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| − | If <math>a_1,a_2,\dots,a_n</math> positive real numbers,then
| + | 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. |
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| − | <math>(1+a_1)(1+a_2)\dots(1+a_n)>1+a_1+a_2+\dots+a_n</math>
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| − | Further if <math>a_1,a_2,\dots,a_n<1,</math>then
| + | ''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> |
| − | <math>(1-a_1)(1-a_2)\dots(1-a_n)>1-a_1-a_2-\dots-a_n</math> | + | 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>. |
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| − | Every bounded [[sequence]] of reals contains a convergent subsequence.
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| − | == Proof ==
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| − | Lemma 1: A bounded increasing sequence of reals converges to its least upper bound.
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| − | 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.
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| − | We can prove analogously that a bounded decreasing sequence of reals converges to its greatest lower bound.
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| − | Lemma 2: Every sequence of reals has a monotone subsequence.
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| − | Proof: Given a sequence <math>(p_n)</math> of reals, we consider two cases:
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| − | * 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.
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| − | In this case, we build a strictly increasing subsequence <math>(i_n)</math>.
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| − | 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.
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| − | 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>.
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| − | We can continue in this fashion to construct a strictly increasing subsequence of <math>(p_n)</math>.
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| − | * Case 2: Case 1 fails.
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| − | 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>.
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| − | 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.
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| − | 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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| − | == See also ==
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| − | [[Category:Analysis]]
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| − | [[Category:Theorems]]
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| − | {{stub}}
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contains a convergent subsequence.
. Bisect the closed interval ![$[-M,M]$](http://latex.artofproblemsolving.com/4/4/0/44064907e31ae52f0521fec6f33aea0046963443.png)
into two intervals ![$[-M,0]$](http://latex.artofproblemsolving.com/6/a/7/6a77ef67f80eb3f0e15694bff2ab032d7460faa0.png)
and ![$[0,M]$](http://latex.artofproblemsolving.com/9/f/1/9f12856c2421502bd82a7148518f8ef138bf48bc.png)
. Let ![$I_1=[-M,0]$](http://latex.artofproblemsolving.com/a/5/8/a587e29b7d0899424f5d511c00631ace6c5d1142.png)
. Take some point 
. Bisect 
into two new intervals, and label the rightmost interval 
. Since there are infinite points in 
, and continue this process by picking some 
. We show that the sequence 
is convergent. Consider the chain
![\[\ldots\subseteq I_k\subseteq I_{k-1}\subseteq\ldots\subseteq I_2\subseteq I_1.\]](http://latex.artofproblemsolving.com/e/5/3/e533ff0fc37da82371c4bde3016cd66744ca9065.png)
contained in each interval. We claim 
. Let 
be arbitrary. The length 
for each 
is, by construction, 
which converges to 
. Choose 
such that for each 
that 
. Since 
and 
are contained in each interval, it follows that 
.
