus, \(L-\epsilon\geq \sup(S)=L\text{,}\) which is a contradiction. This proves the claim, and thus \(L\) is a limit point.