As I know the definition of supremum for set $ S$ is the lowest number that is greater or equal than all the members of $S$. This means : $\forall \: m\in R \:\:m<\sup S,\exists\: s\in S \rightarrow s> x$. Based on this definition we have supremum for all finite subset of $\mathbb N$. For example the $\sup S$ for $S = \{1,2,\dots, k\}$ is $k$.

Is this right?

  • 1
    $\begingroup$ The supremum of any finite set is it maximum $\endgroup$ Nov 2, 2020 at 12:58
  • $\begingroup$ @MauroALLEGRANZA $\mathbb R$ is also infinite, why it has $\sup$? $\endgroup$ Nov 2, 2020 at 13:00
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    $\begingroup$ $\mathbb R$ has no sup. You can prove that by contradiction easily. $\endgroup$
    – player3236
    Nov 2, 2020 at 13:00
  • $\begingroup$ Then what is the definition of a complete set? as I know field F is complere if every upper bounded subset of F has $\sup$ , this condition is correct for natural numbers too but they're not complete. $\endgroup$ Nov 2, 2020 at 13:06
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    $\begingroup$ The set of natural numbers is not a field ! $\endgroup$
    – Fred
    Nov 2, 2020 at 13:11

1 Answer 1


Let's review several senses in which $\Bbb R$ is complete:

  • Each set with an upper bound has a least upper bound This works in $\Bbb N$ because $\Bbb N\subseteq\Bbb R$, so if $S\subseteq\Bbb N$ then $S\subseteq\Bbb R$. Narrowly interpreted, your question boils down to overlooking the bold part; broadly interpreted viz. the comments, other respects in which $\Bbb R$ is complete are worth comparing with $\Bbb N$.
  • Each Dedekind cut is generated by a real number Of course, they're not in general generated by natural numbers. (You could, I suppose, think of naturals as something similar to Dedekind cuts on themselves, but you'd gain nothing from it.)
  • Each Cauchy sequence converges to a value in $\Bbb R$ Any Cauchy sequence in $\Bbb N$ is eventually constant, converging to some natural number.
  • Intersection of interval nested, in the way the nested intervals theorem specifies, is a $1$-element set Since naturals differ by at least $1$ (crucial also in the below discussion of IVT), you eventually can't nest proper-subset intervals further.
  • Monotone convergence In analogy with the least-upper-bound point above, this follows because every sequence in $\Bbb N$ is also in $\Bbb R$. Bolzano-Weierstrass Ditto.
  • Intermediate value theorem This fails for $\Bbb N$ because if $f(a)<0<f(b)$ then there might not be values between $a,\,b$.

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