Difference between revisions of "Bolzano-Weierstrass Theorem"

Latest revision as of 16:18, 12 June 2022

The Bolzano-Weierstrass Theorem is a result in analysis that states that every bounded sequence of real numbers $(a_n)$ contains a convergent subsequence.

Proof: Since $(a_n)$ is assumed to be bounded we have $|a_n|\le M$ . Bisect the closed interval $[-M,M]$ into two intervals $[-M,0]$ and $[0,M]$ . Let $I_1=[-M,0]$ . Take some point $a_{n_1}\in I_1$ . Bisect $I_1$ into two new intervals, and label the rightmost interval $I_2$ . Since there are infinite points in $I_2$ we can pick some $a_{n_2}\in I_2$ , and continue this process by picking some $a_{n_k}\in I_k$ . We show that the sequence $(a_{n_k})$ is convergent. Consider the chain

$\ldots\subseteq I_k\subseteq I_{k-1}\subseteq\ldots\subseteq I_2\subseteq I_1.$

By the Nested Interval Property we know that there is some $x\in\mathbb{R}$ contained in each interval. We claim $\lim a_{n_k}=x$ . Let $\epsilon>0$ be arbitrary. The length $\mathcal{L}(I_k)$ for each $I_k$ is, by construction, $\mathcal{L}(I_k)=M(1/2)^{k-1}$ which converges to $0$ . Choose $N$ such that for each $k\ge N$ that $\mathcal{L}(I_k)<\epsilon$ . Since $a_{n_k}$ and $x$ are contained in each interval, it follows that $|a_{n_k}-x|<\epsilon$ .

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-Every bounded [[sequence]] of reals contains a convergent subsequence.
+The '''Bolzano-Weierstrass Theorem''' is a result in [[analysis]] that states that every bounded [[sequence]] of real numbers <math>(a_n)</math> contains a convergent subsequence.
-== Proof ==
-Lemma 1: A bounded increasing sequence of reals converges to its least upper bound.
-Proof: Suppose <math>(p_n)</math> is a bounded increasing sequence of reals, so <math>p_1\leq p_2\leq p_3\leq\cdots</math>. Let <math>p=\sup(\{p_1,p_2,p_3,\ldots\})</math>. We want to show that <math>(p_n)\rightarrow p</math>. Assume <math>\epsilon>0</math>. By the definition of convergence, we need to produce some <math>n\in\mathbb{N}</math> such that for all <math>m\geq n</math>, <math>|p_m-p|<\epsilon</math>. Since <math>p</math> is the least upper bound of the sequence, there exists an <math>n\in\mathbb{N}</math> such that <math>p_n>p-\epsilon</math>, and since <math>(p_n)</math> is increasing, we know that for all <math>m\geq n</math>, <math>p_m\geq p_n>p-\epsilon</math>. Therefore for all <math>m\geq n</math>, <math>|p_m-p|<\epsilon</math>. This shows that <math>(p_n)</math> converges to <math>p</math>. This completes the proof of Lemma 1.
+''Proof'': Since <math>(a_n)</math> is assumed to be bounded we have <math>|a_n|\le M</math>.  Bisect the closed interval <math>[-M,M]</math> into two intervals <math>[-M,0]</math> and <math>[0,M]</math>.  Let <math>I_1=[-M,0]</math>.  Take some point <math>a_{n_1}\in I_1</math>.  Bisect <math>I_1</math> into two new intervals, and label the rightmost interval <math>I_2</math>.  Since there are infinite points in <math>I_2</math> we can pick some <math>a_{n_2}\in I_2</math>, and continue this process by picking some <math>a_{n_k}\in I_k</math>.  We show that the sequence <math>(a_{n_k})</math> is convergent.  Consider the chain
+<center><cmath>\ldots\subseteq I_k\subseteq I_{k-1}\subseteq\ldots\subseteq I_2\subseteq I_1.</cmath></center>
-We can prove analogously that a bounded decreasing sequence of reals converges to its greatest lower bound.
+By the Nested Interval Property we know that there is some <math>x\in\mathbb{R}</math> contained in each interval.  We claim <math>\lim a_{n_k}=x</math>.  Let <math>\epsilon>0</math> be arbitrary.  The length <math>\mathcal{L}(I_k)</math> for each <math>I_k</math> is, by construction, <math>\mathcal{L}(I_k)=M(1/2)^{k-1}</math> which converges to <math>0</math>.  Choose <math>N</math> such that for each <math>k\ge N</math> that <math>\mathcal{L}(I_k)<\epsilon</math>.  Since <math>a_{n_k}</math> and <math>x</math> are contained in each interval, it follows that <math>|a_{n_k}-x|<\epsilon</math>.
-Lemma 2: Every sequence of reals has a monotone subsequence.
-Proof: Given a sequence <math>(p_n)</math> of reals, we consider two cases:
-* Case 1: Every infinite subsequence of <math>(p_n)</math> contains a term that is strictly smaller than infinitely many later terms in the subsequence.
-In this case, we build a strictly increasing subsequence <math>(i_n)</math>.
-Note that we can apply the assumption of Case 1 to <math>(p_n)</math> itself. Thus there exists some <math>p_a</math> such that there exist infinitely many <math>b>a</math> such that <math>p_b>p_a</math>. Let <math>i_1=p_a</math>, and consider the subsequence <math>(q_n)</math> where <math>q_1=p_a</math>, and <math>q_k</math> is the <math>k-1</math>th term in <math>(p_n)</math> after <math>p_a</math> that is greater than <math>p_a</math>. This exists by the assumption.
-Now we can apply the assumption of Case 1 to <math>(q_n)</math>. Thus there exists some <math>q_c</math> such that there exist infinitely many <math>d>c</math> such that <math>q_d>q_c</math>. Since <math>(q_n)</math> is a subsequence of <math>(p_n)</math>, there exists an <math>e\in\mathbb{N}</math> such that <math>p_e=q_c</math>. We then let <math>i_2=p_e</math>, and let <math>(r_n)</math> be a subsequence of <math>(q_n)</math> such that <math>r_1=q_c</math>, and <math>r_k</math> is the <math>k-1</math>th term in <math>(q_n)</math> after <math>q_c</math> that is gerater tha <math>q_c</math>. Note that <math>(r_n)</math> is a subsequence of <math>(p_n)</math>.
-We can continue in this fashion to construct a strictly increasing subsequence of <math>(p_n)</math>.
-* Case 2: Case 1 fails.
-In other words, there exists a subsequence <math>(q_n)</math> of <math>(p_n)</math> such that every term in <math>(q_n)</math> is at least as large as some later term in <math>(q_n)</math>. In this case, we can build a decreasing subsequence <math>(d_n)</math>.
-Consider such a sequence <math>(q_n)</math>. Let <math>d_1=q_1</math>. By the assumption of Case 2, there exists some <math>q_a\leq q_1</math>. Let <math>d_2=q_a</math>. By the assumption again, there exists some <math>q_a\leq q_b</math>. Let <math>d_3=q_b</math>. We can continue in this fashion to build a decreasing subsequence of <math>(q_n)</math>. Since <math>(q_n)</math> is a subsequence of <math>(p_n)</math>, we have that <math>(d_n)</math> is a decreasing subsequence of <math>(p_n)</math>. 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.
-== See also ==
-[[Category:Analysis]]
-[[Category:Theorems]]
-{{stub}}

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Difference between revisions of "Bolzano-Weierstrass Theorem"

Latest revision as of 16:18, 12 June 2022