Intro Real Analysis: Closed and Open sets Of R. Help with Problem

[MidgetDwarf]

Homework Statement

Let $$A= \{(-1)^n + \frac {2} {n} : n = 1, 2, 3,...\} $$ and $$ B =\{x \in ℚ: 0<x<1 \}.$$

What are the limits points of A and B

Relevant Equations

Definition of a limit point : A point x is a limit point of a set A if $\forall$ epsilon neighborhood of x intersects the set A at some point other than x.

Theorem 1: A point x is a limit point of a set iff $x= \lim a_n$ for some sequence $\{a_n)\}$ satisfying $a_n = x ~\forall n \in N$.

Theorem 2: Density of the Rational Numbers of Q in R.

For the set A:

Note that if n is odd, then $A = \{ -1 + \frac {2} {n} : \text{n is an odd integer} \}$ . If n is even, A = $\{1 + ~ \frac {2} {n} : \text{ n is an even integer} \}$ .
By a previous exercise, we know that $\frac {1} {n}$ -> 0. Let $A_1$ be the sequence when n is odd and $A_2$ be the sequence when n is even. By the Algebraic Limit Theorem, Lim $A_1$ = -1 and Lim $A_2$ = 1. Since -1 is not an element of A, then -1 is a limit point of A. Since 1 is never a term of $A_2$ , then 1 is a limit point of A. ( By Theorem 2).

Therefore, the limit points of A are -1 and 1.

For the set B:

I know that the set of Limit points of Q is R. Since we are only working with members of Q in the set B. I know that the following two sequences $\frac {1} {n}$ where n is equal to or greater than 2 and $\frac {n} {n+1}$ reside in B, and they converge to 0 and 1, respectively. Since both 0 and 1 are not members of their respective sequences, then 0 and 1 are limit points of B.

Do I have this correct so far?

But I am really unsure of B. Since by the Density Theorem of Q in R we know that for every real number there exist a sequence of rational numbers that converge to y. So by this Theorem, the limit points of B is the interval [0,1] ?

Sorry for the sloppy LaTex. This is my first time using it.

 

[MidgetDwarf]

sorry. I tried fixing the code, but i still cannot locate the problem. Wondering if a member can help me also with the syntax.

 

[FactChecker]

I think you should have a double '#' in "never a term of # A_2".
The single '#' did not turn on LaTex and things went wrong from there on.

From what I can see, your logic is correct.

 

[fresh_42]

It does not matter for a limit point whether it is part of the set or not.
Thus there are more limit points in B than the two you mentioned, and your wording for set A should be corrected accordingly.

 

[Mark44]

I tried fixing the code, but i still cannot locate the problem. Wondering if a member can help me also with the syntax.

I fixed things. Take a look at our tuturial -- the link to it is in the bottom left corner of the input pane.

 

[vela]

In your statement of theorem 1, did you mean $a_n \ne x$ for all $n \in \mathbb{N}$?

 

[MidgetDwarf]

It does not matter for a limit point whether it is part of the set or not.
Thus there are more limit points in B than the two you mentioned, and your wording for set A should be corrected accordingly.

Yes thank you. I wrote it in a weird way and I think that is what Mark mentioned. I tried to say that, since 1 is not an element of the sequence that converges to 1. Then the conditions you listed are automatically satisfied.

I could have just listed what you wrote and be done. Thank you .

 

[MidgetDwarf]

I fixed things. Take a look at our tuturial -- the link to it is in the bottom left corner of the input pane.

Thank you. For some reason I missed the tutorial when typing.

 

[MidgetDwarf]

It does not matter for a limit point whether it is part of the set or not.
Thus there are more limit points in B than the two you mentioned, and your wording for set A should be corrected accordingly.

Aww. Thank you. I some how mixed some definitions up. Ie., closed. Where A set is called closed if it contains its limit points.