Closure of A consist of every point of A including the set of all limits of convergent subsequences of A.

Question

$(x_n)_n$ is a sequence in $\mathbb{R}$. $A$ is a set consisting of all terms of that sequence $$A=\{ x_n | n \in \mathbb{N} \}.$$ Show that $\bar{A}$ exists of all points of A including all limits of all convergent subsequences van $(x_n)_n$.

It is trivial that every point in $x_n$ is in $\bar{A}$. To show that the limits are in $\bar{A}$. We take a subsequence of $x_n$ so that we can form a region around the limit and there is one point in the region that is in $A$. Then we take a subsequence smaller than the previous one and do that infinite times untill the limit is in $\bar{A}$.

Is this the right way to do it? How do you write it in a formal proof?

Thanks in advance

Answer 1

Assume $C $ is a closed set that contains $A $, but such that there exists a limit $x $ of a subsequence of $(x_n) $ that is not in $C $.

Then $C$ can not be closed since $\Bbb R \setminus C $ is not open : there is no neighbourhood of $x$ that doesnt meet $C $.

This shows that $\overline {A} $ contains $A $ and its limit points.

Now let $A'$ be the union of $A $ and its limit points. $A'$ is closed because every convergent sequence of $A'$ has its limit in $A'$.

And thus $A'=\overline {A} $.

Answer 2

When you write that it is trivial that every point in $x_n$ is in $A$, my guess is that you meant $\overline A$.

If $a$ is the limit of a subsequence of $(x_n)_{n\in\mathbb N}$, then, by definition of limit, there are $x_n$'s as close to $a$ as you wish. Therefore $a\in\overline A$.

All this proves that $A$ together with the limit points is a subset of $\overline A$. All that remains to be done is that $\overline A$ is the union of $A$ with the set of limit points.