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?

  • 1
    $\begingroup$ Of course because this is the only number with this properties. $\endgroup$
    – YCB
    Commented Dec 26, 2017 at 6:24

1 Answer 1


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.


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