# Thread: Sequences and closed sets

1. ## Sequences and closed sets

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

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

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

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