From Calculus by Tom Apostol:

DEFINITION OF LEAST UPPER BOUND. A number $B$ is called a least Upper bound of a nonempty set $S$ if $B$ has the following two properties: (a) $B$ is an Upper bound for $S$. (b) No number less than $B$ is an Upper bound for $S$

THEOREM 1.32. Let $h$ be a given positive number and let $S$ be a set of real numbers. If $S$ has a supremum, then for some $x$ in $S$ we have $x\gt\ \sup S\space -h$

Let be the set $A=\{1\}$ $A$ is bounded above and is nonempty, therefore it has least upper bound. The least upper bound is $1$.

Now, let $h = 0,1$ $SupA - h = 0,9$ but $0,9 \notin A$ contradicting the theorem.

In the book, there is no asumption that the set should contain infinite elements or to be dense.

Can you clarify this confusion? Thanks.

  • $\begingroup$ $1 > 0.9$, and clearly $1\in A$. Nowhere is said that $\sup S-h\in S$. $\endgroup$ – celtschk Aug 2 '16 at 16:11
  • $\begingroup$ Thanks. I got it all wrong here. I would like to delete the post. $\endgroup$ – Carlitos_30 Aug 2 '16 at 16:21

If $h=0.1$, then $\sup A-h=1-0.1=0.9$. Now the theorem guarantees an $x\in A$ such that $x>\sup A-h=0.9$. And here, $x=1\in A$ fits the bill. Your mistake was assuming that the theorem guarantees an element in $A$ to be equal to $\sup A-h$.

  • $\begingroup$ Carlitos_30 means $h=0.9$, using the notation used in Germany (and probably also some other countries) where the decimal separator is a comma, not a point. $\endgroup$ – celtschk Aug 2 '16 at 16:15
  • $\begingroup$ Yes, I understand the theorem now. Thanks. $\endgroup$ – Carlitos_30 Aug 2 '16 at 16:22
  • $\begingroup$ Thanks. I updated my answer. $\endgroup$ – ervx Aug 2 '16 at 16:22

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