Eventually we know that the length of is . Then let , and choose an such that . So . And which shows that is a limit point of .
But why can't we say that both and are both infinite? Why at least?
Prove that a set which is both bounded and infinite has a limit point. For , let and . Since is bounded, then there is an interval containing . Now if is infinite, then does this imply that both and are both infinite?
In other words, are and both infinite?
First you cannot just . You can say that is a bounded infinite set.
There is a point such there are infinitely many points in either greater than or less that .
We will say greater than .
Define a set as follows: there are infinitely many terms of greater than .
1. is not empty. (WHY?)
2. T has an upper bound. (WHY?)
3. T has a least upper bound, . (WHY?)
4. For each there are infinitely many terms of that are greater that . (WHY?)
5. That means that is a limit point of . (WHY)?
Why can't you say ? It is compact and infinite right? You could have an infinite bounded set that is just constant (e.g. ). Actually, that was purely notational. I did suppose that is a bounded infinite set such that contains . In this case, and . And I split up the interval into 2 sub intervals: and and continued in this way, getting a series of nested intervals.
So
1. is not empty, because for , by Archmidean property.
2. has an upper bound because for , then for all .
3. By least upper bound property, (an ordered set), is not empty, and is bounded above. Thus exists in .
4. is an infinite set.
no, just at least one..
Let us consider the set . Since it is bounded, there is an such that .
Bisect into and .
At least one of these has the properties:
1. intersection with is not empty; and (WHY?)
2. Infinite (WHY?)
Name that set . Again, bisect into and . again, at least one of these has the properties:
1. intersection with is not empty; and (WHY?)
2. Infinite (WHY?)
Name that set ..
Do this continuously and form the sequence of nested intervals .
Use the Property of Nested interval, such that for all .
... now, can you do the rest? you only need to show that every neighborhood of contains infinitely many points from and hence from . (maybe, two steps or three more.. Ü)
1. Consider . Then we have . Then and . Both of these have a nonempty intersection with . I don't see a case where one interval has an empty intersection with and another interval has a nonempty intersection with .
2. Closed intervals are infinite.
3. Choose . Repeat step 1. Do this continuously. As we continue, the length of the intervals, .