# Rudin's Theorem 3.17 Uniqueness of $\limsup s_n$

Let $$\{s_n \}$$ be a sequence of real numbers. Let $$E$$ be the set of all subsequential limits of $$\{s_n \}$$ and $$s^* = \sup E = \limsup\limits_{n\to\infty}s_n$$. 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 $$s_*$$.

Here's Rudin'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 $$s_{n_k} \rightarrow +\infty$$.

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 $$s_n \rightarrow -\infty$$.

This establishes (a) in all cases.

(b) Suppose there is a number $$x > s^*$$ such that $$s_n \geq x$$ for infinitely many values of $$n$$. In that case, there is a number $$y \in E$$ such that $$y \geq x > s^*$$, contradicting the definition of $$s^*$$.

Thus $$s^*$$ satisfies both (a) and (b).

To show the uniqueness, suppose there are two numbers, $$p$$ and $$q$$, which satisfy (a) and (b), and suppose $$p < q$$. Choose $$x$$ such that $$p < x < q$$. Since $$p$$ satisfies (b), we have $$s_n < x$$ for $$n \geq N$$. But then $$q$$ cannot satisfy (a).

In the uniqueness part, I can't see why the last statement implies that $$q$$ cannot satisfy (a). I tried proving by contradition but getting nowhere.

• I see that for $n\ge N$, $|s_n - q| \le |s_n| + |q| < 2|q|$. So we always have $|s_n - q| < 2|q|$ for all but finitely many $s_n$ so $q$ can't be a limit point of $E$ and then no subsequence of $\{s_n\}$ can converge to $q$. Is this right? Commented May 25, 2017 at 14:18
• What happens is that since $n \ge N$ implies $s_n < x$, any subsequential limit of $\{s_n\}$ cannot exceed $x$. But $q$ is such a limit, so $q \in E$ implies $q \le x$. Commented May 25, 2017 at 14:50
• Ok, so explicitly the contradition you get is $x < q \le x$? Commented May 25, 2017 at 15:38
• Correct.$\phantom{}$ Commented May 25, 2017 at 17:23

Suppose there are two numbers, $p$ and $q$, which satisfy (a) and (b), and suppose $p<q$. Choose $x$ such that $p<x<q$. Since $p$ satisfies (b), we have $s_n<x$ for all $n≥N$. This means that for $n≥N$, $|q-s_n|\geq |q-x|$. But then $q$ cannot be a subsequential limit.
Suppose that there are two numbers, $$p$$ and $$q$$, which satisfy $$(a)$$ and $$(b)$$. (Notice that by "numbers", we allow $$p$$, $$q$$ to be $$+\infty$$, $$-\infty$$, or real numbers.) Without loss of generality, let $$p < q$$. Then choose a real number $$x$$ such that $$p (This is always possible). Since $$p$$ satisfies $$(b)$$, we know there exists an integer $$N$$ such that $$n \geq N$$ implies $$s_{n} < x$$. Then $$q$$ cannot be in $$E$$.
To see the last argument, suppose we have $$q \in E$$. Notice that $$\{s_{n}\}$$ is bounded above by $$\max \{s_{1}, s_{2}, \dots, s_{N}, x\}$$, then we cannot have a subsequence of $$\{s_{n}\}$$ that goes to $$+\infty$$. In other words, $$q \ne +\infty$$. It follows that $$q$$ can only possibly be a real number ($$q \ne -\infty$$ since $$q > x$$ by assumption). But we cannot have a subsequence converges to a real $$q$$ also. (Since there is a $$N$$ after which all the $$s_{n}$$'s are less than $$x$$. Suppose there is a subsequence converging to $$q$$, then let $$\delta \in (0, q-x)$$, then there is an $$M$$ such that for all $$n_{k} \geq M$$, $$|s_{n_{k}}-q| < \delta$$, by definition. But that implies $$s_{n_{k}} > x$$ if $$n_{k} \geq M$$, a contradiction.) Thus we cannot have $$q \in E$$, which finishes our proof.