- #1

MidgetDwarf

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- 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.

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.

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