# Bolzano-Weierstrass Theorem

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• Jul 8th 2008, 01:38 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?
• Jul 8th 2008, 02:18 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?
• Jul 8th 2008, 02:19 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)?
• Jul 8th 2008, 02:45 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)?

• Jul 8th 2008, 02:57 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.
• Jul 8th 2008, 03:14 PM
particlejohn
yeah that was just a notational thing. I had assumed a bounded infinite set contained in $[-B,B]$.
• Jul 8th 2008, 04:54 PM
particlejohn
The key question if the following: if $[-L,0]$ and $[0,L ]$ are they both infinite?
• Jul 8th 2008, 08:22 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)
• Jul 9th 2008, 08:31 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)

• Jul 9th 2008, 11:21 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 $I$?
• Jul 9th 2008, 11:26 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.
• Jul 9th 2008, 11:28 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 $I$?

• Jul 9th 2008, 11:30 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 $S$.

• Jul 9th 2008, 11:35 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..
• Jul 9th 2008, 11:37 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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