In Rudin's Principles of Mathematical Analysis, in section 1.8 he gives a definition of the supremum of a set $E$ which is bounded above.

But then he goes in section 1.10 to state the least-upper-bound property giving the same constraints on the set $E$ as in section 1.8, namely that $E \subset S$, $E$ is not empty and $E$ is bounded above, and then stating that the sup $E$ exists in $S$.

Don't we already know from 1.8 that the sup $E$ exists in $S$ if $E \subset S$, $E$ is not empty and $E$ is bounded above? Why state it again?

  • 2
    $\begingroup$ No. The set $\{q\in \mathbb{Q}: q^2<2\}$ doesn't have a least upper bound in $\mathbb{Q}$. That's what Rudin was trying to show with his example on the first page. $\endgroup$ – Qiyu Wen Jun 14 '16 at 17:00
  • $\begingroup$ @QiyuWen, understood $\endgroup$ – M.K. Jun 14 '16 at 17:20

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