Question

I need some help with limit points and neighborhoods.

Answer 1

I'm sending my self in circles. So if a point \(p\) is a limit point of \(E\), then every neighborhood of \(p\) contains a point \(p^{\prime}(\not=p)\) that is also in \(E\).

Answer 2

But if \(p^{\prime}\) is in a neighborhood, then it has to be in every larger neighborhood, right?

Answer 3

So shouldn't some point \(p^{\prime}\) be in EVERY neighborhood of \(p\)?

Answer 4

No no. Only p is in every neighborhood.

Answer 5

That's kind of what I figured had to be the case.

Answer 6

If p is a limit point, then it means that no matter what tiny neighborhood of p you give me, I can give you a point p-prime that is in that neighborhood and also in E.

Now suppose you give me some tiny neighborhood. I give you a p-prime that works for that neighborhood. If you give me a larger neighborhood, then yes, just like you said, p-prime will still be in that larger neighborhood.

Answer 7

(Because suppose it's in every neighborhood, then it isn't in the neighborhood of size \(\lvert p - p^{\prime}\rvert\).)

Answer 8

If you give me a smaller neighborhood though, my p-prime might not be in it anymore. HOWEVER, if p is a limit point, then there IS some other point that works.

Answer 9

In general, there won't be any point that is in every neighborhood.

Answer 10

Only p itself is in every epsilon-neighborhood of p.

Answer 11

Right, and that's the whole idea of letting \(\varepsilon \rightarrow 0\) implying if it's\( < p + \varepsilon\) then it \(= p\).

Answer 12

I didn't word that very well, but you know the kind of proofs I'm talking about.

Answer 13

I think so, yes. Is there a specific problem you're having trouble with?

Answer 14

If it's \(< \varepsilon\) for every \(\varepsilon\) then it's 0. That's what I was going for.

Answer 15

Yeah, there is, "Is there a nonempty perfect set in \(\mathbb{R}^{1}\) which contains no rational?"

Answer 16

no rational elements?

Answer 17

Um... sure. The set of irrational numbers.

Answer 18

Or the set {pi}

Answer 19

But the set of irrational numbers would have a rational number as a limit point, right?

Answer 20

Yes. It would.

Answer 21

But it has to be a perfect set. So it has to equal the set of it's limit points.

Answer 22

Okay. So the qeustion is a nonempty set in R1 with no rational limit points.

Answer 23

But the limit points of \(\mathbb{R} - \mathbb{Q}\) is \(\mathbb{R}\).

Answer 24

Oh. Nonempty and perfect. Remind me what perfect means? Contains all limit points?

Answer 25

Every point is a limit point.

Answer 26

and closed.

Answer 27

Right. Let me think on it a moment.

Answer 28

No rational limit points?

Answer 29

Well if it has a rational limit point, then that point is in the set, contradiction.

Answer 30

Oh right. That was super simple.

Answer 31

But I want to get back to your earlier question. You said something like:
p is a limit point of E.
So for every epsilon-neighborhood of p, there exists a p' in that neighborhood such that p'=p and p' is in E.
If a point p' is in some epsilon neighborhood of p, then it is in every larger epsilon neighborhood of p.
So there must be some point that is in every epsilon neighborhood of p.

Answer 32

Okay.

Answer 33

Now all of that is true until the last statement.

Answer 34

Right.

Answer 35

Okay. I need an example. 0 is a limit point of [0,1].

Answer 36

Why? Well. I need to be able to give a suitable p' for any epsilon neighborhood.

Answer 37

1/n

Answer 38

By the archimedean property.

Answer 39

So fix epsilon >0.
epsilon/2 works as my point for any given epsilon.

Answer 40

True, that would be another route.

Answer 41

I has the set 1/n on the brain instead of all of [0, 1]

Answer 42

had*

Answer 43

Yes. Both work.
Buy you would say, "Hey isn't there some point that is in every epsilon neighborhood?" And I would say, "Give me any point p, and it is not in the neighborhood for epsilon = p/2."

Answer 44

Yeah.

Answer 45

Basically, you can half any epsilon I give you to get a point that is still in [0,1] and also in that neighborhood, but given any point, I can still get closer to 0 and use that as my new epsilon neigborhood.

Answer 46

So I never get a point close enough to be able to say, "This is in every epsilon neighborhood."
In fact, for any point, regardless of how close, there will always be infinitely many epsilon neighborhoods of p that do not include that point.