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Circle is a set of a discontinuities?

  1. Apr 27, 2014 #1
    Why is the characteristic function* of a ball in Rn continuous everywhere except on its surface?My lecturer said that a circle is a 'set of discontinuities' - what exactly does that mean?

    (some context: we're looking at how we can integrate over a ball. Previously we've only looked at Riemann integration over rectangles - for that, he also introduced the theory by first defining integrals for the characteristic function over rectangles).

    *The characteristic function of any set in Rn was defined to be 1 if x was a point inside the set and 0 if t was outside
     
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  3. Apr 28, 2014 #2

    HallsofIvy

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    The characteristic function of set A is defined as "f(x)= 1 if x is in A, 0 other wise". It should be obvious that if we take x to be a point on the surface of a ball, then the limit of the characteristic function is 1 as we approach x along a line inside the ball and 0 as we approach along a line outside the ball.
     
  4. May 1, 2014 #3
    I'm not sure I understand - what about a rectangle? Why is the characteristic function over that not discontinuous?
     
  5. May 1, 2014 #4

    pasmith

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    Exactly what it says: it's a set of points at which a function fails to be continuous.

    It is discontinuous. But consider the following:

    The function [itex]f: \mathbb{R^2} \to \mathbb{R}[/itex] defined by [tex]
    f(x,y) = \begin{cases} 1 & \mbox{if $x \in [a,b]$ and $y \in [c,d]$} \\
    0 & \mbox{otherwise} \end{cases}
    [/tex] is discontinuous at (for example) [itex](b,d)[/itex], because [itex]f(b + \epsilon, d +\delta) = 0[/itex] for every [itex]\epsilon > 0[/itex] and [itex]\delta > 0[/itex].

    The function [itex]g : [a,b] \times [c,d] \to \mathbb{R}[/itex] defined by [tex]
    g(x,y) = 1
    [/tex] is continuous because it is constant.

    If we integrate [itex]f[/itex] over [itex][a,b]\times[c,d][/itex] then we don't care what [itex]f[/itex] does outside that rectangle, so we may as well be dealing with the continuous function [itex]g[/itex].
     
  6. May 1, 2014 #5
    So then the rectangle is an infinite set of discontinuous points as well, but we can integrate over it because we just use g(x,y) instead?

    Why can't we do the same thing with a circle by changing to polar coordinates so that the domain can be defined like the rectangle in Cartesian? (I think we can't do it in Cartesian because the domain is 'harder' to define in that system)
     
  7. May 10, 2014 #6

    HallsofIvy

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    Perhaps the problem is that you aren't clear on what these words you are using mean! A "sphere" is the outer surface of a "ball", not the ball itself. The "characteristic function" of a sphere is 1 on the surface of a ball and 0 otherwise. If you approach a point on the surface of a ball along, say the center of the ball, the characteristic function is always 0. So the limit is 0. But the value of the characteristic function is 1. That is why the sphere is a "set of continuities" for the characteristic function on it. The characteristic function of a sphere is discontinuous exactly on that sphere, continuous everywhere of the sphere.

    Similarly, a "rectangle" in two dimensions is the boundary of a "box". It consists of the four lines forming the boundary, not the interior. The "characteristic function" of the rectangle is 1 on those four lines, 0 off. If you approach any point on the rectangle along a line from an exterior or interior point, the characteristic function is 0 at every point so the limit is 0 but the characteristic function has value 1. So the characteristic function is discontinuous precisely at every point on the rectangle.
     
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