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Bolzano–Weierstrass theorem

If you can’t be kind, at least be vague, Judith Martin

Recall

Definition. In mathematics, a sequence is an enumerated collection of objects in which repetitions are allowed and order matters. Sequences are denoted by $\{a_n\}_{n=1}^\infty$

Examples: {0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …}, {1, 3, 5, 7, 9, …}, i.e., {2n -1}, {(-1)n+1n} that is 1, −2, 3, −4, 5, −6, 7, −8, 9, −10, . ., etc. The limit of a sequence is the value that the terms of a sequence “tend to”, and is denoted as $\lim_{n \to ∞} a_n = A$. If such a limit exists, the sequence is called convergent. A sequence that does not converge is said to be divergent.

More formally, $\lim_{n \to ∞} {a_n}_{n=1}^\infty = A$ if for all ε > 0, there exists N ∈ ℤ such that ∀n ≥ N: |an - A| < ε.

Definition. A subsequence of a sequence is a sequence that can be derived or extracted from the original sequence by deleting some of the terms without changing the order of the remaining terms or choosing infinitely many values from the given sequence, where each successive choice is made by taking a larger index (and by doing so, looking farther away in the sequence). Given $\{a_n\}_{n=1}^\infty ⊆ ℝ$,

$\{n_k\}_{k=1}^\infty ⊆ ℕ$,

the sequence $\{a_{n_k}\}_{k=1}^\infty ⊆ ℝ$ is called a subsequence.

Theorem. Every subsequence of a convergent sequence converges and to the same limit.

Proof. Suppose $\{a_n\}_{n=1}^\infty,$

$\lim_{n \to ∞} a_n = A$ and suppose an arbitrary subsequence, $\{a_{n_k}\}_{k=1}^\infty$

Given ε > 0 ⇒ ∃N ∈ ℕ: ∀n ≥ N |an-L| < ε

By definition of subsequence, $\{n_k\}_{k=1}^\infty ⊆ ℕ$, n1 < n2 < ··· hence, it is unbounded.

Since {nk} is strictly increasing ⇒ ∃K ∈ ℕ such that ∀k ≥ K ⇒ nk ≥ N ⇒ |ank-L| < ε ⇒ $\lim_{n_k \to ∞} a_{n_k} = L$ ∎

Bolzano–Weierstrass theorem

Bolzano–Weierstrass theorem states that each infinite bounded sequence in ℝn has a convergent subsequence.

Proof.

Suppose $\{a_n\}_{n=1}^\infty$ is bounded, then |an| < M for all n ∈ ℕ ↭ an ∈ [-M, M] = [-M, 0] ∪ [0, M] ⇒ there are infinite-many an in [-M, 0] ∪ [0, M] ⇒ there are infinite-many an in [-M, 0] or (maybe there are infinite-many in both) [0, M].

Let’s call it I1 the one that has infinite-many an, and pick an element in $\{a_n\}_{n=1}^\infty ∩ I_1$ and call it an1.

We iterate this process, let’s split I1 into two equal parts, repeat the process and call I2 the one that has infinite-many an and pick an element and call it an2.

Basically, we are constructing a sequence I1 ⊇ I2 ⊇ ··· ⊇ Ik ⊇ ··· where an1∈ I1, an2∈ I2, and so on ⇒[By the nested interval theorem for the real numbers. It states that if you have a sequence of nested, bounded, nonempty, closed intervals, then their intersection is nonempty.] $\bigcap^∞_{i=1} {I_i} ≠ ∅$, length(I1) = M.

Besides, length(I2) = $\frac{M}{2}, ···, length(I_k) = \frac{M}{2^{k-1}}$

$\bigcap^∞{i=1} {I_i} ≠ ∅$, pick an element L ∈ $\bigcap^∞{i=1} {I_i} ≠ ∅$. Claim: $\lim_{k \to ∞} a_{n_k} = L.$

Let ε > 0, we need to choose K, such that ∀k ≥ K ⇒ $|a_{n_k} - L| < ε$

We know that ∀k ≥ K, ank∈ Ik and $length(I_k) = \frac{M}{2^{k-1}}$.

We know that $\lim_{k \to ∞} \frac{1}{2^{k-1}} = 0$. Given ε/M > 0, we choose K > 0, such that ∀k ≥ K, $|\frac{1}{2^{k-1}} - 0|< \frac{ε}{M} ↭ \frac{1}{2^{k-1}} < \frac{ε}{M} ↭ \frac{M}{2^{k-1}}$ <[🚀] ε

In conclusion, let ε > 0, we choose K > 0, such that ∀k ≥ K ⇒ ank∈ Ik, L ∈ Ik and $length(I_k) = \frac{M}{2^{k-1}}$ ⇒ $|a_{n_k} - L| < \frac{M}{2^{k-1}}$ <[🚀] ε ∎

Examples

However, the subsequence $\{(-1)^{2n}\}_{n=1}^\infty = (1, 1, 1, 1, ···)$ converges to 1.

Besides, $\{(-1)^{2n+1}\}_{n=1}^\infty = (-1, -1, -1, -1, ···)$ converges to -1.

Recall. A continued fraction is an expression obtained through an iterative process of representing a number as the sum of its integer part and the reciprocal of another number, then writing this other number as the sum of its integer part and another reciprocal, and so on.

We could use the continued fraction representation of 2π to find a increasing sequence of integers such that sin(n) approaches sin(2π)=0.


This is how you find the continued fraction for any number at all, say x0.

First, let a0 be the largest integer that does not exceed x0, that is, a0=⌊x0⌋. And let b0 be the fractional part of x0, that is b0 = x0 −a0. In our example, x0 = π, a0 = 3, b0 = 0.14159…

Now, we can write the following expression, x 0 = a0 + b0 = a0 + $\frac{1}{x_1}, x_1 = \frac{1}{b_0} = \frac{1}{0.14159…} = 7.0625···$

Then, repeat, let a1= ⌊x1⌋, b1 = x1 −a2, and $x_2 = \frac{1}{b_1}$, so we have a1 = 7, b1 ≈ 0.0625···, and x2 = $\frac{1}{0.0625…} = 15.996···$

π = $3 + \frac{1}{7+\frac{1}{15.996···}}$, and in general, x0 = $a_0 + \frac{1}{a_1+\frac{1}{x_2}}$

Continued fraction for π: (π = [3;7,15,1,292,1,1,1,2,1,3,1,…])

$\pi = 3 + \frac{1}{7 + \frac{1}{15 + \frac{1}{1 + \frac{1}{292 + \frac{1}{1 + \frac{1}{1 + \frac{1}{1 + \dots}}}}}}}$

As it was previously stated, we could use the continued fraction representation of 2π to find a increasing sequence of integers such that sin(n) approaches sin(2π)=0

2·π = 6.283185307179586 = 6 + 0.283185307179586, a1 = 6, x1 = 1/0.283185307179586 = 3.3531256652965528, a2 = 3, x2 = 1/3531256652965528

$2\pi = 6 + \frac{1}{3 + \frac{1}{1 + \frac{1}{1 + \frac{1}{7 + \frac{1}{2 + \frac{1}{146 + \frac{1}{3 + \dots}}}}}}}$

First examples of finite simple continuos fraction, a0 = 6, $a_0+\frac{1}{a_1} = 6 + \frac{1}{3} = \frac{19}{3}, a_0+\frac{1}{a_1+\frac{1}{a_2}} = 6 + \frac{1}{3+\frac{1}{1}} = 6 + \frac{1}{4} = \frac{25}{4},$ and so on.

The series of 2π is 6, $6, \frac{19}{3}, \frac{25}{4}, \frac{44}{7}, \frac{333}{53}, \frac{710}{113}, \frac{103993}{16551}, \frac{312689}{49766}, \frac{1980127}{315147}, \frac{2292816}{364913}, ···$

Ignoring the denominators (it will not make any difference at infinite), we have the subsequence:

sin(6) = −0.279415, sin(19) = 0.149877, sin(25) = −0.132352, sin(44) =0.0177019, sin(333) = −0.00882117, sin(710) =0.0000602887, sin(103993) = −0.0000191293, sin(312689) = 2.900699×106, etc. This subsequence is obviously convergent to zero.

$\{((-1)^{2n+1}+1)(2n+1)\}_{n=1}^\infty= (0, 0, 0, 0, ···)$.

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
  8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
  9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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