Is it possible to find the supremum of $\{+\infty\}$?

Question

EDIT:

I wonder how the supremum of $\{+\infty\}$ is determined. In fact, Baby Rudin suggests that $s^{*}\in E$ even if $s^{*}=+\infty$ (Theorem 3.17):

If $s^{*}=+\infty$, then $E$ is not bounded above; hence $\{s_{n}\}$ is not bounded above, and there is a subsequences $\{s_{n_{k}}\}$ such that $s_{n_{k}}\to+\infty$.

Does it suggest that $+\infty\le+\infty$ (hence $+\infty$ qualifies as a upper bound)? If so, what's the reason that a set with its supremum is not bounded above, which totally makes no sense.

Answer 1

It may be that you have gotten to used to thinking of $\infty$ as merely a concept: you have taken to heart that $\infty$ is not a number, but rather a stand-in for some limit of some sort. Therefore, let me stress that in this context $\infty$ is a number, specifically an "extended real number", and all the usual rules of an ordering on numbers apply.

Formally, the extended real numbers are given by the set $$ \overline{\mathbb{R}} = \mathbb{R} \cup \{-\infty, \infty\}. $$ While addition and multiplication are only partially defined on this set, the ordering $\le$ is fully defined:

• $-\infty \le x$ for all $x \in \overline{\mathbb{R}}$;

• $x \le \infty$ for all $x \in \overline{\mathbb{R}}$; and

• If $x, y \in \mathbb{R}$, with $x$ less than or equal to $y$ under the usual ordering on $\mathbb{R}$, then $x \le y$. (Therefore we do not need two different symbols for $\le$ of the reals and $\le$ of the extended reals.)

This makes $\le$ into a total order on $\overline{\mathbb{R}}$. In particular, $\infty$ and $\infty$ are the same extended real number, so $\infty \le \infty$.

The useful thing is that any set of real numbers (or even any set of extended real numbers) has an infimum or supremum in $\overline{\mathbb{R}}$. In particular, $\infty$ is a perfectly valid least upper bound for a set.

Now, let me address the lingering problem: what does bounded above mean? In analysis, "bounded above" will always be short for "bounded above by a real number" and not "bounded above by an extended real number". The latter notion would not be so useful, because every set of real numbers is bounded above by $\infty$. So we say that $S$ bounded above if there exists $r \in \mathbb{R}$ such that $s \le r$ for all $s \in S$.

Addendum:

In your excerpt from Rudin, $E$ is the set of limit points of a sequence $s_n$, and $s^*$ is the supremum of $E$. Let me translate / explain the passage:

If $s^{*}=+\infty$, then $E$ is not bounded above;

This is true for the following reason: the least upper bound of $E$ is $\infty$, which in particular means that any finite extended real number $r$ is not an upper bound of $E$. So there is no $r \in \mathbb{R}$, hence $E$ is not "bounded above" by the above definition.

hence $\{s_{n}\}$ is not bounded above, and there is a subsequences $\{s_{n_{k}}\}$ such that $s_{n_{k}}\to+\infty$.

If $s_n$ were bounded above by $r \in \mathbb{R}$, then all limits would be $\le r$, so this cannot be the case. Thus $s_n$ is not "bounded above" (again by the above definition), and there must be a subsequence which goes to $\infty$.