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How to prove an interval is open?

  1. Dec 12, 2010 #1
    1. The problem statement, all variables and given/known data
    I can repeat the proof, but then I won't understand it.
    How do I prove that an open interval like (1,2) is open?

    2. Relevant equations
    I know it has something to do with neighbourhood with a given epsilon, but I dont really get what a neighbourhood is, I have a maybe rough picture of it but not sufficient enough
    it'd be nice if you could explain in some sort of analogy or plain english :)

    3. The attempt at a solution
    let x be in (1,2)
    |x-1|>=e |x-2|=>e
    then neighbouthood will be inside (1,2)
    and I just get lost from there
  2. jcsd
  3. Dec 12, 2010 #2
    Here's some plain English:

    A set is open if for any point inside the set, you can always find more points between here and the edge. So (1,2) is open because if you are at 1.99, you can find 1.999, and so on. But [1,2] is closed because if you are at 2, then there are no more points between here and the edge.
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