Difference between revisions of "Theorem:Bolzano-Weierstrass"

Latest revision as of 12:26, 8 November 2021

Theorem 1 (Bolzano-Weierstrass): Let $(a_{n})$ be a bounded sequence. Then there exists a subsequence of $(a_{n})$ , call it $(a_{n_{k}})$ that is convergent.

Proof 1: Let $(a_{n})$ be a bounded sequence, that is the set $\{a_{n}:n\in \mathbb {N} \}$ is bounded. Suppose that $a$ is a lower bound for this set, and $b$ is an upper bound for this set, and so $a\leq a_{n}\leq b_{n}$ for all $n\in \mathbb {N}$ . Therefore the set $\{a_{n}:n\in \mathbb {N} \}$ is contained in the interval $I_{1}=[a,b]$ .

Let's take $n_{1}=1$ , i.e., let the first term of the subsequence $(a_{n_{k}})$ be equal to the first term of the parent sequence $(a_{n})$ .

We will now take the interval $I_{1}$ and divide into two subintervals of equal length. We will call these intervals $I_{1}'$ and $I_{1}''$ . We will also divide the set of indices into two sets. Let $A_{1}:=\{n\in \mathbb {N} :n>n_{1},a_{n}\in I_{1}'\}$ and let $B_{1}=\{n\in \mathbb {N} :n>n_{1},a_{N}\in I_{1}''\}$ . We note that $A_{1}$ is the set of indices $n$ greater than the first index in our subsequence, $n_{1}$ , such that $a_{n}$ is in the interval $I_{1}'$ . Similarly, $B_{1}$ is the set of indices $n$ greater than the first index in our subsequence, $n_{1}$ such that $a_{N}$ is in the interval $I_{1}''$ .

We note that both of the sets $A_{1}$ and $B_{1}$ cannot be finite since $A_{1}$ and $B_{1}$ contain all of the indices of the sequence $(a_{n})$ which has infinitely many indices since it is an infinite sequence. If $A_{1}$ is infinite, then we let $I_{2}=I_{1}'$ , and we will let $n_{2}$ be the smallest natural number index in the set $A_{1}$ , which we are ensured to have by the Well Ordering Principle. If $A_{1}$ is finite, then $B_{1}$ is infinite and we let $I_{2}=I_{1}''$ and we let $n_{2}$ be the smallest natural number index in the set $B_{1}$ , which once again, we are ensured to have by the Well Ordering Principle.

We will now take the interval $I_{2}$ and subdivide it into two subintervals of equal length, call them $I_{2}'$ and $I_{2}''$ . Once again, we will subdivide the set of indices into two sets. Let $A_{2}=\{n\in \mathbb {N} :n>n_{2},a_{n}\in I_{2}'\}$ and let $B_{2}=\{n\in \mathbb {N} :n>n_{2},a_{n}\in I_{2}''\}$ .

Once again, if $A_{2}$ is infinite, then we will let $I_{3}=I_{2}'$ , and let $n_{3}$ be the smallest natural number index of $A_{2}$ . If $A_{2}$ is finite, then $B_{2}$ is infinite and let $n_{3}$ be the smallest natural number index of $B_{2}$ .

We will then take the interval $I_{3}$ and subdivide it into two subintervals of equal length, call them $I_{3}'$ and $I_{3}''$ . We will also subdivide the set of indices into two sets. $A_{3}=\{n\in \mathbb {N} :n>n_{3},a_{n}\in I_{3}'\}$ and $B_{3}=\{n\in \mathbb {N} :n>n_{3},a_{n}\in I_{3}''\}$

We will continue this pattern to obtain a sequence of nested intervals $I_{1}\supseteq I_{2}\supseteq ...\supseteq I_{k}\supseteq ...$ . Furthermore, we will also have a subsequence $(a_{n_{k}})$ of $(a_{n})$ where $a_{n_{k}}\in I_{k}$ for every $k\in \mathbb {N}$ .

We also note that the length of any interval $I_{k}$ is precisely equal to ${\frac {b-a}{2^{k-1}}}$ . 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 $A'$ . 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:

The Bolzano-Weierstrass Theorem under a CC BY-SA license

@@ 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 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>
-<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 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>
-<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 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>
-<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>

Difference between revisions of "Theorem:Bolzano-Weierstrass"

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