Bolzano-Weierstrass Theorem

--Bhaskaracharya 05:41, 16 October 2013 (EDT)== Statement == If $a_1,a_2,\dots,a_n$ positive real numbers,then

$(1+a_1)(1+a_2)\dots(1+a_n)>1+a_1+a_2+\dots+a_n$

Further if $a_1,a_2,\dots,a_n<1,$ then

$(1-a_1)(1-a_2)\dots(1-a_n)>1-a_1-a_2-\dots-a_n$

Every bounded sequence of reals contains a convergent subsequence.

Proof

Lemma 1: A bounded increasing sequence of reals converges to its least upper bound.

Proof: Suppose $(p_n)$ is a bounded increasing sequence of reals, so $p_1\leq p_2\leq p_3\leq\cdots$ . Let $p=\sup(\{p_1,p_2,p_3,\ldots\})$ . We want to show that $(p_n)\rightarrow p$ . Assume $\epsilon>0$ . By the definition of convergence, we need to produce some $n\in\mathbb{N}$ such that for all $m\geq n$ , $|p_m-p|<\epsilon$ . Since $p$ is the least upper bound of the sequence, there exists an $n\in\mathbb{N}$ such that $p_n>p-\epsilon$ , and since $(p_n)$ is increasing, we know that for all $m\geq n$ , $p_m\geq p_n>p-\epsilon$ . Therefore for all $m\geq n$ , $|p_m-p|<\epsilon$ . This shows that $(p_n)$ converges to $p$ . This completes the proof of Lemma 1.

We can prove analogously that a bounded decreasing sequence of reals converges to its greatest lower bound.

Lemma 2: Every sequence of reals has a monotone subsequence.

Proof: Given a sequence $(p_n)$ of reals, we consider two cases:

Case 1: Every infinite subsequence of $(p_n)$ contains a term that is strictly smaller than infinitely many later terms in the subsequence.

In this case, we build a strictly increasing subsequence $(i_n)$ .

Note that we can apply the assumption of Case 1 to $(p_n)$ itself. Thus there exists some $p_a$ such that there exist infinitely many $b>a$ such that $p_b>p_a$ . Let $i_1=p_a$ , and consider the subsequence $(q_n)$ where $q_1=p_a$ , and $q_k$ is the $k-1$ th term in $(p_n)$ after $p_a$ that is greater than $p_a$ . This exists by the assumption.

Now we can apply the assumption of Case 1 to $(q_n)$ . Thus there exists some $q_c$ such that there exist infinitely many $d>c$ such that $q_d>q_c$ . Since $(q_n)$ is a subsequence of $(p_n)$ , there exists an $e\in\mathbb{N}$ such that $p_e=q_c$ . We then let $i_2=p_e$ , and let $(r_n)$ be a subsequence of $(q_n)$ such that $r_1=q_c$ , and $r_k$ is the $k-1$ th term in $(q_n)$ after $q_c$ that is gerater tha $q_c$ . Note that $(r_n)$ is a subsequence of $(p_n)$ .

We can continue in this fashion to construct a strictly increasing subsequence of $(p_n)$ .

Case 2: Case 1 fails.

In other words, there exists a subsequence $(q_n)$ of $(p_n)$ such that every term in $(q_n)$ is at least as large as some later term in $(q_n)$ . In this case, we can build a decreasing subsequence $(d_n)$ .

Consider such a sequence $(q_n)$ . Let $d_1=q_1$ . By the assumption of Case 2, there exists some $q_a\leq q_1$ . Let $d_2=q_a$ . By the assumption again, there exists some $q_a\leq q_b$ . Let $d_3=q_b$ . We can continue in this fashion to build a decreasing subsequence of $(q_n)$ . Since $(q_n)$ is a subsequence of $(p_n)$ , we have that $(d_n)$ is a decreasing subsequence of $(p_n)$ . This completes the proof of Lemma 2.

The Bolzano-Weierstrass Theorem follows immediately: every bounded sequence of reals contains some monotone subsequence by Lemma 2, which is in turn bounded. This subsequence is convergent by Lemma 1, which completes the proof.

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Bolzano-Weierstrass Theorem

Proof

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