Bolzano–Weierstrass theorem

[hy23]

Homework Statement

This is not a homework question. I just don't quite understand the proof behind Bolzano–Weierstrass theorem. First of all, what are subsequences? And what is the proof trying to say when it's talking about the peaks?

Thanks for any help in advance, and please dumb it down for me!

Homework Equations

The Attempt at a Solution

 

[╔(σ_σ)╝]

What do you already know about subsequences? Also there are definitions of subsequence online. What don't you understand about the definitions?

 

[hy23]

ok so I have read a lot of definitions of subsequences but still it's kind of fuzzy in my head, like if I have a bounded sequence, say sin x, where x is all natural numbers, then can I say that sin n, where n is all even numbers, is a subsequence of sin x? And how in the world is that subsequence converging, it oscillates just like the sequence.

I guess to understand the theorem it would help if you had a good example. Thanks

 

[HallsofIvy]

Yes, if your original sequence is sin(1), sin(2), sin(3), sin(4), sin(5), sin(6), ... then sin(2), sin(4), sin(6), ... is a subsequence precisely because it is a subset of the first sequence- that's all 'subsequence' means. A subsenquence is is just a subset of the original sequence.

 

[Office_Shredder]

ok so I have read a lot of definitions of subsequences but still it's kind of fuzzy in my head, like if I have a bounded sequence, say sin x, where x is all natural numbers, then can I say that sin n, where n is all even numbers, is a subsequence of sin x? And how in the world is that subsequence converging, it oscillates just like the sequence.

I guess to understand the theorem it would help if you had a good example. Thanks

It only says there exists a single convergent subsequence. Usually if you just pick a random one it's not going to be the one that converges.

For example, suppose we just had the sequence -1,1,-1,1,-1,1,-1,1,... This is a bounded sequence, and it has a convergent subsequence: 1,1,1,1,1,1,... But it also has diverging subsequences, for example I could skip every other -1 and get 1,-1,1,1,-1,1,1,-1,1,1,... which diverges still

 

[╔(σ_σ)╝]

ok so I have read a lot of definitions of subsequences but still it's kind of fuzzy in my head, like if I have a bounded sequence, say sin x, where x is all natural numbers, then can I say that sin n, where n is all even numbers, is a subsequence of sin x? And how in the world is that subsequence converging, it oscillates just like the sequence.

I guess to understand the theorem it would help if you had a good example. Thanks

Just to add to what is already mentioned...

Let $$x_n$$be a sequence a subsequence is any sequence $$x_{g(n)}$$ where $$g(n)$$ is a increasing function from N to a subset of N.

Basically, you take your regular suquence and remove terms in a non-random order.

With your example sin(n) is the orginal sequence.

g(n) = 2n . This function is increasing and the range is a subset of n.

Your original sequence could be listed as follows

sin(1), sin(2),sin(3),sin(4),...,sin(n-1),sin(n)

A subsequence can be gotten from the original sequence by remmoving all the odd terms; which gives
sin(2),sin(4)...,sin(2(n-1)),sin(2n).

bolzano Weierstrass (BW)
The monotone subsequence theorem states that every sequence in R has a monotone subsequence.

The bolzano weierstrass theorem goes a little further to say that if the sequence is bounded it has a convergence subsequence.


The idea of the peak proof is to find a monotone subsequence and then apply the monotone convergence theorem.

 

[hy23]

ok so I'm a bit more clear on the definition of subsequences now, but how does the BW theorem work: is the idea of the peak basically - a point is a peak if it is larger than all previous points, and if we can find an infinite number of peaks then the subsequence is monotone increasing and because the sequence is bounded, the subsequence converges?

and in the case of sin x, would a convergent subsequence be sin($$\pi$$x) since it converges to 0? but how would the proof work because where are the peaks when all terms are the same?


p.s. sry the pi is not suppose to be superscripted, idk why it's like that

 

[╔(σ_σ)╝]

ok so I'm a bit more clear on the definition of subsequences now, but how does the BW theorem work: is the idea of the peak basically - a point is a peak if it is larger than all previous points, and if we can find an infinite number of peaks then the subsequence is monotone increasing and because the sequence is bounded, the subsequence converges?

and in the case of sin x, would a convergent subsequence be sin($$\pi$$x) since it converges to 0? but how would the proof work because where are the peaks when all terms are the same?p.s. sry the pi is not suppose to be superscripted, idk why it's like that

Actually, your definition of peak is opposite to what it should be but ,to be honest, it does not really matter. The standard definition of a peak at n is $$x_n > x_m$$ for m >n. So the peak is greater than the sucessor terms.

The idea is that the peaks form a monotone decreasing sequence ( it could also be monotone increasing with your definition of peak).

If there are infinitely many peaks then we can form a subsequence of thoes peaks and if that subsequence is bounded then it must converge.

However, a sequence need not have infinite number of peaks. i.e { 100, 99, 899,60, 4,4,4,4,4,4,4,4,...}. This sequence has peaks 100,899 and 60 ( by my definition of peak ).

The idea is that if a sequence has only a finite amount of peaks it means that after a while the sequence either doesn't "move" around (i.e it is eventually constant) or it is increasing.

Also, a sequence need not have peaks at all. Your example is a very good example of this.

If a sequence has no peaks then it is either constant or increasing.

In the constant case the sequence converges.

In the increasing case it converges if it is bounded.

 

[hy23]

wait, in your case how is it that 100 is a peak when 899 is greater than 100? Or do you mean that 100 only has to be bigger than the one subsequent term (99)?

the other thing is I get what you are saying about how after a while, the sequence either flattens out and I can see that clearly it will converge, but you also say that if it doesn't stay constant it increases and if it has an upper bound then it converges, this is what I don't understand because how do you know that there won't be more peaks, that after it increases it might decrease in an oscillatory fashion

do you have any good examples of a sequence that is not a trig function that has a convergent subsequence, an example that really demonstrates the proof nicely might put my brain to rest

thanks!

 

[╔(σ_σ)╝]

wait, in your case how is it that 100 is a peak when 899 is greater than 100? Or do you mean that 100 only has to be bigger than the one subsequent term (99)?

Great! You caught me napping :tongue:.

I made a mistake sorry for the confusion.

the other thing is I get what you are saying about how after a while, the sequence either flattens out and I can see that clearly it will converge, but you also say that if it doesn't stay constant it increases and if it has an upper bound then it converges, this is what I don't understand because how do you know that there won't be more peaks, that after it increases it might decrease in an oscillatory fashion

What I was saying is that if it has finitely many peaks then it is either evetually constant or eventually increasing. The word finite is very important because sequences are infinite in nature.

If the sequence decreases later in an oscilatory fashion then you have peaks! If the oscillatory decrease continues "forever" then you have infinitly many peaks. If it stops after a while, it means that ( i.e assume the oscillation occurs from n =1000 to 50,000 , after 50,000 the sequence is well behaved) after a while the sequence behaves itself.

Remember in the study of sequences the behaviour as n goes to infinity is what matters, not what happens at between say 1,000,000 and 24,000,000.

do you have any good examples of a sequence that is not a trig function that has a convergent subsequence, an example that really demonstrates the proof nicely might put my brain to rest
thanks!

Trig are nice examples because of the oscillation but consider the sequence...

$$s_n = \frac{1}{n} \quad for \quad odd \quad n \quad divisible \quad by \quad 5$$

$$s_n = -4 + \frac{1}{\sqrt{n}} \quad for \quad odd \quad n \quad not \quad divisible \quad by \quad 5$$
$$s_n= -e^{ \frac{1}{n} } \quad for \quad even \quad n$$

This sequence has 3 convergence subsequences but there are weird twist and turns everywhere.