Question

Construct a subset A of the real numbers such that the set of all limit points of A = the natural numbers

Answer 1

Construct a subset A of R with
\[A' = \mathbb{N} \]

Answer 2

I THINK I understand the concept of limit points, but getting a function that has limit points which are only natural numbers and not anything else I'm not sure of how to do.

Answer 3

so you need a bijection from S subset R to N?

Answer 4

I would think? I never thought of it in terms of bijective. I'm just unsure how to get the set of limit points to contain ONLY natural numbers.

Answer 5

If your set A were already the natural numbers, then \(A^\prime =\mathbb N \). Since the natural numbers are already closed (with respect to the standard topology on \(\mathbb R\), which I assume is what we are using here).

Answer 6

Actually, we haven't mentioned anything about closure and topology. Thank you for mentioning that, though, it helps me solve my problem. Is there a way to come to that conclusion without having that knowledge, though?

Answer 7

@nerdguy2535 Whenever you aren't busy :)

Answer 8

Hmm, can you tell me the definitions we are working with? That way I can try to help construct a better solution. How are you defining the limit points of a set?

Answer 9

So, the way we have it defined:

"A point p element of the real numbers is said to be a limit point of E if every for every epsilon > 0, the deleted neighborhood of p intersect E is non empty.

"A point p element of E is an isolated point of E if there exists epislon > 0 such the epsilon neighborhood p intersect E = {p}"

As for other notes we have (sorry if these don't help, thisis just listing all I have to work with)

"Let E be a subset of R:
a) If p is an element of E', then every neighborhood of p contains infinitely many points of E
b) If p is an element of E', then there exists a sequence {p_{n}} in E with p_{n} not equal to p for all n element of the natural numbers, such that p_{n} -> infinity = p"

We have that Q' = R. It also saysthat a finite set has no limit points.

Then we have an example that proves E' = {0} for E = {1/n | n is an element of N}
That's all I have to work with.

Answer 10

From my understanding, p is a limit point of a set E if there's another point within E for every neighborhood of p. But I guess that hasn't been enough to help me construct a subset per se.

Answer 11

Alright, this is enough information. Everything is here, you just have to be slightly clever. Do you understand the example that showed E'={0} if \(E=\{\frac{1}{n}\mid n\in \mathbb N\}\)?

Answer 12

Well....it went through cases where p was 0 (I'm guessing because the example already mentioned 0 was a limit point), where p = 1/n, p was an element of (0,1), p < 0, and p >1. So I suppose all the typical cases you would need.

To show p = 0, it just used the archmiedian property, so straight forward. For p = 1/n, it let epsilon = 1/n - 1/(n+1) > 0 and said that the intersection of E and the neighborhood of p = {p}. I'd have to stare at that for a bit, though, lol.

For p <0, it said let epsilon = -p > 0 and said that the intersection of the neighborhood of p and E was the empty set. And chose epsilon = p - 1 > 0 for the p > 1 case.

For the last 3, it used the neighborhood and not the deleted neighborhood. I guess if you can show something is an isolated point, then that shows it isn't a limit point. But for the last two, saying the intersection of the neighboor of p and E is empty, I'm not sure if that says anythign about the deleted neighborhood without thinking about it more.

Either way, this is using a bunch of cases to show 0 is a limit point. I'm not sure how you would do an argument for saying E' is all the natural numbers, though. By contradiction?

What about the proof should I be keying in on? xD

Answer 13

Well, lets not try to jump straight from a set where E'={0} all the way to ALL of the natural numbers. Lets take baby steps. Do you think you could construct a set E where E'={1}? Using the same idea as the one to get E'={0}? (You dont have to write it all out, just tell me if you think you could do it or not).

Answer 14

Well, I could be wrong, but it looks like you could link finding limit points to finding limits or finding subsequential limits. So if there is that connection, Could I say E = { n/(n+1) | n element of the natural numbers} ? If so, then I would need a case where p = 1, p > 1, and p < 1, right? Would I need a case like in the example where I have something like from (0,1) as well?

Answer 15

Yep, that would work. I was thinking \(1+\frac{1}{n}\), but your idea works too :) Now how about E'={2}? or E'={3}? How about E'={m} for any natural number m? Do you see the pattern?

Answer 16

So does that mean n + 1/n works?

As a side note, how do you use the equation editor to type 1 + 1/n without making the text go to a separate line? Whenever I click the equation editor, everything I type goes to a different line and makes it all choppy :(

Answer 17

You don't want \(n+\frac{1}{n}\), since this goes to infinity as n goes to infinity. You want to have another variable m totally separate from n, so something like this: $$m+\frac{1}{n}.$$Then as n goes to infinity, this sequence converges to m.

To make equations, type "\ (" "\ )" (without the quotes and space inbetween).

Answer 18

Thanks ^_^ So if I have something \(m + {1/n}\) that converges to m, what do I have to say about m? Just say that m is an element of n?

Answer 19

You want to put all these elements together. Create the set: $$E=\{m+\frac{1}{n}\mid m,n\in \mathbb N \}.$$ This set will have all the natural numbers as limit points.

Answer 20

Alrighty. So, are we assuming m goes to infinity as well, or is it like a constant term and only n goes to infinity?

Answer 21

While m can go to infinity, that doesn't really help us. You really want to think of it as, "For a fixed m, as n goes to infinity, we have a sequence that converges to m". Since m can be any natural number, we conclude the set of limit points is \(\mathbb N\).

Answer 22

I see. What concerns me now then is showing they're the only limit points. Like, if m = 1 and n goes to infinity, couldn't you say something like 1.1 was a limit point since you can always find a deleted neighboorhood of 1.1 that contains an element of E for some given n? If that question makes sense.

Answer 23

Your question does make sense, and it is something we have to be careful about. However, using the exact same ideas from the proof of the E'={0} case, I claim we can indeed show that \(\mathbb N\) are the only limit points.

Answer 24

So for example, you brought up 1.1. This is \(1+\frac{1}{10}\). This is going to be an isolated point of our set, since the closest points to it are \(1+\frac{1}{9}\) and \(1+\frac{1}{11}\). We will definitely be able to find an epsilon neighborhood around 1.1 that doesn't contain any points from our set in it.

Answer 25

Pictures really help. Here is the E'={0} case: (Yay for bad pictures). If E is just the set 1/n, it looks something like this. You were able to show that 0 is the only limit point, so all the other points in E are isolted.

Answer 26

Here is our set: \(E=\{m+\frac {1}{n}\mid m,n\in \mathbb N\}\)
It does the same thing as your E' = {0} set, just at every natural number. So its going to behave the exact same way. Every point will be isolated except for the natural numbers.