Explanation of the Bolzano-Weierstrass theorem proof

[BifSlamkovich]

Homework Statement

Every bounded sequence has a convergent subsequence.

Homework Equations

Suppose that closed intervals I_0 $\supset$ I_1$\supset$ ... $\supset$ I_m and natural numbers n$_{1}$ < n$_{2}$ <
... < n$_{m}$ have been chosen such that for each 0 $\leq$ k $\leq$ m,
(2)
|I$_{k}$| = $b-a/2^{k}$, x$_{n}_{k}$$\in$I$_{k}$n and x$_{n}$ $\in$ I$_{k}$ for infinitely many n.

The Attempt at a Solution

My question is what is the guarantee that x_n will be in I_k?

 

[voko]

Each subsequent interval I is chosen so that it contains infinitely many members of the sequence. This is always possible because the main sequence contains infinitely many elements.

 

[BifSlamkovich]

Each subsequent interval I is chosen so that it contains infinitely many members of the sequence. This is always possible because the main sequence contains infinitely many elements.

It seems that I do not understand the notation for

the statement:

x$_{n}$ $\in$ I$_{k}$ for inifinitely many n.


Does it mean x$_{1}$, x$_{2}$, x$_{3}$, x$_{4}$ have to all be in the chosen interval?

 

[BifSlamkovich]

How does this theorem apply to the sequence:

$x$$_{n}$ = ($-$$1$)$^{n}$ * $1$-$1/n$?

 

[voko]

x$_{n}$ $\in$ I$_{k}$ for inifinitely many n.

This is some non-standard notation (or perhaps you misrepresented it). What is meant is that every time the interval is divided into two subintervals, at least one of them must contain an infinite number of the sequence's members. Non necessarily all members, of course.

 

[voko]

How does this theorem apply to the sequence:

$x$$_{n}$ = ($-$$1$)$^{n}$ * $1$-$1/n$?

The theorem applies to this sequence just fine. It contains, for example, this converging sequence: $1 - 1/(2n)$.

 

[BifSlamkovich]

The theorem applies to this sequence just fine. It contains, for example, this converging sequence: $1 - 1/(2n)$.

Suppose that closed intervals I_0 ⊃ I_1⊃ ... ⊃ I_m and natural numbers n1 < n2 <
... < nm have been chosen such that for each 0 ≤ k ≤ m,
(2)
|Ik| = b−a/2k, x_{n}_{k}∈Ikn and xn ∈ Ik for infinitely many n.

So, in trying to apply the above proof to this particular sequence, when we choose the subintervals of the interval [-1,1], they would have to be the ones that contain infinitely many x_n's, right? For instance, I couldn't "choose" the subinterval [0,1/2] because that contains only finitely many x_n's, but if I choose the subinterval [1/2,1], that contains infinitely many x_n's? I just want to clarify that this is what they mean.

 

[BifSlamkovich]

x_n = [(−1)^n] * [1-{1/n}]

Each side of 0 is going to contain x_n for infinitely many x_n's, supposedly. But does it really, because I thought infinitely many x_n's meant for all x_n's?

 

[voko]

The sequence is contained in [-2, 1]. Now consider [-2, -0.5) and [-0.5, 1]. Both intervals contain infinitely many elements of the sequence, but you have to choose one. Let's choose [-0.5, 1] and split it again: [-0.5, 0.25) and [0.25, 1]. The first one contains no elements, the second contains infinitely many of them, so choose it and split it again into [0.25, 0.625) and [0.625, 1]. The first one contains only one element, the second contains infinitely many, choose it and split: [0.625, 0.8125) and [0.8125, 1]. Again the second one contains infinitely many members. This pattern continues infinitely.

 

[HallsofIvy]

It seems that I do not understand the notation for

the statement:

x$_{n}$ $\in$ I$_{k}$ for inifinitely many n.


Does it mean x$_{1}$, x$_{2}$, x$_{3}$, x$_{4}$ have to all be in the chosen interval?

No, it means, as it says, that "infinitely many" of the numbers are in the interval. That might be "all except the first four terms" or "all except the first 10 billion terms" or "all even terms with even indices", etc.