# Understanding Baby Rudin theorem 3.17

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 $$s_*$$.

Question : Can theorem 3.17 be taken as an equivalent definition of 3.16? i.e., if I have a sequence $$(s_n)$$ and suppose I find a number $$z$$ satisfying the properties $$(a)$$ and $$(b),$$ does that mean I have found the limit superior?

• Of course because this is the only number with this properties.
– YCB
Commented Dec 26, 2017 at 6:24

Yes. Since $s^*$ is the only number with properties (a) and (b), any number with those properties is equal to $s^*$, so you could instead define $s^*$ as the unique number with those properties.