Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The circle as a set closed and bounded

  1. Oct 31, 2012 #1
    Hi guys,

    I would like to understand why a circle (and in general a n-sphere) as a subset of R^2 (in general R^(n+1)) with the standard topolgy is considered a closed and a bounded set.

    I think that this can be a closed set because its complement (the interior of the circle and the rest of the plane) is open. And could be bounded because it has a finite extension (but ths is very intuitive). I cannot figure out what is the interior, the closure or the boundary of this set.

    Thank you for your help.
  2. jcsd
  3. Oct 31, 2012 #2
    Consider the function

    [tex]f:\mathbb{R}^2\rightarrow \mathbb{R}:(x,y)\rightarrow \sqrt{x^2+y^2}[/tex]

    Then the circle is [itex]f^{-1}(1)[/itex] and thus closed.
    Now, the closure is easy to find, no?

    For the interior, you have to find out for which points p on the circle, there exists an open ball (=disk) that is contains in the circle. What are those points?
    Last edited: Nov 2, 2012
  4. Nov 1, 2012 #3
    Hi micromass.
    Thanks for your reply.
    I think you mean the function [itex](x,y) → x^{2}+y^{2}-a^{2}[/itex], being a the radius of the circle. I think you said that this is a closed set because is the preimage of the zero point that is a closed set on the reals.
    I don't figure out why you said that now the closure is easy to find, I guess you mean that if the closure of the image is zero, the closure of the preimage is the circle itself.
    I think that this set hasn't interior, because there isn't a open ball contained in the circle, all the disks that intersect the circle has elements that don't belong to the circle.
    I would prefer to consider the circle as a subset of R^2, I think that it isn't needed to introduce the function "f". What do you think?
    Thank you
  5. Nov 1, 2012 #4


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    The maximum distance between two points on a circle or a sphere is the diameter. So it is bounded.

    Your statement that the circle is the complement of an open set is correct an is not just intuitive. Try to find a proof using this idea.

    The function X^2 + y^2 is continuous and the circle is the inverse image of a point, other than zero. E.G. the circle of radius 1 is the inverse image of 1. But a point on the real line is closed and the inverse image of a closed set under a continuous map is also closed. This proof buys you nothing over the direct proof that the circle is the complement of an open set because the proof that a point is closed is the same.

  6. Nov 1, 2012 #5
    Thank you lavinia.
    I think I got it.
  7. Nov 2, 2012 #6


    User Avatar
    Science Advisor

    Edit: there are other approaches/arguments that make the proof for a single point easier than that for S^1.
    Last edited: Nov 2, 2012
  8. Nov 2, 2012 #7
    What exactly was wrong in lavinia's explanation?
  9. Nov 2, 2012 #8


    User Avatar
    Science Advisor

    I'm not saying the explanation is wrong, only that the statement is not accurate IMHO:

    S/he said that the proof of a point being closed is just as hard, or almost the same as

    the proof of S1 being closed. I understood that the proof consisted of showing

    the complement is open (in which case both arguments would be equivalent).

    But if we use the limit point approach, the fact that a point is closed is immediate, as

    any ball about a point is not contained in the point.

    Basically, I object to the statement " .......because the proof that a point is closed is the same" . It is the same if we use the complement-is-open

    approach, but not otherwise.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook