# 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 $x \in \mathbb{R}$. – twnly Jan 20 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 – Ross Millikan Jan 20 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. – Ross Millikan Jan 20 at 3:20
• Thank you very much, Ross Millikan. I am greatly relieved at your comment. – tchappy ha Jan 20 at 3:25