Theorem:Bolzano-Weierstrass - Revision history

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Lila at 21:12, 14 October 2021

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Lila: Created page with "===Theorem (Bolzano—Weierstrass)===
Every bounded sequence of real numbers contains a conve..."

2021-10-14T21:11:55Z

Created page with "===Theorem (Bolzano—Weierstrass)=== <blockquote style="background: white; border: 1px solid black; padding: 1em;">Every bounded sequence of real numbers contains a conve..."

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===Theorem (Bolzano—Weierstrass)===
<blockquote style="background: white; border: 1px solid black; padding: 1em;">Every bounded sequence of real numbers contains a convergent subsequence.</blockquote>

Let (''x''''n'') be a sequence of real numbers bounded by a real number ''M'', that is |''x''''n''| < ''M'' for all ''n''. We define the set ''A'' by ''A'' = {''r'' | |''r''| ≤ ''M'' and ''r'' < ''x''''n'' for infinitely many ''n''}. We note that ''A'' is non-empty since it contains −''M'' and ''A'' is bounded above by ''M''. Let ''x'' = sup ''A''.

We claim that, for any ε > 0, there must be infinitely many points of ''x''''n'' in the interval (''x'' − ε, ''x'' + ε). Suppose not and fix an ε > 0 so that there are only finitely many values of ''x''''n'' in the interval (''x'' − ε, ''x'' + ε). Either ''x'' ≤ ''x''''n'' for infinitely many ''n'' or ''x'' ≤ ''x''''n'' for at most only finitely many ''n'' (possibly no ''n'' at all). Suppose ''x''< ''x''''n'' for infinitely many ''n''. Clearly in this case ''x'' ≠ ''M''. If necessary restrict ε so that ''x'' + ε ≤ ''M''. Set ''r'' = ''x'' + ε/2 we have that ''r'' < ''x''''n'' for infinitely many ''n'' because there are only finitely many ''x''''n'' in the set [''x'',''r''] and ''x'' must be less than infinitely many ''x''''n'', furthermore |''r''| < ''M''. Thus ''r'' is in ''A'', which contradicts that ''x'' is an upper bound for ''A''. Now suppose ''x''< ''x''''n'' for at most finitely many ''n''. Set ''y'' = ''x'' − ε/2. Then there are at most only finitely man ''n'' so that ''x''''n'' ≥ ''y''. Thus, if ''r'' < x''n'' for infinitely many ''n'', we have that ''r'' ≤ ''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 > 0, there must be infinitely many points of ''x''''n'' in the interval (''x'' − ε, ''x'' + ε).

Now we show there is a subsequence that converges to ''x''. We define the subsequence inductively, choose any ''x''n1 from the interval (''x'' − 1, ''x'' + 1). Assuming we have chosen ''x''''n''1, …, ''x''''n''''k''−1, choose ''x''''n''''k'' to be an element in the interval (''x'' − 1/k, ''x'' + 1/k) so that ''n''''k''∉{''n''1, …, ''n''k−1}, this is possible as there are infinitely many elements of (''x''''n'') in the interval. Notice that for this choice of ''x''''n''''k'' we have that |''x'' − ''x''''n''''k''|<1/k. Hence for any ε>0, if we take any ''k'' > 1/ε, then |''x''''n''''k''-x| < ε. That is the subsequence (''x''''n''k) → ''x''. <math>\Box</math>

@@ Line 5: / Line 5: @@
 <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>
 [[File:Bolzano-Weierstrass 1.png|center|Bolzano-Weierstrass 1]]
 <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>
@@ Line 12: / Line 14: @@
 <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>
 [[File:Bolzano-Weierstrass 2.png|center|Bolzano-Weierstrass 2]]
 <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>
@@ Line 19: / Line 23: @@
 <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>
@@ Line 26: / Line 32: @@
 <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>

@@ Line 5: / Line 5: @@
 <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>
-<div class="image-container aligncenter"><img src="http://mathonline.wdfiles.com/local--files/the-bolzano-weierstrass-theorem/Screen%20Shot%202014-10-22%20at%2011.36.10%20PM.png" alt="Screen%20Shot%202014-10-22%20at%2011.36.10%20PM.png" class="image" /></div>
+[[File:Bolzano-Weierstrass 1.png|center|Bolzano-Weierstrass 1]]
 <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>
@@ Line 12: / Line 12: @@
 <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>
-<div class="image-container aligncenter"><img src="http://mathonline.wdfiles.com/local--files/the-bolzano-weierstrass-theorem/Screen%20Shot%202014-10-23%20at%2012.00.55%20AM.png" alt="Screen%20Shot%202014-10-23%20at%2012.00.55%20AM.png" class="image" /></div>
+[[File:Bolzano-Weierstrass 2.png|center|Bolzano-Weierstrass 2]]
 <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>
@@ Line 19: / Line 19: @@
 <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>
-<div class="image-container aligncenter"><img src="http://mathonline.wdfiles.com/local--files/the-bolzano-weierstrass-theorem/Screen%20Shot%202014-10-23%20at%2012.14.38%20AM.png" alt="Screen%20Shot%202014-10-23%20at%2012.14.38%20AM.png" class="image" /></div>
+[[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>
@@ Line 26: / Line 26: @@
 <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>
-<div class="image-container aligncenter"><img src="http://mathonline.wdfiles.com/local--files/the-bolzano-weierstrass-theorem/Screen%20Shot%202014-10-23%20at%2012.23.20%20AM.png" alt="Screen%20Shot%202014-10-23%20at%2012.23.20%20AM.png" class="image" /></div>
+[[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>
-<div class="image-container aligncenter"><img src="http://mathonline.wdfiles.com/local--files/the-bolzano-weierstrass-theorem/Screen%20Shot%202014-10-23%20at%2012.30.02%20AM.png" alt="Screen%20Shot%202014-10-23%20at%2012.30.02%20AM.png" class="image" /></div>
+[[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>

← Older revision		Revision as of 18:19, 8 November 2021
Line 53:		Line 53:
	<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>		<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>		</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

← Older revision		Revision as of 21:12, 14 October 2021
Line 7:		Line 7:

	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'' − 1, ''x'' + 1). Assuming we have chosen ''x''<sub>''n''<sub>1</sub></sub>, …, ''x''<sub>''n''<sub>''k''−1</sub></sub>, choose ''x''<sub>''n''<sub>''k''</sub></sub> to be an element in the interval (''x'' − 1/k, ''x'' + 1/k) so that ''n''<sub>''k''</sub>∉{''n''<sub>1</sub>, …, ''n''<sub>k−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'' − ''x''<sub>''n''<sub>''k''</sub></sub>\|<1/k. Hence for any ε>0, if we take any ''k'' > 1/ε, then \|''x''<sub>''n''<sub>''k''</sub></sub>-x\| < ε. That is the subsequence (''x''<sub>''n''<sub>k</sub></sub>) → ''x''. <math>\Box</math>		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'' − 1, ''x'' + 1). Assuming we have chosen ''x''<sub>''n''<sub>1</sub></sub>, …, ''x''<sub>''n''<sub>''k''−1</sub></sub>, choose ''x''<sub>''n''<sub>''k''</sub></sub> to be an element in the interval (''x'' − 1/k, ''x'' + 1/k) so that ''n''<sub>''k''</sub>∉{''n''<sub>1</sub>, …, ''n''<sub>k−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'' − ''x''<sub>''n''<sub>''k''</sub></sub>\|<1/k. Hence for any ε>0, if we take any ''k'' > 1/ε, then \|''x''<sub>''n''<sub>''k''</sub></sub>-x\| < ε. That is the subsequence (''x''<sub>''n''<sub>k</sub></sub>) → ''x''. <math>\Box</math>
		+
		+	==Resources==
		+	* [https://en.wikibooks.org/wiki/Real_Analysis/Sequences Sequences], Wikibooks: Real Analysis

Theorem:Bolzano-Weierstrass - Revision history

Lila at 18:26, 8 November 2021

Lila at 18:25, 8 November 2021

Lila at 18:19, 8 November 2021

Lila at 18:19, 8 November 2021

Lila at 21:12, 14 October 2021

Lila: Created page with "===Theorem (Bolzano—Weierstrass)=== Every bounded sequence of real numbers contains a conve..."

Lila: Created page with "===Theorem (Bolzano—Weierstrass)===
Every bounded sequence of real numbers contains a conve..."