Difference between revisions of "Theorem:Bolzano-Weierstrass"

From Department of Mathematics at UTSA
Jump to navigation Jump to search
 
(3 intermediate revisions by the same user not shown)
Line 1: Line 1:
−
===Theorem (Bolzano—Weierstrass)===
+
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
−
<blockquote style="background: white; border: 1px solid black; padding: 1em;">Every bounded sequence of real numbers contains a convergent subsequence.</blockquote>
+
<td><strong>Theorem 1 (Bolzano-Weierstrass):</strong> Let <span class="math-inline"><math>(a_n)</math></span> be a bounded sequence. Then there exists a subsequence of <span class="math-inline"><math>(a_n)</math></span>, call it <span class="math-inline"><math>(a_{n_k})</math></span> that is convergent.</td>
 +
</blockquote>
 +
<ul>
 +
<li><strong>Proof 1:</strong> Let <span class="math-inline"><math>(a_n)</math></span> be a bounded sequence, that is the set <span class="math-inline"><math>\{ a_n : n \in \mathbb{N} \}</math></span> is bounded. Suppose that <span class="math-inline"><math>a</math></span> is a lower bound for this set, and <span class="math-inline"><math>b</math></span> is an upper bound for this set, and so <span class="math-inline"><math>a \leq a_n \leq b_n</math></span> for all <span class="math-inline"><math>n \in \mathbb{N}</math></span>. Therefore the set <span class="math-inline"><math>\{ a_n : n \in \mathbb{N} \}</math></span> is contained in the interval <span class="math-inline"><math>I_1 = [a, b]</math></span>.</li>
 +
</ul>
  
−
Let (''x''<sub>''n''</sub>) be a sequence of real numbers bounded by a real number ''M'', that is  |''x''<sub>''n''</sub>|&nbsp;&lt;&nbsp;''M'' for all ''n''. We define the set ''A'' by ''A''&nbsp;=&nbsp;{''r''&nbsp;|&nbsp;|''r''|&nbsp;&le;&nbsp;''M'' and ''r''&nbsp;&lt;&nbsp;''x''<sub>''n''</sub> for infinitely many ''n''}.  We note that ''A'' is non-empty since it contains &minus;''M'' and ''A'' is bounded above by ''M''.  Let ''x''&nbsp;=&nbsp;sup&nbsp;''A''. 
+
[[File:Bolzano-Weierstrass 1.png|center|Bolzano-Weierstrass 1]]
  
−
We claim that, for any ε&nbsp;&gt;&nbsp;0, there must be infinitely many points of ''x''<sub>''n''</sub> in the interval (''x''&nbsp;&minus;&nbsp;ε,&nbsp;''x''&nbsp;+&nbsp;ε). Suppose not and fix an ε&nbsp;&gt;&nbsp;0 so that there are only finitely many values of ''x''<sub>''n''</sub> in the interval (''x''&nbsp;&minus;&nbsp;ε,&nbsp;''x''&nbsp;+&nbsp;ε). Either ''x''&nbsp;&le;&nbsp;''x''<sub>''n''</sub> for infinitely many ''n'' or ''x''&nbsp;&le;&nbsp;''x''<sub>''n''</sub> for at most only finitely many ''n'' (possibly no ''n'' at all).  Suppose ''x''&lt;&nbsp;''x''<sub>''n''</sub> for infinitely many ''n''. Clearly in this case ''x''&nbsp;&ne;&nbsp;''M''.  If necessary restrict ε so that ''x''&nbsp;+&nbsp;ε&nbsp;&le;&nbsp;''M''.  Set ''r''&nbsp;=&nbsp;''x''&nbsp;+&nbsp;ε/2 we have that ''r''&nbsp;&lt;&nbsp;''x''<sub>''n''</sub> for infinitely many ''n'' because there are only finitely many ''x''<sub>''n''</sub> in the set [''x'',''r''] and ''x'' must be less than infinitely many ''x''<sub>''n''</sub>, furthermore |''r''|&nbsp;&lt;&nbsp;''M''. Thus ''r'' is in ''A'', which contradicts that ''x'' is an upper bound for ''A''. Now suppose ''x''&lt;&nbsp;''x''<sub>''n''</sub> for at most finitely many ''n''. Set ''y''&nbsp;=&nbsp;''x''&nbsp;&minus;&nbsp;ε/2.  Then there are at most only finitely man ''n'' so that ''x''<sub>''n''</sub>&nbsp;&ge;&nbsp;''y''.  Thus, if ''r''&nbsp;&lt;&nbsp;x<sub>''n''</sub> for infinitely many ''n'', we have that ''r''&nbsp;&le;&nbsp;''y''. This means that ''y'' is an upper bound for ''A'' that is less than ''x'', contradicting that ''x'' wast the ''least'' upper bound of ''A''. In either case we arrive at a contradiction, thus we must have that for any ε&nbsp;&gt;&nbsp;0, there must be infinitely many points of ''x''<sub>''n''</sub> in the interval (''x''&nbsp;&minus;&nbsp;ε,&nbsp;''x''&nbsp;+&nbsp;ε).
+
<ul>
−
   
+
<li>Let's take <span class="math-inline"><math>n_1 = 1</math></span>, i.e., let the first term of the subsequence <span class="math-inline"><math>(a_{n_k})</math></span> be equal to the first term of the parent sequence <span class="math-inline"><math>(a_n)</math></span>.</li>
−
Now we show there is a subsequence that converges to ''x''.  We define the subsequence inductively, choose any ''x''<sub>n<sub>1</sub></sub> from the interval (''x''&nbsp;&minus;&nbsp;1,&nbsp;''x''&nbsp;+&nbsp;1).  Assuming we have chosen ''x''<sub>''n''<sub>1</sub></sub>,&nbsp;…,&nbsp;''x''<sub>''n''<sub>''k''&minus;1</sub></sub>, choose ''x''<sub>''n''<sub>''k''</sub></sub> to be an element in the interval (''x''&nbsp;&minus;&nbsp;1/k,&nbsp;''x''&nbsp;+&nbsp;1/k) so that ''n''<sub>''k''</sub>&notin;{''n''<sub>1</sub>,&nbsp;…,&nbsp;''n''<sub>k&minus;1</sub>}, this is possible as there are infinitely many elements of (''x''<sub>''n''</sub>) in the interval. Notice that for this choice of ''x''<sub>''n''<sub>''k''</sub></sub> we have that |''x''&nbsp;&minus;&nbsp;''x''<sub>''n''<sub>''k''</sub></sub>|<1/k.  Hence for any ε>0, if we take any ''k''&nbsp;&gt;&nbsp;1/ε, then  |''x''<sub>''n''<sub>''k''</sub></sub>-x|&nbsp;&lt;&nbsp;ε.  That is the subsequence (''x''<sub>''n''<sub>k</sub></sub>)&nbsp;→&nbsp;''x''. <math>\Box</math>
+
</ul>
 +
<ul>
 +
<li>We will now take the interval <span class="math-inline"><math>I_1</math></span> and divide into two subintervals of equal length. We will call these intervals <span class="math-inline"><math>I_1'</math></span> and <span class="math-inline"><math>I_1''</math></span>. We will also divide the set of indices into two sets. Let <span class="math-inline"><math>A_1 := \{ n \in \mathbb{N} : n > n_1, a_n \in I_1' \}</math></span> and let <span class="math-inline"><math>B_1 = \{ n \in \mathbb{N} : n > n_1, a_N \in I_1'' \}</math></span>. We note that <span class="math-inline"><math>A_1</math></span> is the set of indices <span class="math-inline"><math>n</math></span> greater than the first index in our subsequence, <span class="math-inline"><math>n_1</math></span>, such that <span class="math-inline"><math>a_n</math></span> is in the interval <span class="math-inline"><math>I_1'</math></span>. Similarly, <span class="math-inline"><math>B_1</math></span> is the set of indices <span class="math-inline"><math>n</math></span> greater than the first index in our subsequence, <span class="math-inline"><math>n_1</math></span> such that <span class="math-inline"><math>a_N</math></span> is in the interval <span class="math-inline"><math>I_1''</math></span>.</li>
 +
</ul>
  
−
==Resources==
+
[[File:Bolzano-Weierstrass 2.png|center|Bolzano-Weierstrass 2]]
−
* [https://en.wikibooks.org/wiki/Real_Analysis/Sequences Sequences], Wikibooks: Real Analysis
+
 
 +
<ul>
 +
<li>We note that both of the sets <span class="math-inline"><math>A_1</math></span> and <span class="math-inline"><math>B_1</math></span> cannot be finite since <span class="math-inline"><math>A_1</math></span> and <span class="math-inline"><math>B_1</math></span> contain all of the indices of the sequence <span class="math-inline"><math>(a_n)</math></span> which has infinitely many indices since it is an infinite sequence. If <span class="math-inline"><math>A_1</math></span> is infinite, then we let <span class="math-inline"><math>I_2 = I_1'</math></span>, and we will let <span class="math-inline"><math>n_2</math></span> be the smallest natural number index in the set <span class="math-inline"><math>A_1</math></span>, which we are ensured to have by the Well Ordering Principle. If <span class="math-inline"><math>A_1</math></span> is finite, then <span class="math-inline"><math>B_1</math></span> is infinite and we let <span class="math-inline"><math>I_2 = I_1''</math></span> and we let <span class="math-inline"><math>n_2</math></span> be the smallest natural number index in the set <span class="math-inline"><math>B_1</math></span>, which once again, we are ensured to have by the Well Ordering Principle.</li>
 +
</ul>
 +
<ul>
 +
<li>We will now take the interval <span class="math-inline"><math>I_2</math></span> and subdivide it into two subintervals of equal length, call them <span class="math-inline"><math>I_2'</math></span> and <span class="math-inline"><math>I_2''</math></span>. Once again, we will subdivide the set of indices into two sets. Let <span class="math-inline"><math>A_2 = \{ n \in \mathbb{N} : n > n_2, a_n \in I_2' \}</math></span> and let <span class="math-inline"><math>B_2 = \{ n \in \mathbb{N} : n > n_2, a_n \in I_2'' \}</math></span>.</li>
 +
</ul>
 +
 
 +
[[File:Bolzano-Weierstrass 3.png|center|Bolzano-Weierstrass 3]]
 +
 
 +
<ul>
 +
<li>Once again, if <span class="math-inline"><math>A_2</math></span> is infinite, then we will let <span class="math-inline"><math>I_3 = I_2'</math></span>, and let <span class="math-inline"><math>n_3</math></span> be the smallest natural number index of <span class="math-inline"><math>A_2</math></span>. If <span class="math-inline"><math>A_2</math></span> is finite, then <span class="math-inline"><math>B_2</math></span> is infinite and let <span class="math-inline"><math>n_3</math></span> be the smallest natural number index of <span class="math-inline"><math>B_2</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li>We will then take the interval <span class="math-inline"><math>I_3</math></span> and subdivide it into two subintervals of equal length, call them <span class="math-inline"><math>I_3'</math></span> and <span class="math-inline"><math>I_3''</math></span>. We will also subdivide the set of indices into two sets. <span class="math-inline"><math>A_3 = \{ n \in \mathbb{N} : n > n_3, a_n \in I_3' \}</math></span> and <span class="math-inline"><math>B_3 = \{ n \in \mathbb{N} : n > n_3, a_n \in I_3'' \}</math></span></li>
 +
</ul>
 +
 
 +
[[File:Bolzano-Weierstrass 4.png|center|Bolzano-Weierstrass 4]]
 +
 
 +
<ul>
 +
<li>We will continue this pattern to obtain a sequence of nested intervals <span class="math-inline"><math>I_1 \supseteq I_2 \supseteq ... \supseteq I_k \supseteq ...</math></span>. Furthermore, we will also have a subsequence <span class="math-inline"><math>(a_{n_k})</math></span> of <span class="math-inline"><math>(a_n)</math></span> where <span class="math-inline"><math>a_{n_k} \in I_k</math></span> for every <span class="math-inline"><math>k \in \mathbb{N}</math></span>.</li>
 +
</ul>
 +
 
 +
[[File:Bolzano-Weierstrass 5.png|center|Bolzano-Weierstrass 5]]
 +
 
 +
<ul>
 +
<li>We also note that the length of any interval <span class="math-inline"><math>I_k</math></span> is precisely equal to <span class="math-inline"><math>\frac{b - a}{2^{k-1}}</math></span>. By the nested intervals theorem there exists a common point <span class="math-inline"><math>\xi \in I_k</math></span> for all <span class="math-inline"><math>k \in \mathbb{N}</math></span>, and so we have that:</li>
 +
</ul>
 +
<div style="text-align: center;"><math>\begin{align} \mid a_{n_k} - \xi \mid \leq \frac{b - a}{2^{k-1}} \end{align}</math></div>
 +
<ul>
 +
<li>Therefore the subsequence <span class="math-inline"><math>(a_{n_k})</math></span> converges to <span class="math-inline"><math>\xi</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
 +
</ul>
 +
<ul>
 +
<li><strong>Proof 2:</strong> Let <span class="math-inline"><math>(a_n)</math></span> be a bounded sequence. By the <a href="/the-monotone-subsequence-theorem">The Monotone Subsequence Theorem</a>, every sequence of real numbers has a monotonic subsequence. Let <span class="math-inline"><math>(a_{n_k})</math></span> be this monotonic subsequence. Since <span class="math-inline"><math>(a_n)</math></span> is bounded, it follows that <span class="math-inline"><math>(a_{n_k})</math></span> is also bounded since the values <span class="math-inline"><math>a_{n_k}</math></span> are contained in the sequence <span class="math-inline"><math>(a_n)</math></span>. Since <span class="math-inline"><math>(a_{n_k})</math></span> is both bounded and monotonic, then by <a href="/the-monotone-convergence-theorem">The Monotone Convergence Theorem</a>, <span class="math-inline"><math>(a_{n_k})</math></span> is a convergent subsequence. <span class="math-inline"><math>\blacksquare</math></span></li>
 +
</ul>
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
<td><strong>Theorem 2:</strong> If <span class="math-inline"><math>(a_n)</math></span> is a bounded sequence such that every convergent subsequence <span class="math-inline"><math>(a_{n_k})</math></span> of <span class="math-inline"><math>(a_n)</math></span> converges to <span class="math-inline"><math>L</math></span> then <span class="math-inline"><math>(a_n)</math></span> converges to <span class="math-inline"><math>L</math></span>.</td>
 +
</blockquote>
 +
<ul>
 +
<li><strong>Proof:</strong> Suppose that <span class="math-inline"><math>A = (a_n)</math></span> does not converge to <span class="math-inline"><math>L</math></span>. Then there exists a subsequence <span class="math-inline"><math>A' = (a_{n_k})</math></span> and an <span class="math-inline"><math>\epsilon_0 > 0</math></span> such that <span class="math-inline"><math>\mid a_{n_k} - L \mid \geq \epsilon_0</math></span> for all <span class="math-inline"><math>k \in \mathbb{N}</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li>Since <span class="math-inline"><math>A</math></span> is bounded, any subsequence of <span class="math-inline"><math>A</math></span> is also bounded, and so the subsequence <span class="math-inline"><math>A'</math></span> is bounded. Since <span class="math-inline"><math>A'</math></span> is bounded, then by the Bolzano-Weierstrass theorem there exists a convergent subsequence <span class="math-inline"><math>A''</math></span> of <span class="math-inline"><math>A'</math></span>. Since <span class="math-inline"><math>A''</math></span> is a convergent subsequence of <span class="math-inline"><math>A'</math></span>, then it is also a convergent sequence of <span class="math-inline"><math>A</math></span> by transitivity, and by the hypothesis, <span class="math-inline"><math>A''</math></span> converges to <span class="math-inline"><math>L</math></span>.</li>
 +
</ul>
 +
<ul>
 +
<li>But then convergence of <span class="math-inline"><math>A''</math></span> to <span class="math-inline"><math>L</math></span> contradicts the assumption that <span class="math-inline"><math>\mid a_{n_k} - L \mid \geq \epsilon_0</math></span>, and so the assumption that <span class="math-inline"><math>A</math></span> did not converge to <span class="math-inline"><math>L</math></span> was false. Therefore <span class="math-inline"><math>(a_n)</math></span> converges to <span class="math-inline"><math>L</math></span>. <span class="math-inline"><math>\blacksquare</math></span></li>
 +
</ul>
 +
 
 +
==Licensing==
 +
Content obtained and/or adapted from:
 +
* [http://mathonline.wikidot.com/the-bolzano-weierstrass-theorem The Bolzano-Weierstrass Theorem] under a CC BY-SA license

Latest revision as of 12:26, 8 November 2021

Theorem 1 (Bolzano-Weierstrass): Let be a bounded sequence. Then there exists a subsequence of , call it that is convergent.

  • Proof 1: Let be a bounded sequence, that is the set is bounded. Suppose that is a lower bound for this set, and is an upper bound for this set, and so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a \leq a_n \leq b_n} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \in \mathbb{N}} . Therefore the set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{ a_n : n \in \mathbb{N} \}} is contained in the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_1 = [a, b]} .
Bolzano-Weierstrass 1
  • Let's take Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_1 = 1} , i.e., let the first term of the subsequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_{n_k})} be equal to the first term of the parent sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} .
  • We will now take the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_1} and divide into two subintervals of equal length. We will call these intervals Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_1'} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_1''} . We will also divide the set of indices into two sets. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_1 := \{ n \in \mathbb{N} : n > n_1, a_n \in I_1' \}} and let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_1 = \{ n \in \mathbb{N} : n > n_1, a_N \in I_1'' \}} . We note that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_1} is the set of indices Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} greater than the first index in our subsequence, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_1} , such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_n} is in the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_1'} . Similarly, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_1} is the set of indices Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} greater than the first index in our subsequence, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_1} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_N} is in the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_1''} .
Bolzano-Weierstrass 2
  • We note that both of the sets Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_1} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_1} cannot be finite since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_1} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_1} contain all of the indices of the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} which has infinitely many indices since it is an infinite sequence. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_1} is infinite, then we let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_2 = I_1'} , and we will let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_2} be the smallest natural number index in the set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_1} , which we are ensured to have by the Well Ordering Principle. If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_1} is finite, then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_1} is infinite and we let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_2 = I_1''} and we let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_2} be the smallest natural number index in the set Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_1} , which once again, we are ensured to have by the Well Ordering Principle.
  • We will now take the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_2} and subdivide it into two subintervals of equal length, call them Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_2'} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_2''} . Once again, we will subdivide the set of indices into two sets. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_2 = \{ n \in \mathbb{N} : n > n_2, a_n \in I_2' \}} and let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_2 = \{ n \in \mathbb{N} : n > n_2, a_n \in I_2'' \}} .
Bolzano-Weierstrass 3
  • Once again, if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_2} is infinite, then we will let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_3 = I_2'} , and let be the smallest natural number index of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_2} . If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_2} is finite, then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_2} is infinite and let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_3} be the smallest natural number index of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_2} .
  • We will then take the interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_3} and subdivide it into two subintervals of equal length, call them Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_3'} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_3''} . We will also subdivide the set of indices into two sets. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_3 = \{ n \in \mathbb{N} : n > n_3, a_n \in I_3' \}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B_3 = \{ n \in \mathbb{N} : n > n_3, a_n \in I_3'' \}}
Bolzano-Weierstrass 4
  • We will continue this pattern to obtain a sequence of nested intervals Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_1 \supseteq I_2 \supseteq ... \supseteq I_k \supseteq ...} . Furthermore, we will also have a subsequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_{n_k})} of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{n_k} \in I_k} for every Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \in \mathbb{N}} .
Bolzano-Weierstrass 5
  • We also note that the length of any interval Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I_k} is precisely equal to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{b - a}{2^{k-1}}} . By the nested intervals theorem there exists a common point Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \xi \in I_k} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \in \mathbb{N}} , and so we have that:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \mid a_{n_k} - \xi \mid \leq \frac{b - a}{2^{k-1}} \end{align}}
  • Therefore the subsequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_{n_k})} converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \xi} . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}
  • Proof 2: Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} be a bounded sequence. By the <a href="/the-monotone-subsequence-theorem">The Monotone Subsequence Theorem</a>, every sequence of real numbers has a monotonic subsequence. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_{n_k})} be this monotonic subsequence. Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} is bounded, it follows that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_{n_k})} is also bounded since the values Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{n_k}} are contained in the sequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} . Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_{n_k})} is both bounded and monotonic, then by <a href="/the-monotone-convergence-theorem">The Monotone Convergence Theorem</a>, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_{n_k})} is a convergent subsequence. Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}

Theorem 2: If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} is a bounded sequence such that every convergent subsequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_{n_k})} of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} .

  • Proof: Suppose that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A = (a_n)} does not converge to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} . Then there exists a subsequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A' = (a_{n_k})} and an Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_0 > 0} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mid a_{n_k} - L \mid \geq \epsilon_0} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k \in \mathbb{N}} .
  • Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} is bounded, any subsequence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} is also bounded, and so the subsequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A'} is bounded. Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A'} is bounded, then by the Bolzano-Weierstrass theorem there exists a convergent subsequence Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A''} of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A'} . Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A''} is a convergent subsequence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A'} , then it is also a convergent sequence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} by transitivity, and by the hypothesis, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A''} converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} .
  • But then convergence of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A''} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} contradicts the assumption that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mid a_{n_k} - L \mid \geq \epsilon_0} , and so the assumption that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} did not converge to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} was false. Therefore Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_n)} converges to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} . Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \blacksquare}

Licensing

Content obtained and/or adapted from:

Retrieved from "https://mathresearch.utsa.edu/wiki/index.php?title=Theorem:Bolzano-Weierstrass&oldid=3630"