The Bolzano Weirstrass Proof by Construction

[Bachelier]

Is this good? thanks

Proof:

let $$S$$ be an infinite bounded set in $$\Re$$

then $$\exists$$ real numbers a, b such that
$$S$$ is in [a, b]

One of the intervals, [a, (a+b)/2] or [(a+b)/2, b] contains an infinite set of members of S. Let's choose one with this property and call it $$[a_1, b_1]$$

Continuing in this fashion we obtain for each positive integer n the closed interval $$[a_n, b_n]$$ with the following properties:

1. $$d(a_n, b_n) \leq |a-b|/2^n$$

2. $$[a_n, b_n]$$ contains infinitely many points

3. and $$[a_n, b_n] \subset [a, b]$$​

so $$[a_n, b_n]$$ is bounded. let $$P= \{a_n|n \in\mathbb{N} \}$$

So P is bounded. let x= sup P

Surely the sequence $$a_n$$ is increasing.

Claim: $$a_n$$ converges to x

now $$\forall \epsilon > 0 \ \exists N \in \mathbb{N}$$ with $$N > |a-b|/2^n \ \forall \ n>N$$

then $$|a_n - x| \leq |a-b|/2^n < |a-b|/2^N< |a-b|/N = \epsilon$$

 

[Stephen Tashi]

If you are proving that each bounded sequence has a convergent subsequence, you must also deal with the case that the bounded sequence takes on only a finite number of values.

 

[Bachelier]

If you are proving that each bounded sequence has a convergent subsequence, you must also deal with the case that the bounded sequence takes on only a finite number of values.

Isn't it that in that case just select one of those finite values and thenceforth you have a convergent subsequence?

 

[Stephen Tashi]

Isn't it that in that case just select one of those finite values and thenceforth you have a convergent subsequence?

Yes

Surely the sequence $$a_n$$ is increasing.

The sequence is nondecreasing if that's how you define "increasing".

let $$P= \{a_n \ | \ n \in {N} \}$$

What is $${N}$$?

You conclude by claiming that an arbitrary epsilon is equal to |a-b| divided by an integer, which can't be right.

 

[Bachelier]

What is $${N}$$?

Sorry that was $$\mathbb{N}$$

My browser was playing tricks with my code.

You conclude by claiming that an arbitrary epsilon is equal to |a-b| divided by an integer, which can't be right.

Actually I found a large enough natural number N such that the distance from a_n to sup(P) is less than $$\epsilon \ (\forall \epsilon > 0)$$

 

[Bachelier]

The question is, do I have to include the b_ns as well?

 

[Stephen Tashi]

now $$\forall \epsilon > 0 \ \exists N \in \mathbb{N}$$ with $$N > |a-b|/2^n \ \forall \ n>N$$

then $$|a_n - x| \leq |a-b|/2^n < |a-b|/2^N< |a-b|/N = \epsilon$$

That makes no sense to me. What statement is the "if" that goes before the "then"?

Are you saying that the last relation should be $$\leq \epsilon$$ instead of = $$\epsilon$$ ?

(And do you find the way LaTex shows up on the forum to be buggy? I admit that I'm out of practice with LaTex, but I can cut and paste things other people have written and it doesn't show up correctly. Sometimes what shows up are symbols from parts of posts that I haven't even pasted. I've tried both the Opera and Firefox browsers.
Edit: The problem is discussed in this post: https://www.physicsforums.com/showthread.php?t=482096 )

 

[Bachelier]

OK, take a look at this:

$$\forall \epsilon > 0 \ , \ (\exists N \in \mathbb{N}) \ with \ [N > |a-b|/\epsilon] \ such \ that \ \forall \ n>N \ \mbox{implies that}$$

$$|a_n - x| \leq |a-b|/2^n < |a-b|/2^N \ \mbox{(because N is strictly less than n so 1/2^n < 1/2^N) \ (note: \ 2^N > N)}$$

$$Hence |a_n - x| < |a-b|/N = \epsilon \ \mbox{after replacing N}$$

 

[Bachelier]

$$should \ have \ divided \ by \ \epsilon \ in \ my \ original \ post. \$$