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Analysis question regarding the relationship between open and closed intervals

  1. Dec 12, 2012 #1
    Let a and b be real numbers with a<b, and let x be a real number. Suppose that for each ε>0, the number x belongs to the open interval (a-ε, b+ε). Prove that x belongs to the interval [a, b].
     
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  3. Dec 12, 2012 #2

    SammyS

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    Use double $ signs for LaTeX . Double # signs for inline LaTeX .

    What have you tried?

    Where are you stuck?
     
  4. Dec 12, 2012 #3
    I tried showing that if y was in the compliment of the set then the closed interval would follow. For example, y≤a-ε and y+ε≤a. Since ε>0, we know that y<a. Thus x≥a. I used a similar proof for the upper bound. Is there a better way to do this?
     
  5. Dec 12, 2012 #4

    pasmith

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    I think your argument works, but the one which certainly works is this:

    If [itex]x < a[/itex], then there exists some [itex]\epsilon > 0[/itex] such that [itex]x \notin (a - \epsilon, b + \epsilon)[/itex]. For example, if [itex]\epsilon = (a - x)/2[/itex] then [itex]a - \epsilon = (a + x)/2 > x[/itex].

    Similarly, if [itex]x > b[/itex] then [itex]x \notin (a -\epsilon, b + \epsilon)[/itex] when [itex]\epsilon = (x - b)/2 > 0[/itex].

    Hence if [itex]x \in (a -\epsilon, b + \epsilon)[/itex] for all [itex]\epsilon > 0[/itex] then [itex]x \geq a[/itex] and [itex]x \leq b[/itex]. Thus [itex]x \in [a,b][/itex].
     
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