Bolzano-Weierstrass Theorem

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July 8th 2008, 12:38 PM
particlejohn

Bolzano-Weierstrass Theorem

Prove that a set which is both bounded and infinite has a limit point. For $I = [a,b]$ , let $I^{L} = [(a+b)/2, b]$ and $I^{R} = [a, (a+b)/2]$ . Since $S$ is bounded, then there is an interval $I_{1} = [-B, B]$ containing $S$ . Now if $S$ is infinite, then does this imply that both $I_{1}^{R} \cap S$ and $I_{1}^{L} \cap S$ are both infinite?

In other words, are $[0,B] \cap S$ and $[-B,0] \cap S$ both infinite?
July 8th 2008, 01:18 PM
particlejohn

Eventually we know that the length of $I_n$ is $B/2^{n-2} \to 0$ . Then let $\varepsilon > 0$ , and choose an $n \in \bold{N}$ such that $B/2^{n-2} < \varepsilon$ . So $x_0 \in I_n \subset (x_{0}- \varepsilon, x_{0} + \varepsilon)$ . And $S \cap (x_{0}- \varepsilon, x_{0}+ \varepsilon) \backslash \{x_0 \} \neq \emptyset$ which shows that $x_0$ is a limit point of $S$ .

But why can't we say that both $[-B,0] \cap S$ and $[0,B] \cap S$ are both infinite? Why at least?
July 8th 2008, 01:19 PM
Plato

First you cannot just $I = \left[ {a,b} \right]$ . You can say that $I$ is a bounded infinite set.
There is a point $x_0 \in I$ such there are infinitely many points in $I$ either greater than or less that $x_0$ .
We will say greater than $x_0$ .
Define a set $T$ as follows: $x \in T\quad iff$ there are infinitely many terms of $I$ greater than $x$ .
1. $T$ is not empty. (WHY?)
2. T has an upper bound. (WHY?)
3. T has a least upper bound, $\gamma$ . (WHY?)
4. For each $\delta > 0$ there are infinitely many terms of $I$ that are greater that $\gamma - \delta$ . (WHY?)
5. That means that $\gamma$ is a limit point of $I$ . (WHY)?
July 8th 2008, 01:45 PM
particlejohn

Why can't you say $I = [a,b]$ ? It is compact and infinite right? You could have an infinite bounded set that is just constant (e.g. $\{1,1,1, \ldots \}$ ). Actually, that was purely notational. I did suppose that $S$ is a bounded infinite set such that $[-B,B]$ contains $S$ . In this case, $a = -B$ and $b = B$ . And I split up the interval into 2 sub intervals: $[-B, 0]$ and $[0, B ]$ and continued in this way, getting a series of nested intervals.

So $T = \{x : \ \text{there are infinitely many terms of} \ I \ \text{greater than} \ x \}$

1. $T$ is not empty, because for $x = a$ , $y = \frac{a+b}{2} > x$ by Archmidean property.

2. $T$ has an upper bound because for $\iota > b$ , then $\iota > x$ for all $x \in [a,b]$ .

3. By least upper bound property, $T \subset \bold{R}$ (an ordered set), $T$ is not empty, and $T$ is bounded above. Thus $\sup T$ exists in $\bold{R}$ .

4. $(\gamma - \delta, \gamma) \cap T$ is an infinite set.

Quote:

Originally Posted by Plato

First you cannot just $I = \left[ {a,b} \right]$ . You can say that $I$ is a bounded infinite set.
There is a point $x_0 \in I$ such there are infinitely many points in $I$ either greater than or less that $x_0$ .
We will say greater than $x_0$ .
Define a set $T$ as follows: $x \in T\quad iff$ there are infinitely many terms of $I$ greater than $x$ .
1. $T$ is not empty. (WHY?)
2. T has an upper bound. (WHY?)
3. T has a least upper bound, $\gamma$ . (WHY?)
4. For each $\delta > 0$ there are infinitely many terms of $I$ that are greater that $\gamma - \delta$ . (WHY?)
5. That means that $\gamma$ is a limit point of $I$ . (WHY)?
July 8th 2008, 01:57 PM
Plato

What makes you think that $I$ is compact?
It does not say that $I$ is closed, only inifinite and bounded.
We could have $I = \left\{ {1 - \frac{1}{n}:n \in Z^ + } \right\}$
That is infinite and bounded but not compact.
July 8th 2008, 02:14 PM
particlejohn

yeah that was just a notational thing. I had assumed a bounded infinite set contained in $[-B,B]$ .
July 8th 2008, 03:54 PM
particlejohn

The key question if the following: if $[-L,0]$ and $[0,L ]$ are they both infinite?
July 8th 2008, 07:22 PM
kalagota

no, just at least one..

Let us consider the set $I$ . Since it is bounded, there is an $L$ such that $I = [-L,L]$ .

Bisect $I$ into ${I_1}' = [-L,0]$ and ${I_1}'' = [0,L]$ .
At least one of these has the properties:
1. intersection with $I$ is not empty; and (WHY?)
2. Infinite (WHY?)

Name that set $I_1$ . Again, bisect $I_1$ into ${I_2}'$ and ${I_2}''$ . again, at least one of these has the properties:
1. intersection with $I_1$ is not empty; and (WHY?)
2. Infinite (WHY?)

Name that set $I_2$ ..

Do this continuously and form the sequence of nested intervals $\{I_n\}$ .

Use the Property of Nested interval, $\exists \ x$ such that $x \in I_n$ for all $n$ .

... now, can you do the rest? you only need to show that every neighborhood of $x$ contains infinitely many points from $I_n$ and hence from $I$ . (maybe, two steps or three more.. Ü)
July 9th 2008, 07:31 PM
particlejohn

1. Consider $L = 1$ . Then we have $I = [-1,1]$ . Then $I_{1}' = [-1,0]$ and $I_{1}'' = [0,1]$ . Both of these have a nonempty intersection with $I$ . I don't see a case where one interval has an empty intersection with $I$ and another interval has a nonempty intersection with $I$ .

2. Closed intervals are infinite.

3. Choose $I_1 = I_{1}'$ . Repeat step 1. Do this continuously. As we continue, the length of the intervals, $L/2^{n-2} \to 0$ .

Quote:

Originally Posted by kalagota

no, just at least one..

Let us consider the set $I$ . Since it is bounded, there is an $L$ such that $I = [-L,L]$ .

Bisect $I$ into ${I_1}' = [-L,0]$ and ${I_1}'' = [0,L]$ .
At least one of these has the properties:
1. intersection with $I$ is not empty; and (WHY?)
2. Infinite (WHY?)

Name that set $I_1$ . Again, bisect $I_1$ into ${I_2}'$ and ${I_2}''$ . again, at least one of these has the properties:
1. intersection with $I_1$ is not empty; and (WHY?)
2. Infinite (WHY?)

Name that set $I_2$ ..

Do this continuously and form the sequence of nested intervals $\{I_n\}$ .

Use the Property of Nested interval, $\exists \ x$ such that $x \in I_n$ for all $n$ .

... now, can you do the rest? you only need to show that every neighborhood of $x$ contains infinitely many points from $I_n$ and hence from $I$ . (maybe, two steps or three more.. Ü)
July 9th 2008, 10:21 PM
kalagota

Quote:

Originally Posted by particlejohn

1. Consider $L = 1$ . Then we have $I = [-1,1]$ . Then $I_{1}' = [-1,0]$ and $I_{1}'' = [0,1]$ . Both of these have a nonempty intersection with $I$ . I don't see a case where one interval has an empty intersection with $I$ and another interval has a nonempty intersection with $I$ .

2. Closed intervals are infinite.

3. Choose $I_1 = I_{1}'$ . Repeat step 1. Do this continuously. As we continue, the length of the intervals, $L/2^{n-2} \to 0$ .

no! what if your $I=\left\{2 + \frac{1}{n} : n\in \mathbb{N}\right\}$

the questions states that for a given bounded and infinite interval.. Show that it has a limit point..

what you did is to take the $L=1$ without the interval..
so how would you know if this $L$ is sufficient to bound your arbitrary interval $I$ ?
July 9th 2008, 10:26 PM
kalagota

Quote:

Originally Posted by particlejohn

1. Consider $L = 1$ . Then we have $I = [-1,1]$ . Then $I_{1}' = [-1,0]$ and $I_{1}'' = [0,1]$ . Both of these have a nonempty intersection with $I$ . I don't see a case where one interval has an empty intersection with $I$ and another interval has a nonempty intersection with $I$ .

it really doesn't matter whether both intervals satisfy the conditions since you can do it simultaneously.. your job is to show the existence of a limit point.
July 9th 2008, 10:28 PM
particlejohn

I was just illustrating the fact that why do we have at least one interval whose intersection with $S$ is nonempty? This implies that the other intervals intersection with $S$ is empty. But I cant find a case for that.

Quote:

Originally Posted by kalagota

no! what if your $I=\left\{2 + \frac{1}{n} : n\in \mathbb{N}\right\}$

the questions states that for a given bounded and infinite interval.. Show that it has a limit point..

what you did is to take the $L=1$ without the interval..
so how would you know if this $L$ is sufficient to bound your arbitrary interval $I$ ?
July 9th 2008, 10:30 PM
particlejohn

Quote:

Originally Posted by particlejohn

Eventually we know that the length of $I_n$ is $B/2^{n-2} \to 0$ . Then let $\varepsilon > 0$ , and choose an $n \in \bold{N}$ such that $B/2^{n-2} < \varepsilon$ . So $x_0 \in I_n \subset (x_{0}- \varepsilon, x_{0} + \varepsilon)$ . And $S \cap (x_{0}- \varepsilon, x_{0}+ \varepsilon) \backslash \{x_0 \} \neq \emptyset$ which shows that $x_0$ is a limit point of $S$ .

I have already done that.
July 9th 2008, 10:35 PM
kalagota

I see, your question is why at least, right?

because the union of a finite and an infinite sets is infinite..
however, union of finite sets is still finite..
July 9th 2008, 10:37 PM
particlejohn

can you give me an example of where one's intersection is empty and another intersection is not?

Quote:

Originally Posted by kalagota

I see, your question is why at least, right?

because the union of a finite and an infinite sets is infinite..
however, union of finite sets is still finite..

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