# double closure of A = closure of A

prove: $\bar A = \bar{\bar A}$ ,

I can't quite latex it, but I think that I am asked to show that closure of A is the closure of the closure of A. The exact question is:" double bar A = bar A".

Is this too easy?

i) $\bar A$ is defined to be $A \cup A'$ where $A'$ is the set of all limit points of $A$.

ii) $A = \bar A$ if and only if $A$ is closed.

We know $\bar A$ is definitely closed since its defined to have all of the limit points of $A$. so when we use the closure function to $\bar A$ it should do nothing since the set is already closed. so $\bar{\bar A} = \overline{(\bar A )}$ but since the inside, $\bar A$ is already closed, this returns itself. so $\overline{(\bar A)} = \bar A$.

• You've got it! I don't see anything wrong with your reasoning. – Tony Mottaz Feb 19 '16 at 9:06
• Another definition (that I prefer) is the intersection of all closed sets containing the set in question. With this definition, the result is immediate. – copper.hat Feb 19 '16 at 9:08

From ii) (your ii)) if A is closed so $A=\bar A$, since $\bar A$ is closed so it is equals to it's closure "the double bar of A" : $\bar A=\bar {\bar A}$
Your reasoning is wrong for $\overline{A}$: it's defined to have all the limit points of $A$, not necessarily for $\overline{A}$. And any set $B$ is closed iff it contains its (i.e. $B$'s!) limit points (not only those of a smaller set).
So you still have to show that if $x$ is a limit point of $\overline{A} = A \cup A'$, then $x$ is in $\overline{A}$. If this holds, $\overline{A}$ is indeed closed.
So let $x$ be such a limit point of $\overline{A}$, and suppose (for a contradiction) that $x \notin \overline{A} = A \cup A'$. In particular, $x \notin A$ and $x$ is not a limit point of $A$, so there is some open neighbourhood / ball /set (whatever you like, depending on your text) $O$ that contains $x$ and does not intersect $A$ (it contains no points of $A$ except possibly $x$, by being a non-limit point, but we also assume $x \notin A$). But then no point of $O$ can be a limit point of $A$ either (nor is it a point of $A$). So $O$ is disjoint from $A \cup A' = \overline{A}$, and so $x$ cannot be a limit point of $\overline{A}$, which is a contradiction.
This shows that $\overline{A}$ is closed and then (ii) does the rest to see that $\overline{\overline{A}} = \overline{A}$ as you state.