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Topology (showing set is not open)

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Show [0,1] is not open in ℝ

    2. Relevant equations

    [0,1] is open if and only if ℝ\[0,1] is closed.

    3. The attempt at a solution
    ℝ\[0,1] = (-∞,0) U (1,∞), this set is open. Despite the if and only if statement this is enough to say that [0,1] is not open in ℝ.

    Is this correct?
     
    Last edited: Aug 28, 2012
  2. jcsd
  3. Aug 28, 2012 #2
    That is not enough, unless you say that the only clopen subsets in ℝ are ℝ and ∅, in which case that plus what you said above should be fine (if you can't say that, then the fact that a set has an open complement doesn't necessarily mean that it is not open, only that it is closed).

    So, you can show that any basis element of ℝ in the standard topology containing 0 must contain an element outside of [0,1], which means that [0,1] is not open.
     
  4. Aug 28, 2012 #3
    we haven't gotten to basis elements, we JUST started talking about a topology.

    Could I assume that [0,1] is open and then show that because ℝ\[0,1] is not closed (it is a union of open sets) then it must be true that [0,1] is not open in the first place.
     
  5. Aug 28, 2012 #4
    "Show [0,1] is not open in ℝ"


    Polamaluisraw, remember, for a set U to be open it must be a neighborhood of each of its points. Can you see any points in [0,1] for which [0,1] cannot be a neighborhood? And, if so, why?
     
  6. Aug 28, 2012 #5
    You seem to be assuming that "not closed" is the same thing as "open". Here's a counterexample: [itex](0,1][/itex]. This set is not closed because its complement is not open. Indeed, [itex]\mathbb{R}-(0,1]=(-\infty,0]\cup (1,\infty)[/itex] is not open, since the left interval is not a union of open balls. However, the interval [itex](0,1][/itex] it is not open, because again, it is not a union of open balls either.

    Try assuming it is open. Can you prove that if an interval of the real line is open, then each of its points is the center of an open ball which is contained in the interval? If you can (or if you have already done this), then you're done, because it's not possible to put an open ball at 0, and have the entire thing sit inside [0,1].

    Edit: jmjlt88 has phrased my suggestion a bit more elegantly in terms of neighborhoods. I like it.
     
  7. Aug 28, 2012 #6
    First off thank you for the replies.

    I found this and wanted to show you all to see if this might be the problem..
    http://science.kennesaw.edu/~plaval/math4381/openclosed.pdf [Broken]

    at the bottom it has "techniques to remember"

    the way they are listed I though it was sufficient enough to simply show that the complement
    was not closed.
     
    Last edited by a moderator: May 6, 2017
  8. Aug 28, 2012 #7
    yes, either of the end point since they are included in the set.

    if you were to place an n-ball centered at either of those points it would leave the set?
     
  9. Aug 28, 2012 #8
    By definition of an open set we can show [0,1] is not open.

    since 1 is contained in [0,1] then for [0,1] to be open there must exist an ε>0 such that Bε(1) is contained in [0,1] for any ε>0. it is clear that Bε(1) is not contained in [0,1] because for any ε>0 1+ε is not contained in the set. since the definition is not satisfied [0,1] is not open.
     
  10. Aug 28, 2012 #9

    Dick

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    Really close. But technically not quite right. It's not 'for any ε>0'. You said it right the first time, 'it's for some ε>0'. But if Bε(1) is an open ball, 1+ε isn't in the open ball either. Can you think of a better point in the open ball?
     
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