Difference between revisions of "Bolzano-Weierstrass Theorem"
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− | + | 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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− | + | ''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> | |
− | + | 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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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 contains a convergent subsequence.
Proof: Since is assumed to be bounded we have . Bisect the closed interval into two intervals and . Let . Take some point . Bisect into two new intervals, and label the rightmost interval . Since there are infinite points in we can pick some , and continue this process by picking some . We show that the sequence is convergent. Consider the chain
By the Nested Interval Property we know that there is some 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 .