0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $1 > 0.9$, 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
3
$\begingroup$

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 $\sup A-h$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.