Difference between revisions of "The Nested Interval Theorem in Higher Dimensions"
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We have just looked at what exactly a Nested Interval is, and we are about to look at a critically important theorem in Real Analysis. Before we look at the Nested Intervals Theorem let's first look at the following important lemma that will be used to prove the Nested Intervals Theorem. | We have just looked at what exactly a Nested Interval is, and we are about to look at a critically important theorem in Real Analysis. Before we look at the Nested Intervals Theorem let's first look at the following important lemma that will be used to prove the Nested Intervals Theorem. | ||
+ | |||
+ | <blockquote style="background: white; border: 1px solid black; padding: 1em;"> | ||
+ | :'''Lemma 1:''' Let <math>a < b</math> and let <math>c < d</math> and let <math>I = [a, b]</math> and <math>J = [c, d]</math>. Then <math>I \subseteq J</math> if and only if <math>c \leq a < b \leq d</math>. | ||
+ | </blockquote> | ||
+ | |||
+ | Now let's look at the Nested Intervals theorem. | ||
+ | |||
+ | <blockquote style="background: white; border: 1px solid black; padding: 1em;"> | ||
+ | '''Theorem 1:''' If the interval <math>I_n = [a_n, b_n]</math> for <math>n \in \mathbb{N}</math> is a sequence of closed bounded nested intervals then there exists a real number <math>\xi = \sup \{ a_n : n \in \mathbb{N} \}</math> such that <math>\xi \in \bigcap_{n=1}^{\infty} I_n</math>. | ||
+ | </blockquote> | ||
+ | |||
+ | *'''Proof of Theorem:''' We note that by the definition of nested intervals that <math>I_n \subseteq I_1</math> for all <math>n \in \mathbb{N}</math> so then <math>a_n \leq b_1</math>. | ||
+ | |||
+ | *Now consider the nonempty set <math>A = \{ a_n : n \in \mathbb{N} \}</math> that is bounded above by <math>b_1</math>. Thus this set has a supremum in the real numbers and denote it <math>\sup A = \xi</math> so that <math>a_n \leq\xi</math> for all <math>n \in \mathbb{N}</math>. | ||
+ | |||
+ | *We now want to show that <math>\xi \leq b_n</math> for all <math>n \in \mathbb{N}</math>. We will do this by showing that <math>a_n \leq b_k</math> for all <math>n, k \in \mathbb{N}</math>. | ||
+ | |||
+ | *First consider the case where <math>n \leq k</math>. We thus have that <math>I_n \supseteq I_k</math> by the definition of nested intervals and so by lemma 1 we get that <math>a_n \leq a_k \leq b_k \leq b_n</math>. Here we see that <math>a_n \leq b_k</math>. | ||
+ | |||
+ | *Now consider the case where <math>n > k</math>. We thus have that <math>I_k \supseteq I_n</math> by the definition of nested intervals and so by lemma 1 once again we have that <math>a_k \leq a_n \leq b_n \leq b_k</math>. Once again we have that <math>a_n \leq b_k</math> | ||
+ | |||
+ | *So then <math>a_n \leq b_k</math> for all <math>n, k \in \mathbb{N}</math>, and so <math>b_k</math> is an upper bound to the set <math>A</math> and so <math>\sup A = \xi \leq b_k</math> for all <math>k \in \mathbb{N}</math>. Furthermore we have that <math>a_k \leq \xi</math> for all <math>k \in \mathbb{N}</math>, and so <math>\xi \in I_k</math> for every <math>k \in \mathbb{N}</math> and thus <math>\xi \in \bigcap_{n=1}^{\infty} I_n</math> and so the set theoretic union is nonempty. <math>\blacksquare</math> | ||
+ | |||
+ | <blockquote style="background: white; border: 1px solid black; padding: 1em;"> | ||
+ | '''Theorem 2:''' If the interval <math>I_n = [a_n, b_n]</math> for <math>n \in \mathbb{N}</math> is a sequence of closed bounded nested intervals then there exists a real number <math>\eta = \inf \{ b_n : n \in \mathbb{N} \}</math> such that <math>\eta \in \bigcap_{n=1}^{\infty} I_n</math>. | ||
+ | </blockquote> | ||
+ | |||
+ | *'''Proof:''' We note that by the definition of nested intervals that <math>I_n \subseteq I_1</math> for all <math>n \in \mathbb{N}</math> so then <math>a_1 \leq b_n</math>. | ||
+ | |||
+ | *Now consider the nonempty set <math>B = \{ b_n : n \in \mathbb{N} \}</math> that is bounded below by <math>a_1</math>. Thus this set has an infimum in the real numbers and denote it <math>\inf B = \eta</math> so that <math>\eta \leq b_n</math> for all <math>n \in \mathbb{N}</math>. | ||
+ | |||
+ | *We now want to show that <math>a_n \leq \eta</math> for all <math>n \in \mathbb{N}</math>. We will do this by showing that <math>a_k \leq b_n</math> for all <math>n, k \in \mathbb{N}</math>. | ||
+ | |||
+ | *First consider the case where <math>n \leq k</math>. We thus have that <math>I_n \supseteq I_k</math> by the definition of nested intervals and so by lemma 1 we get that <math>a_n \leq a_k \leq b_k \leq b_n</math>. Here we see that <math>a_k \leq b_n</math>. | ||
+ | |||
+ | *Now consider the case where <math>n > k</math>. We thus have that <math>I_k \supseteq I_n</math> by the definition of nested intervals and so by lemma 1 once again we have that <math>a_k \leq a_n \leq b_n \leq b_k</math>. Once again we have that <math>a_k \leq b_n</math> | ||
+ | |||
+ | *So then <math>a_k \leq b_n</math> for all <math>n, k \in \mathbb{N}</math>, and so <math>a_k</math> is an upper bound to the set <math>A</math> and so <math>a_k \leq \eta = \inf B</math> for all <math>k \in \mathbb{N}</math>. Furthermore we have that <math>\eta \leq b_k</math> for all <math>k \in \mathbb{N}</math>, and so <math>\eta \in I_k</math> for every <math>k \in \mathbb{N}</math> and thus <math>\eta \in \bigcap_{n=1}^{\infty} I_n</math>. <math>\blacksquare</math> | ||
+ | |||
+ | <blockquote style="background: white; border: 1px solid black; padding: 1em;"> | ||
+ | :'''Theorem 3:''' Let <math>A := \{ a_n : n \in \mathbb{N} \}</math> and <math>B := \{ b_n : n \in \mathbb{N} \}</math>. If <math>\sup A = \xi</math> and <math>\inf B = \eta</math> then if the interval <math>I_n = [a_n, b_n]</math> is a sequence of closed bounded nested intervals then <math>[\xi, \eta] = \bigcap_{n=1}^{\infty} I_n</math>. | ||
+ | </blockquote> | ||
+ | |||
+ | *'''Proof:''' We first prove that <math>[\xi, \eta] \subseteq \bigcap_{n=1}^{\infty} I_n</math>. Now let <math>x \in [\xi, \eta] = \{ x \in \mathbb{R} : \xi \leq x \leq \eta</math>. Therefore <math>\xi \leq x \leq \eta</math>. But we know that <math>a_n \leq \xi</math> for all <math>n \in \mathbb{N}</math> and we know that <math>\eta \leq b_n</math> for all <math>n \in \mathbb{N}</math> and so <math>a_n \leq \xi \leq x \leq \eta \leq b_n</math> for all <math>n \in \mathbb{N}</math>. Therefore <math>x \in [a_n, b_n]</math> for all <math>n \in \mathbb{N}</math> or in other words <math>x \in \bigcap_{n=1}^{\infty} I_n</math>. Therefore <math>[\xi, \eta] \subseteq \bigcap_{n=1}^{\infty} I_n</math>. | ||
+ | |||
+ | *We will now prove that <math>\bigcap_{n=1}^{\infty} I_n \subseteq [\xi, \eta]</math>. Let <math>x \in \bigcap_{n=1}^{\infty} I_n</math>. Then <math>a_n \leq x \leq b_n</math> for all <math>n \in \mathbb{N}</math>. We also know that <math>a_n \leq \xi \leq \eta \leq b_n</math> for all <math>n \in \mathbb{N}</math>. Suppose that <math>x</math> is such that <math>a_n \leq x \leq \xi</math>. Since <math>\xi</math> is the supremum of the set <math>A</math> then there exists an element <math>a_x \in \{ a_n : n \in \mathbb{N} \}</math> such that <math>x < a_x</math> and so <math>x \not \in [a_x, b_x]</math> and therefore <math>x \not \in \bigcap_{n=1}^{\infty} I_n</math>, a contradiction. Now suppose that <math>x</math> is such that <math>\eta \leq x \leq b_n</math> for all <math>n \in \mathbb{N}</math>. Since <math>\eta</math> is the infimum of the set <math>B</math> then there exists an element <math>b_x \in \{ b_n : n \in \mathbb{N} \}</math> such that <math>b_x < x</math> and so <math>x \not \in [a_x, b_x]</math> and so <math>x \not \in \bigcap_{n=1}^{\infty} I_n</math>, once again, a contradiction. We must therefore have that <math>a_n \leq \xi \leq x \leq \eta \leq b_n</math> and so <math>x \in [\xi, \eta]</math> and so <math>\bigcap_{n=1}^{\infty} I_n \subseteq [\xi, \eta]</math>. | ||
+ | |||
+ | *Since <math>[\xi, \eta] \subseteq \bigcap_{n=1}^{\infty} I_n</math> and <math>\bigcap_{n=1}^{\infty} I_n \subseteq [\xi, \eta]</math> we have that <math>[\xi, \eta] = \bigcap_{n=1}^{\infty} I_n</math>. <math>\blacksquare</math> | ||
== Licensing == | == Licensing == | ||
Content obtained and/or adapted from: | Content obtained and/or adapted from: | ||
* [http://mathonline.wikidot.com/nested-intervals Nested Intervals, mathonline.wikidot.com] under a CC BY-SA license | * [http://mathonline.wikidot.com/nested-intervals Nested Intervals, mathonline.wikidot.com] under a CC BY-SA license | ||
+ | * [http://mathonline.wikidot.com/the-nested-intervals-theorem The Nested Intervals Theorem, mathonline.wikidot.com] under a CC BY-SA license |
Latest revision as of 16:29, 13 November 2021
Nested Intervals
- Definition: A sequence of intervals where is said to be nestedif .
For example, consider the interval We note that , , , … As we can see and so the sequence of intervals is nested. The diagram above illustrates this specific nesting of intervals.
Sometimes a nested interval will have a common point. In this specific example, the common point is since for all . We denote the set theoretic intersection of all these intervals to be the set of common points in a nested set of intervals:
Sometimes a set of nested intervals does not have a common point though. For example consider the set of intervals . Clearly since , , … However, there is no common point for these intervals.
Example 1
Determine the set of points to which the set of nested intervals have in common.
We first note that , , , … We conjecture that the set of points are contained within all the intervals. This can be informally deduced since and .
The Nested Intervals Theorem
We have just looked at what exactly a Nested Interval is, and we are about to look at a critically important theorem in Real Analysis. Before we look at the Nested Intervals Theorem let's first look at the following important lemma that will be used to prove the Nested Intervals Theorem.
- Lemma 1: Let and let and let and . Then if and only if .
Now let's look at the Nested Intervals theorem.
Theorem 1: If the interval for is a sequence of closed bounded nested intervals then there exists a real number such that .
- Proof of Theorem: We note that by the definition of nested intervals that for all so then .
- Now consider the nonempty set that is bounded above by . Thus this set has a supremum in the real numbers and denote it so that for all .
- We now want to show that for all . We will do this by showing that for all .
- First consider the case where . We thus have that by the definition of nested intervals and so by lemma 1 we get that . Here we see that .
- Now consider the case where . We thus have that by the definition of nested intervals and so by lemma 1 once again we have that . Once again we have that
- So then for all , and so is an upper bound to the set and so for all . Furthermore we have that for all , and so for every and thus and so the set theoretic union is nonempty.
Theorem 2: If the interval for is a sequence of closed bounded nested intervals then there exists a real number such that .
- Proof: We note that by the definition of nested intervals that for all so then .
- Now consider the nonempty set that is bounded below by . Thus this set has an infimum in the real numbers and denote it so that for all .
- We now want to show that for all . We will do this by showing that for all .
- First consider the case where . We thus have that by the definition of nested intervals and so by lemma 1 we get that . Here we see that .
- Now consider the case where . We thus have that by the definition of nested intervals and so by lemma 1 once again we have that . Once again we have that
- So then for all , and so is an upper bound to the set and so for all . Furthermore we have that for all , and so for every and thus .
- Theorem 3: Let and . If and then if the interval is a sequence of closed bounded nested intervals then .
- Proof: We first prove that . Now let . Therefore . But we know that for all and we know that for all and so for all . Therefore for all or in other words . Therefore .
- We will now prove that . Let . Then for all . We also know that for all . Suppose that is such that . Since is the supremum of the set then there exists an element such that and so and therefore , a contradiction. Now suppose that is such that for all . Since is the infimum of the set then there exists an element such that and so and so , once again, a contradiction. We must therefore have that and so and so .
- Since and we have that .
Licensing
Content obtained and/or adapted from:
- Nested Intervals, mathonline.wikidot.com under a CC BY-SA license
- The Nested Intervals Theorem, mathonline.wikidot.com under a CC BY-SA license