1. ## 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?

2. 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?

3. 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)?

4. 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.

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)?

5. 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.

6. yeah that was just a notational thing. I had assumed a bounded infinite set contained in $[-B,B]$.

7. The key question if the following: if $[-L,0]$ and $[0,L ]$ are they both infinite?

8. 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)

9. 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, $L/2^{n-2} \to 0$.

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)

10. 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 $I$?

11. 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.

12. 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.

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 $I$?

13. 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 $S$.

14. 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..

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

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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