2.5 Subsequences and cluster points  (Page 2/4)

We give next what is probably the most useful fundamental result about sequences, the Bolzano-Weierstrass Theorem. It is this theorem that willenable us to derive many of the important properties of continuity, differentiability, and integrability.

Bolzano-weierstrass

Every bounded sequence $\left\{a_n\right\}$ of real or complex numbers has a cluster point. In other words, every bounded sequence has a convergent subsequence.

The Bolzano-Weierstrass Theorem is, perhaps not surprisingly, a very difficult theorem to prove. We begin with a technical, but very helpful, lemma.

Let $\left\{a_n\right\}$ be a bounded sequence of real numbers; i.e., assume that there exists an $M$ such that $|a_n|\le M$ for all $n.$ For each $n\ge 1,$ let $S_n$ be the set whose elements are $\{a_n,a_{n+1},a_{n+2},...\}.$ That is, $S_n$ is just the elements of the tail of the sequence from $n$ on. Define $x_n=sup{S}_n={sup}_{k\ge n}{a}_k.$ Then

1. The sequence $\left\{x_n\right\}$ is bounded (above and below).
2. The sequence $\left\{x_n\right\}$ is non-increasing.
3. The sequence $\left\{x_n\right\}$ converges to a number $x.$
4. The limit $x$ of the sequence $\left\{x_n\right\}$ is a cluster point of the sequence $\left\{a_n\right\}.$ That is, there exists a subsequence $\left\{b_k\right\}$ of the sequence $\left\{a_n\right\}$ that converges to $x.$
5. If $y$ is any cluster point of the sequence $\left\{a_n\right\},$ then $y\le x,$ where $x$ is the cluster point of part (4). That is, $x$ is the maximum of all cluster points of the sequence $\left\{a_n\right\}.$

Since $x_n$ is the supremum of the set $S_n,$ and since each element of that set is bounded between $-M$ and $M,$ part (1) is immediate.

Since $S_{n+1}\subseteq S_n,$ it is clear that

showing part (2).

The fact that the sequence $\left\{x_n\right\}$ converges to a number $x$ is then a consequence of [link] .

We have to show that the limit $x$ of the sequence $\left\{x_n\right\}$ is a cluster point of $\left\{a_n\right\}.$ Notice that $\left\{x_n\right\}$ may not itself be a subsequence of $\left\{a_n\right\},$ each $x_n$ may or may not be one of the numbers $a_k,$ so that there really is something to prove. In fact, this is the hard part of this lemma.To finish the proof of part (4), we must define an increasing sequence $\left\{n_k\right\}$ of natural numbers for which the corresponding subsequence $\left\{b_k\right\}=\left\{a_{n_k}\right\}$ of $\left\{a_n\right\}$ converges to $x.$ We will choose these natural numbers $\left\{n_k\right\}$ so that $|x-a_{n_k}|<1/k.$ Once we have accomplished this, the fact that the corresponding subsequence $\left\{a_{n_k}\right\}$ converges to $x$ will be clear. We choose the $n_k$ 's inductively. First, using the fact that $x=lim{x}_n,$ choose an $n$ so that $|x_n-x|=x_n-x<1/1.$ Then, because $x_n=sup{S}_n,$ we may choose by [link] some $m\ge n$ such that $x_n\ge a_m>x_n-1/1.$ But then $|a_m-x|<1/1.$ (Why?) This $m$ we call $n_1.$ We have that $|a_{n_1}-x|<1/1.$

Next, again using the fact that $x=lim{x}_n,$ choose another $n$ so that $n>n_1$ and so that $|x_n-x|=x_n-x<1/2.$ Then, since this $x_n=sup{S}_n,$ we may choose another $m\ge n$ such that $x_n\ge a_m>x_n-1/2.$ This $m$ we call $n_2.$ Note that we have $|a_{n_2}-x|<1/2.$

Arguing by induction, if we have found an increasing set $n_1<n_2<...<n_j,$ for which $|a_{n_i}-x|<1/i$ for $1\le i\le j,$ choose an $n$ larger than $n_j$ such that $|x_n-x|<1/(j+1).$ Then, since $x_n=sup{S}_n,$ choose an $m\ge n$ so that $x_n\ge a_m>x_n-1/(j+1).$ Then $|a_m-x|<1/(j+1)$ , and we let $n_{j+1}$ be this $m.$ It follows that $|a_{n_{j+1}}-x|<1/(j+1).$

So, by recursive definition, we have constructed a subsequence of $\left\{a_n\right\}$ that converges to $x,$ and this completes the proof of part (4) of the lemma.

Finally, if $y$ is any cluster point of $\left\{a_n\right\},$ and if $y=lim{a}_{n_k},$ then $n_k\ge k,$ and so $a_{n_k}\le x_k,$ implying that $x_k-a_{n_k}\ge 0.$ Hence, taking limits on $k,$ we see that $x-y\ge 0,$ and this proves part (5).