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Open balls are open sets

  1. Apr 20, 2012 #1
    Could someone please show that an open ball is open where the definition of "open" is: A set is open if for each x in U there is an open rectangle A such that x in A is contained in U. Where an open rectangle is (a_1,b_1)×…×(a_n,b_n). I also realize that one can use rectangles or balls, but I would like to see the proof using rectangles, as this is the definition used in Spivak's calculus on manifolds. Please avoid any reference to putting an open ball in this open ball, that will only "push back" the proof. If someone could give a detailed proof that would be much appreciated. The rectangle need not be a hypercube.
  2. jcsd
  3. Apr 20, 2012 #2


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    Try to do it first in 2 dimensions. Then do it in 3 dimensions. Then you should see how to handle the general case. In all the cases, you only really need pythagoras's thm.
  4. Apr 20, 2012 #3

    Sorry, you aren't telling me anything that I haven't already tried.

    Well I suppose in R^1, in our "open ball" we could pick any point x; then compute the distance to the boundary (let that equal eps), and we can fit an "open rectangle" centered at that point (x-eps/2, x+eps/2) and that would show that the set is open. The reason that I can't figure this out in R^2 is that if we were to repeat this algorithm and make the rectangle go half the distance in the direction to the nearest point on the boundary of the open ball, how far should we go in the other direction (perpendicular to that direction?)... because the limit to the size of the rectangle in that direction would be contingent on the curvature right? If the size of the open ball in R^2 was "large" then we could go far in the other direction, but if its a small circle then we hit the boundary quicker.
    Last edited: Apr 20, 2012
  5. Apr 20, 2012 #4
    OK, how does the proof go in the case n=2 then?
  6. Apr 20, 2012 #5
    What I meant was that I tried, but don't know how the proof goes in R^2.
  7. Apr 21, 2012 #6


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    Draw a circle. Then draw a square inscribed into it with its 4 corners touching the circle. Then ask: knowing the circle has radius R, how can I apply pythagoras to find the lenght of the square?
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