# What does "$E$ is not bounded above" mean? I am confused. "Principles of Mathematical Analysis" by Walter Rudin Theorem 3.17.

I am reading Walter Rudin's "Principles of Mathematical Analysis".

There are the following definition and theorem and its proof in this book.

Definition 3.16:

Let $$\{ s_n \}$$ be a sequence of real numbers. Let $$E$$ be the set of numbers $$x$$ (in the extended real number system) such that $$s_{n_k} \rightarrow x$$ for some subsequence $$\{s_{n_k}\}$$. This set $$E$$ contains all subsequential limits, plus possibly the numbers $$+\infty$$, $$-\infty$$.

Put $$s^* = \sup E,$$ $$s_* = \inf E.$$

Theorem 3.17:

Let $$\{s_n \}$$ be a sequence of real numbers. Let $$E$$ and $$s^*$$ have the same meaning as in Definition 3.16. Then $$s^*$$ has the following two properties:

(a) $$s^* \in E$$.

(b) If $$x> s^*$$, there is an integer $$N$$ such that $$n \geq N$$ implies $$s_n < x$$.

Moreover, $$s^*$$ is the only number with the properties (a) and (b).

Of course, an analogous result is true for $$s_*$$.

Proof:

(a)
if $$s^* = +\infty$$, then $$E$$ is not bounded above; hence $$\{s_n\}$$ is not bounded above, and there is a subsequence $$\{s_{n_k}\}$$ such that $$s_{n_k} \to +\infty$$.

If $$s^*$$ is real, then $$E$$ is bounded above, and at least one subsequential limit exists, so that (a) follows from Theorems 3.7 and 2.28.

If $$s^* = -\infty$$, then $$E$$ contains only one element, namely $$-\infty$$, and there is no subsequential limit. Hence, for any real $$M$$, $$s_n > M$$ for at most a finite number of values of $$n$$, so that $$s_n \to -\infty$$.

This establishes (a) in all cases.

I cannot understand the following argument:

(a)
if $$s^* = +\infty$$, then $$E$$ is not bounded above; hence $$\{s_n\}$$ is not bounded above, and there is a subsequence $$\{s_{n_k}\}$$ such that $$s_{n_k} \to +\infty$$.

What does "$$E$$ is not bounded above" mean?
p.12, Rudin wrote "It is then clear that $$+\infty$$ is an upper bound of every subset of the extended real number system".
And $$E$$ is a subset of the extended real number system.

Does this mean "$$E \cap \mathbb{R}$$ is not bounded in $$\mathbb{R}$$"?

Then, for example,
Let $$\{s_n\}$$ be a sequence such that $$s_n = n$$.
Then $$E = \{+\infty\}$$ and $$s^* = +\infty$$.
And $$E \cap \mathbb{R} = \emptyset$$.
And $$\emptyset$$ is bounded above.

I am very confused.

I wanna change the above proof to the following proof:

(a)
if $$s^* = +\infty$$, then $$\{s_n\}$$ is not bounded above, and there is a subsequence $$\{s_{n_k}\}$$ such that $$s_{n_k} \to +\infty$$.

• Bounded above usually just means $\leq x$ for a finite $x \in \mathbb{R}$. Commented Jan 20, 2019 at 2:56
• @twnly: that would be correct in the usual reals, but in the extended reals $+\infty$ is an upper bound for everything as OP cites from p.12 Commented Jan 20, 2019 at 2:59
• I agree with your change to the proof. I worry about criticizing a classic that has gone through many editions, but I think it is clear you understand what is going on. Commented Jan 20, 2019 at 3:20
• Thank you very much, Ross Millikan. I am greatly relieved at your comment. Commented Jan 20, 2019 at 3:25

Short answer: I think you are slightly unclear on the definition of an upper limit. Basically, though, if $$\sup(E) = +\infty$$, then $$E \cap\mathbb{R}$$ is not bounded above (in $$\mathbb{R}$$). We use this to show that $$\{s_n\}$$ is not bounded above (in $$\mathbb{R}$$, recall $$\{s_n\}$$ is a sequence in $$\mathbb{R}$$, not $$\mathbb{R} \cup\{\pm\infty\}$$). We use the fact that $$\{s_n\}$$ is not bounded above, to show that there exists a subsequence which diverges to $$+\infty$$. So, $$+\infty\in\mathbb{R}.$$

Let $$(a_n)$$ be a sequence in $$\mathbb{R}.$$ Define $$A$$ to be the subset $$A$$ of $$\mathbb{R} \cup\{\pm\infty\}$$ consisting of all subsequential limits of $$(a_n)$$ and, also, $$\pm\infty\in A$$ if there exists a subsequence of $$(a_n)$$ which diverges to $$\pm\infty$$, respectively.
$$a^*$$ is merely a shorthand $$\limsup_{n\to\infty} a_n$$. The definition of $$\limsup_{n\to\infty} a_n$$ is $$\sup(A)$$ where $$A$$ is defined as above. (This is what definition 3.16 means).
Therefore, $$\sup(A) = +\infty$$ means, by definition of supremum, that if $$x\in\mathbb{R} \cup\{\pm\infty\}\ni x<+\infty$$ then $$x$$ is not an upper bound of $$A$$. In other words, no real number is an upper bound of $$A$$. In other words, as subset of $$\mathbb{R}$$, $$A$$ is not bounded above, by the definition of ''bounded above'' (see definition 1.7). Moreover, $$A\cap\mathbb{R}\neq\varnothing,$$ since $$\sup(\varnothing) = -\infty$$ (see here). Therefore, we avoid vacuous statements.
In other words, since $$A$$ is not bounded above in $$\mathbb{R},$$ $$\forall M\in\mathbb{R},\hspace{1mm}\exists\hspace{1mm} a\in A \cap\mathbb{R}\ni a> M.$$ In other words, viewing the definition of $$A$$, for all $$M\in\mathbb{R}$$ there exists a real number $$a>M$$ for which there exists a subsequence of $$(a_n)$$ which converges to $$a$$.
Therefore, fix an arbitrary $$M \in\mathbb{R}$$ and consider the real (i.e., $$a \neq +\infty$$) $$a>M$$ for which there exists a sub-sequence $$(a_{n_k})$$ of $$(a_n)$$ such that $$a_{n_k} \to a$$. Consider, also, this sub-sequence $$(a_{n_k})$$. Using the definition of convergence of a sequence, we have that for the positive quantity $$a -M$$, there exists an $$N\in\mathbb{Z}\ni k \geq N \implies |a - a_{n_k}| . Consider any integer $$k \geq N$$. So, now we have fixed a $$k\in\mathbb{Z}$$ such that $$|a - a_{n_k}| . It folows (since $$a - a_{n_k} \leq |a - a_{n_k}| < a - M$$) that $$a_{n_k} > M$$. Now, forget that we fixed $$M$$. This shows that $$\forall M \in\mathbb{R}$$, some term of $$(a_n)$$ is greater than $$M$$. Hence, $$(a_n)$$ is not bounded above. With this in hand, it is relatively easy to construct a sub-sequence of $$(a_n)$$ which diverges to $$+\infty.$$ This, I leave to you.