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If you consider all the real numbers, is infinity the supremum? What about the maximum?

I know the supremum does not have to be in the set and the maximum does, but I'm confused as to how to answer these questions. Are the real numbers still technically bounded above even though they go on forever?

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    $\begingroup$ We could say $\sup\Bbb R=\infty$, the same way we say $\lim_{n\to\infty}n=\infty$. But $\infty$ is not the maximum since, as you say, it is not in the space. And the reals are not bounded above, as that would require some number $r\in\Bbb R$ such that $x\le r$ for all $x\in\Bbb R$. $\endgroup$ – Stefan Hamcke Sep 7 '15 at 20:05
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    $\begingroup$ This has become a hot question now. 'is infinity a maximum' is a VERY 'accessible' question, I urge you not to upvote it (don't downvote, just hold back) as it's not well researched, nor a good question. $\endgroup$ – ToolPurger Sep 7 '15 at 20:55
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    $\begingroup$ If you make this more rigorous, it's effectively the way of defining infinity. $\endgroup$ – asmeurer Sep 7 '15 at 21:45
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    $\begingroup$ @ToolPurger well, if "not well researched" and "not a good question" aren't good criteria for a downvote, what are? $\endgroup$ – David Z Sep 8 '15 at 5:38
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Let $T$ be a totally ordered set. The supremum of a subset $S \subseteq T$, if it exists, is an element $s \in T$ such that (1) $s$ is an upper bound of $S$, and (2) for all $x$ which are upper bounds of $S$, $x \ge s$. You can prove that if such an $s$ exists then it is unique using the properties of a total order.

It is certainly not guaranteed that a supremum exists. For instance, $\varnothing$ only has a supremum if $T$ has a minimum element. The set $\{x \;:\; x^2 \le 2\}$ has no supremum in $\mathbb{Q}$. And unbounded sets in $\mathbb{R}$ do not have any supremum.

In particular, supremums do not always exist in the real numbers. Only bounded, nonempty subsets of $\boldsymbol{\mathbb{R}}$ have a least upper bound.

However, it is convenient when speaking of the supremum on real number sets to consider subsets of the extended real numbers, $\overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty,\infty\}$ instead of subsets of the real numbers. If you do this, then every subset of $\boldsymbol{\overline{\mathbb{R}}}$ has a least upper bound, and $\sup$ is a well-defined function from $\mathcal{P}(\overline{\mathbb{R}})$ to $\overline{\mathbb{R}}$. For example:

  • $\sup \varnothing = -\infty$,

  • $\sup \{-\infty\} = -\infty$,

  • $\sup \mathbb{R} = \infty$,

and so on.


A maximum of $S \subseteq T$ is simply a supremum of $S$ such that $s \in S$. Often supremums are not maximums; $[0,1)$ has supremum $1$ but no maximum (in $\mathbb{R}$). $\varnothing$ has supremum $-\infty$ but no maximum (in $\overline{\mathbb{R}}$).

Unlike with supremums, whether or not a set has a maximum does not depend on which set it is contained in. So

  • $\mathbb{R}$ has no supremum as a subset of itself, but does have a supremum in $\overline{\mathbb{R}}$.

  • $\mathbb{R}$ has no maximum in either itself or in $\overline{\mathbb{R}}$.

Any set with a maximum has a supremum, so supremum is a strictly more general notion than maximum.

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    $\begingroup$ If OP is confused as to why $\sup\varnothing=-\infty$: Everything is an upper bound of the empty set, so the least upper bound is just the least element in the space. Why is everything an upper bound of $\varnothing$? Because, for any $c$, there's nothing in $\varnothing$ bigger than $c$. (Indeed, there's nothing in $\varnothing$ at all.) So $c$ is an upper bound. $\endgroup$ – Akiva Weinberger Sep 7 '15 at 20:35
  • $\begingroup$ Tiny suggestion: Write element instead of value because this is precisely what you mean. I've looked around a bit and there might be valid reason to write value here (Wikipedia's definition, Rudin if you consider values of sets as values of the identity function), but I wanted to express my concerns considering the intuition of some people expecting a value to be an element of $\mathbb{R}$. Tiny improvement: To provide a complete answer, you probably should cover the $\max$-part of the question with a short sentence (or two). $\endgroup$ – Piwi Sep 8 '15 at 0:11
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    $\begingroup$ @Piwi thanks for the good suggestions, I updated the answer! $\endgroup$ – 6005 Sep 8 '15 at 2:23
  • $\begingroup$ Seems like in the first part you missed mentioning that a supremum of $S$ must also be greater than or equal to all elements of $S$. Otherwise it's not unique. $\endgroup$ – David Z Sep 8 '15 at 5:56
  • $\begingroup$ @DavidZ Wow. So I did. Thanks, fixed. $\endgroup$ – 6005 Sep 8 '15 at 6:09
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Neither the maximum or supremum of a subset are guaranteed to exist. If you consider the real numbers as a subset of itself, there is no supremum. If you consider it a subset of the extended real numbers, which includes infinity, then infinity is the supremum.

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    $\begingroup$ If you allow extended real values, then the supremum of any set of reals (also, any set of extended reals) is guaranteed to exist. (the supremum of the set of reals is $+\infty$ and the supremum of the empty set is $-\infty$) $\endgroup$ – Hurkyl Sep 7 '15 at 22:25
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The supremum of a set $A$ of real numbers can fail to exist for two reasons: Either there is no upper bound at all, or among those upper bounds there is no least upper bound. In the first case it is customary to write $\sup A=+\infty$, in the second case it is customary to write $\sup A=-\infty$. The appearence of the special symbol $\infty$ does not signify an actual supremum here. Compare with $\lim_{x\to a} f(x)=\infty$, which we write when the limit actually fails to exist - in a special way: If this where really a limit, the statement would claim that for each $\epsilon>0$ there exists $\delta>0$ such that $0<|x-a|<\delta$ implies $|f(x)-\infty|<\epsilon$ - which is nonsense.

Then why do we even introduce these weird notations for different cases of non-existen suprema? The advantage is that these symbolic notations allow us to extend the validity of certain theorems. For example, If $A$, $B$ are two sets and $\sup A$, $\sup B$ exist, then $\sup(A\cup B)$ also exists and $\sup(A\cup B)=\max\{\sup A,\sup B\}$. If we employ the intuitive rules for maximum involving $\pm\infty$, then $\sup(A\cup B)=\max\{\sup A,\sup B\}$ continuous to hold even in cases where one or both of $\sup A$, $\sup B$ fails to exist!

So while $\infty$ is - in the sense above - the supremum of for example the set of all real numbers (as well as the supremum of $\mathbb N$ and any other set that is not bounded from above), it is certainly not the maximum of any set: The maximum of a set $A$ is always either an element of $A$ or it does not exist (and as the special symbol $\infty$ is not an element of $\mathbb R$, it cannot happen that $\max A=\infty$ for some $A\subseteq \mathbb R$); we do not introduce some fancy special symbolic notation for the case of nonexistent maximum (and that even though the maximum as even more possibilities to fail to exist than the supremum)! This difference between maximum and supremum may sound surprising at first, but it has turned out to be useful. (For example, when a fucnstion attains its maximum we want to be able to pin down a point where that happens; if there is no maximum it is important to notice that no such point can be pinned down.

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It depends on whether you take the supremum in $\Bbb R$ or in $\Bbb R\cup\{\infty\}$. In the first case, no, because $\infty$ is not a real number. In second case, yes.

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The most commom case is that Sup is are defined in terms of real numbers, $\mathbb R.$ You can, and people have, append $\infty$ to the real numbers. The cost of this convenience is that you have changed the set that you are working on and now you have to check how every definition and theorem you use has been affected by this.

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