# Thread: Sequences and closed sets

1. ## Sequences and closed sets

Let $\displaystyle A \doteq \{x_n\}^\infty_{n \doteq 1}$ be a sequence and let $\displaystyle B$ be the set of all subsequential limits of $\displaystyle A$. I have to show that $\displaystyle A \cup B$ is closed.

I got that $\displaystyle B$ is closed from that theorem in Rudin's textbook that said that the subsequential limits form a closed set which leads that $\displaystyle B$ is closed. Now I figured that I should prove that $\displaystyle A$ is closed as well in order to complete the proof, and that's where I'm stuck at.

2. Originally Posted by darkchibi07
Let $\displaystyle A \doteq \{x_n\}^\infty_{n \doteq 1}$ be a sequence and let $\displaystyle B$ be the set of all subsequential limits of $\displaystyle A$. I have to show that $\displaystyle A \cup B$ is closed.
I got that $\displaystyle B$ is closed from that theorem in Rudin's textbook that said that the subsequential limits form a closed set which leads that $\displaystyle B$ is closed. Now I figured that I should prove that $\displaystyle A$ is closed as well in order to complete the proof, and that's where I'm stuck at.
You cannot prove that $\displaystyle A$ is closed, consider $\displaystyle A = \left\{ {\frac{1}{n}} \right\}$.
But if $\displaystyle a$ is a limit point of $\displaystyle A$ them some subsequence of $\displaystyle A$ converges to $\displaystyle a$. Because $\displaystyle B$ is the set of all subsequential limits of $\displaystyle A$ then it follows that $\displaystyle A \cup B$ is closed. And of course $\displaystyle B$ is closed because any point limit point of $\displaystyle B$ is also a limit point of $\displaystyle A$, so it must belong to $\displaystyle B$ meaning it is closed.

3. Was the subsequence of $\displaystyle A$ you chose just arbituary, or do I have to construct it formally?