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Closed set?

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Is [0, infinity) a closed set?


    2. Relevant equations
    N/A


    3. The attempt at a solution
    It's easy to say that its not. But the solution in my textbook suggests otherwise. Why is this so?

    Thanks!
    M
     
  2. jcsd
  3. Feb 12, 2010 #2

    LCKurtz

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    If it isn't closed, you should be able to find a limit point that isn't in the interval, right? What point would that be?
     
  4. Feb 12, 2010 #3

    vela

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    Is its complement open?
     
  5. Feb 12, 2010 #4
    LCKurtz:

    Ok, I think I understand it for integers. Say, [1,5), where the limit point 5 is not in the interval. But what confuses me is infinity. How is it precisely defined with respect to actual numbers like integers?

    vela:

    What is its complement? R\[0, infinity)?

    Hmm....
     
  6. Feb 12, 2010 #5

    vela

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    Yes, that's the complement.

    Just curious, what's the definition of a closed set that your class uses?
     
  7. Feb 12, 2010 #6
    If its complement is open, then I know that the interval is closed. But why? why is its complement open?

    I appreciate you guys for not giving me the answers directly. I'm really trying hard to understand it.

    M
     
  8. Feb 12, 2010 #7

    vela

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    What's the definition of an open set?
     
  9. Feb 12, 2010 #8
    A set that contains only its interior points....

    Its boundary is contained in its complement...
     
  10. Feb 12, 2010 #9

    vela

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    The complement of [itex][0,\infty)[/itex] is [itex](-\infty,0)[/itex]. Let [itex]x\in(-\infty,0)[/itex]. Is there an open interval centered on x that's contained in [itex](-\infty,0)[/itex]?
     
  11. Feb 12, 2010 #10
    i think that it needs to be open on both sides.
    if it were (0, infinity) it would be open. but since its its [0, infinity), it is a closed interval since you could make x = 0 and have a point, y, that lied to the left of 0, yet still within the Br(0) (ball of radius r about 0).
     
  12. Feb 12, 2010 #11

    LCKurtz

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    Infinity is not part of the real numbers. [0,oo) is just a convenient alternative notation for [itex]x \ge 0[/itex]. You might think of the symbol as indicating the values of x are not bounded above.
     
  13. Feb 12, 2010 #12

    Hurkyl

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    Just FYI -- "closed" only makes sense relative to a space. The right question is
    Is [0, +infinity) a closed subset of the reals​


    It seems to be the habit to introduce it as a sort of "useful fiction". The set of nonnegative reals is quite interval-like, and it's useful to introduce a formal interval-like notation to write such things. Thus "[0, +infinity)".


    However, if you go on, you should learn about the extended real numbers which are, IMO, a much better way to go about doing things. This number system has two additional numbers that the real numbers don't: -infinity and +infinity. And it turns out that every "useful fiction" that you learn in the introductory classes turns out to port to ordinary things in this more sophisticated approach, despite having the exact same notation. e.g. In the extended real numbers, [0,+infinity) is a perfectly ordinary interval, which consists of the nonnegative real numbers. (And the interval [0,+infinity] would consist of all nonnegative extended real numbers)
     
  14. Feb 13, 2010 #13
    It's like putting too much air in a balloon!
     
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