Results 1 to 10 of 10

Math Help - closed sets

  1. #1
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1

    closed sets

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

    is  \overline A closed??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Feb 2008
    Posts
    36
    Thanks
    1
    Quote Originally Posted by xalk View Post
    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).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517

    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.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Nov 2008
    Posts
    394
    Quote Originally Posted by Gamma View Post
    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.
    Last edited by aliceinwonderland; April 21st 2009 at 11:45 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Quote Originally Posted by aliceinwonderland View Post
    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??
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517

    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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Quote Originally Posted by Gamma View Post
    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??
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,395
    Thanks
    1481
    Awards
    1
    Quote Originally Posted by xalk View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,977
    Thanks
    1121
    Quote Originally Posted by Gamma View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517

    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.
    Last edited by Gamma; April 22nd 2009 at 01:09 PM. Reason: forgot a set
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Metric spaces, open sets, and closed sets
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: March 16th 2011, 05:17 PM
  2. Open sets and closed sets
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 30th 2010, 10:05 AM
  3. Approximation of borel sets from the top with closed sets.
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: February 18th 2010, 08:51 AM
  4. Metric Space, closed sets in a closed ball
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 19th 2009, 05:30 PM
  5. Continuity & inverse image of closed sets being closed
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: March 9th 2009, 05:07 PM

Search Tags


/mathhelpforum @mathhelpforum