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Easy Containment question

  1. May 27, 2012 #1
    Prove that Sint[itex]\subseteq[/itex] S

    Ok I thought this was going to be easy, but apparently i'm having issues.

    So I said:

    for all ##x## let x[itex]\in[/itex]Sint

    ==> there exists a B(r, x) s.t with any point a in Sint |X-a| < r.

    But now S is the B(r,x) s.t |x-a| = r.....

    This is where I'm stuck. I know I have to show that the properties of Sint also apply to S. Help
     
  2. jcsd
  3. May 27, 2012 #2
    you should have said,

    If x is in the interior of S, there exists a ball B(x,r) contained in S. Thus, for any a in the entire ambient space, if |a-x|<r, then a is in B(x,r) and thus in S.

    The ball is in S, but why do you think S is in the ball?

    I think the key point is that x itself is in the ball! Thus x is in... where?! (... as desired.)
     
  4. May 27, 2012 #3

    Dick

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    I have no idea what you are talking about. But I am pretty sure you don't know what S^int means. This IS be easy. Review the definition of S^int, ok?
     
  5. May 27, 2012 #4
    That's exactly it, I do know what Sint means: it's the B(r,x) [itex]\subseteq[/itex]S. i.e: The ball of radius "r" about the point "x". contained in S, but then I didn't "prove" anything I just stated the definition.

    Is there some characteristic about S that I should be using to join them? Or is it just that simple as the definition?
     
  6. May 27, 2012 #5

    Dick

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    It's super easy. x is in B(r,x). If B(r,x) is contained in S then x is an element of S.
     
  7. May 27, 2012 #6
    maaaaaaannnn. Is a question like that just made to mess with my head because it is just TOO easy, so you start thinking in circles?
     
  8. May 27, 2012 #7

    Dick

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    If you overthink something that is easy, then you can start thinking in circles. But some things are just easy.
     
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