Let assume that if I write set it means an arbitrary subset of $\mathbb{R}$.

  1. Some sets have a maximum and some do not. They may have (or not) supremum instead.

  2. Please verify my understanding of supremum which is as follows. Assume we have a bounded set $A$ (it means there is an interval $(a,b)$ such that $A \subset (a,b)$). To find the supremum of $A$ you need to find the smallest upper boundary of $A$.

My question is based on the first point - why one can assume that there is the smallest element in the set of all upper boundaries of $A$? This would mean that there is a minimum, i.e. the symmetric situation to the maximum but as I spotted - not every set has a maximum/minimum element.

Is this related to the fact that the set of all upper boundaries has the cardinality of continuum and is compact and thus it must have a minimum element in it?

Thank you for your clarification. Tom

  • 1
    $\begingroup$ I'm not sure if this answers your question, but if $X$ is any subset of $\Bbb R$, then provided that $X$ is bounded above and non-empty, $X$ has a supremum. This is known as the "least upper bound" property of $\Bbb R$. This is either taken as an axiom, or it can be proven, depending on how you construct the reals. Any analysis book like Rudin, Spivak, or Tao, will contain a proof. $\endgroup$
    – Joe
    Aug 30, 2021 at 12:31

2 Answers 2


tl; dr: If $A$ is a non-empty set of real numbers and is bounded above, then the set of upper bounds of $A$ is a closed set of real numbers with a lower bound, so has a smallest element.

As I read the question, the issue is the asymmetry between the set $A$ (which need not have a largest element), and its set of upper bounds (which always has a smallest element).

In the hope an example clarifies, suppose $A = (-\infty, 0)$ is the set of negative real numbers. The set $U$ of upper bounds of $A$ is $[0, \infty)$, the set of non-negative real numbers. Note carefully, however, that $A$ is not the set of lower bounds of $U$. The set of lower bounds of $U$ is $(-\infty, 0]$, the set of non-positive reals.

A bit of thought shows more: If $A$ is a set of real numbers and $U$ is the set of upper bounds of $A$, then exactly one of the following holds: (i) $U = \varnothing$; (ii) $U = (-\infty, \infty)$; (iii) There exists a real number $\alpha$ such that $U = [\alpha, \infty)$.

(The proof is left as an instructive exercise using completeness of the reals.)

Analogous claims hold for lower bounds. In particular, there exists no set of real numbers whose set of lower bounds is $A = (-\infty, 0)$.


I think what you’re asking is why every nonempty bounded subset has a least upper bound. Or if we allow the upper bound $\infty$ why every subset has a least upper bound.

This is either taken as an axiom for the real numbers, or follows from the construction of the real numbers from the rationals. This is usually done either by Dedekind cuts, or Cauchy sequences.

  • $\begingroup$ Every non-empty bounded subset. =) $\endgroup$
    – user21820
    Nov 6, 2021 at 14:50
  • $\begingroup$ @user21820 Any real number $M$ is an upper bound for the empty set, because $x\in\emptyset \implies x\leq M$. $\endgroup$ Nov 6, 2021 at 15:03
  • $\begingroup$ Your comment is true. Your statement in the post is false. ∅ has no least upper bound in ℝ. $\endgroup$
    – user21820
    Nov 6, 2021 at 15:07
  • $\begingroup$ @user21820 Okay, just poorly formulated. It again has a least upper bound if we allow $-\infty$ and $\infty$. This least upper bound for $\emptyset$ then exists and is $-\infty$. $\endgroup$ Nov 6, 2021 at 16:14
  • $\begingroup$ That's right. I personally like having the affinely-extended reals for that reason. I just pointed it out because surely you want to correct your post! $\endgroup$
    – user21820
    Nov 6, 2021 at 16:16

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