1
$\begingroup$

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 $.

$\endgroup$
  • $\begingroup$ You've got it! I don't see anything wrong with your reasoning. $\endgroup$ – Tony Mottaz Feb 19 '16 at 9:06
  • 1
    $\begingroup$ Another definition (that I prefer) is the intersection of all closed sets containing the set in question. With this definition, the result is immediate. $\endgroup$ – copper.hat Feb 19 '16 at 9:08
3
$\begingroup$

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" : $\bar A=\bar {\bar A}$

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.