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Does .99999… = 1?

Example 1.6 on page no 9 in "Elementary Real Analysis" by Bruckner say that " the interval (0,1) has no maximum and minimum as 0 and 1 do not belong to the set." My doubt is if it is an interval, all the numbers between 0 and 1 are included in the set so, .9999..... is maximum, definitely I am missing some point, but don't know where am I going wrong. Pls explain.

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    what you are missing is that 0.99999999..... = 1 – Andrea Mori Mar 28 '11 at 19:24
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    $.9999..... = 1$, and by definition, $1$ is not a member of $(0,1)$. http://math.stackexchange.com/questions/11/does-99999-1/49#49 – Jonas Meyer Mar 28 '11 at 19:25
  • Thanx Andra, Thanx Jonas, –  Mar 29 '11 at 03:38
  • @Andrea Mori, @Jonas Meyer, Thanx Andra, Thanx Jonas, I am still confused, I will put my question this way, if any set is non-empty (which means it has something) then greatest of that something can be treated as "maximum" of that set, correct? So in above example of interval (0,1), if 1 is not included in the set, but definitely something less than 1 is included (as the set is nonempty), then that number which is less than 1 and not equal to 1 is "maximum" for that interval/set, correct? –  Mar 29 '11 at 03:46
  • the problem is that given any $x<1$ you can always find a $y$ such that $x<y<1$. Therefore the interval $(0,1)$ has no maximum (by definition, the maximum of a set is an element of the set which is no less than any other element of the set) – Andrea Mori Mar 29 '11 at 10:26
  • @Andrea Mori, Thanx Andrea, your answer cleared this doubt, I will contact you for more in future, take care. –  Mar 29 '11 at 12:28
  • Andrea, why don't you post your comment as an answer? This way, your answer can be accepted. – JavaMan Mar 29 '11 at 15:27

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