1. ## closed sets

given A={1/n:nεN} prove if A is open or closed set in R (=real Nos).

is $\overline A$ closed??

2. Originally Posted by xalk
given A={1/n:nεN} prove if A is open or closed set in R (=real Nos).

is $\overline A$ closed??

A can't be closed because it does not contain all of its limit points (namely 0). Instead of wondering whether the complement of A is closed, remember that A is open in the reals if for every point of A, there is an open interval which is entirely contained in A. This is impossible since given any point of A, any interval containing this point contains some real numbers which are not in A (for example, any irrational number in that interval).

3. ## Alternatively

It is also pretty easy to see the complement of A is open.
$A^c= (-\infty,0)\cup \bigcup _{n=1}^{\infty} (\frac{1}{n}, \frac{1}{n+1})$ which is a union of open sets which is open.

But yeah mylestone is spot on about the limit point thing too, just thought it might be useful to see the standard definition for when a set is closed if its complement is open.

4. Originally Posted by Gamma
It is also pretty easy to see the complement of A is open.
$A^c= (-\infty,0) \cup \bigcup _{n=1}^{\infty} (\frac{1}{n}, \frac{1}{n+1})$ which is a union of open sets which is open.

But yeah mylestone is spot on about the limit point thing too, just thought it might be useful to see the standard definition for when a set is closed if its complement is open.
$A^c= (-\infty,0) \cup \bigcup _{n=1}^{\infty} (\frac{1}{n}, \frac{1}{n+1})$ should be changed to $A^c= (-\infty,0] \cup \bigcup _{n=1}^{\infty} (\frac{1}{n+1}, \frac{1}{n}) \cup (1, \infty)$

As mylestone said, $A^{c}$ is neither open (its complement is not closed) nor closed in the standard topology because A does not contain 0, which is a limit point of A.

Anyhow, a K-topology on $\mathbb{R}$ allows K={1/n | n=1,2,...} to be closed by giving a basis element having a form (a,b)-K (Remember a basis element is always open in the given topology). The K-topology on $\mathbb{R}$ is an example whose space is Hausdorff but not regular.

5. Originally Posted by aliceinwonderland
$A^c= (-\infty,0) \cup \bigcup _{n=1}^{\infty} (\frac{1}{n}, \frac{1}{n+1})$ should be changed to $A^c= (-\infty,0] \cup \bigcup _{n=1}^{\infty} (\frac{1}{n+1}, \frac{1}{n}) \cup (1, \infty)$

As mylestone said, $A^{c}$ is neither open (its complement is not closed) nor closed in the standard topology because A does not contain 0, which is a limit point of A.

Anyhow, a K-topology on $\mathbb{R}$ allows K={1/n | n=1,2,...} to be closed by giving a basis element having a form (a,b)-K (Remember a basis element is always open in the given topology). The K-topology on $\mathbb{R}$ is an example whose space is Hausdorff but not regular.
Thank you but what about $\overline A$??is it closed or open??

6. ## closure

that notation is the closure of A, it is the smallest closed set containing your set A. It is closed for sure, that is the whole point. You just take your set and adjoin all the limit points so you can be sure your new set contains all of its limit points and therefore is closed. That set I screwed up and posted before is the complement of your closure, not the original set. My bad, and thanks for catching there error alice.

7. Originally Posted by Gamma
that notation is the closure of A, it is the smallest closed set containing your set A. It is closed for sure, that is the whole point. You just take your set and adjoin all the limit points so you can be sure your new set contains all of its limit points and therefore is closed. That set I screwed up and posted before is the complement of your closure, not the original set. My bad, and thanks for catching there error alice.
How do we prove that $\overline A$ is closed??

8. Originally Posted by xalk
How do we prove that $\overline A$ is closed??
In order for $x \in \overline A$ it must be true that $x \in A$ or $x$ is a limit point of $A$.
You want to prove that $\overline {\overline A}=\overline A$.

9. Originally Posted by Gamma
It is also pretty easy to see the complement of A is open.
$A^c= (-\infty,0)\cup \bigcup _{n=1}^{\infty} (\frac{1}{n}, \frac{1}{n+1})$ which is a union of open sets which is open.
No, that is not the complement of A. As myestone pointed out, A does not include 0.
$A^c= (-\infty,0]\cup \bigcup _{n=1}^{\infty} (\frac{1}{n}, \frac{1}{n+1})$
which is not open.

But yeah mylestone is spot on about the limit point thing too, just thought it might be useful to see the standard definition for when a set is closed if its complement is open.
You do understand, then, that you were saying mylestone is wrong? He's not of course. A is NOT closed and its complement is NOT open.

10. ## Missing the point

I am pretty sure you guys are failing to read what I just wrote. I admit I was incorrect in my first post. The union of open intervals that I posted is the complement of $\overline A$. I am saying that mylestone is correct $\overline A = A \cup {0}$ it is A unioned with it's limit points as he said. My proof is for why $\overline A$ is closed.

$\overline A ^c = (-\infty , 0) \cup \bigcup _{n=1}^{\infty}(\frac{1}{n} , \frac{1}{n+1})\cup (1, \infty)$ That is the complement of $\overline A$ can be written as a union of open sets, and is therefore open. So when the complement of a set is open, that means the set is closed, so $\overline A$ is closed.

Which is obvious, you are closing a set A when you write $\overline A$. By definition, $\overline A = \bigcap_{A \subset K, K closed} K$ which is the intersection of all closed sets containing A and the intersection of closed sets is closed.