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 $.