(Baby Rudin, Theorem 3.19) I am trying to prove:

Let $\{s_n \}$ and $\{t_n \}$ be sequences of real numbers. If $s_n \leq t_n$ for $n \geq N$, where $N$ is fixed, then $$\lim_{n\to\infty} \sup s_n \leq \lim_{n\to\infty} \sup t_n.$$

I know this theorem has been proved many times in the past on this website, but it seems like all the proofs that were provided implicitly (and erroneously) assume that both $\lim_{n\to\infty} \sup s_n = s^*$ and $\lim_{n\to\infty} \sup t_n$ are finite. Since this need not necessarily be the case, I thought of asking a new question. Since the finite cases have already been addressed, it remains to deal with the infinite cases:

My attempt at completing the proof: Suppose $t^* = +\infty$. Then, the result clearly follows; so, assume that $t^* < +\infty$. [Then, I prove that this implies that $s^* < +\infty$]. Now, suppose $s^* = -\infty$ and the result clearly follows; so, assume that $s^* > -\infty$. Then, I need to show that $t^* > -\infty$. When this is done, we can assume that both $s^*, t^*$ are finite.

How can I show that $t^* > -\infty$ in the proof above?

Rudin has the following theorems/definitions related to lim-sup and lim-inf: Definition 3.15:

Let $\{s_n \}$ be a sequence of real numbers with the following property: For every real $M$ there is an integer $N$ such that $n \geq N$ implies $s_n \geq M$. We then write $$s_n \rightarrow +\infty.$$ Similarly, if for every real $M$ there is an integer $N$ such that $n \geq N$ implies $s_n \leq M$, we write $$s_n \rightarrow -\infty.$$

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 as defined in Definition 3.5, plus possibly the numbers $+\infty$, $-\infty$.

We now recall Definitions 1.8 and 1.23 and put $$s^* = \sup E,$$ $$s_* = \inf E.$$ The numbers $s^*$, $s_*$ are called the upper and lower limits of $\{s_n \}$; we use the notation $$\lim_{n\to\infty} \sup s_n = s^*, \ \ \ \lim_{n\to\infty} \inf s_n = s_*.$$

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 $x > s^*$, then 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).


2 Answers 2


From your question, it looks like you're only interested in showing that $t^* > -\infty$ when $s^* > -\infty$.

Since $s^* > -\infty$, there exists a subsquence $s_{n_k}$ such that $s_{n_k} \to s$ for some $s > -\infty$.

Now, consider the subsequence $\{t_{n_k}\}$, it must have a (sub-)subsquence that converges to some $t$. However, $t_{n_k} \ge s_{n_k}$ for all sufficiently large $k$ and thus, $t \ge s > -\infty$.

Since $t^*$ is the supremum of all possible subsequential limits, we see that $t^* \ge t > -\infty$.

  • $\begingroup$ Why would $\{t_n_k\}$ have a subsequence that converges to some $t$? $\{t_n_k\}$ need not be bounded, right (not sure why the equations are being boxed automatically in the comment)? $\endgroup$ Commented Jun 30, 2020 at 16:05
  • $\begingroup$ Given any sequence, it has a convergent subsequence in the extended sense. If it's unbounded, you can find one that converges to either $\infty$ or $-\infty$. In this case, $-\infty$ won't be possible due to the lower bound by $s_{n_k}$. $\endgroup$ Commented Jun 30, 2020 at 16:07
  • $\begingroup$ A quick way to see that: Assume wlog that $a_n$ is unbounded above. You can construct a strictly increasing sequence $n_1, n_2, \ldots$ such that $a_{n_1} > 1$, $a_{n_2} > 2$, and so on. Clearly $\lim_k a_{n_k} = \infty$. (The reason you can construct it will follow quite straight-fowardly from the definition of not being bounded above.) $\endgroup$ Commented Jun 30, 2020 at 16:11
  • $\begingroup$ Also, about the boxing, I think it's because you have to put nested subscripts via brackets. Try \{t_{n_k}\} instead to get $\{t_{n_k}\}$. $\endgroup$ Commented Jun 30, 2020 at 16:20
  • 1
    $\begingroup$ Thanks, I really like the simplicity of your proof! $\endgroup$ Commented Jun 30, 2020 at 19:31

You could show that $t_* > -\infty$, or you could simply show that $s_* \le t_*$ by working with the assumption that $t_* < \infty$. This will be just as hard (or as easy) as showing that $t_* > - \infty$ at this point in my opinion. As you already noted, $s_* < \infty$ in this case.

Assume for the sake of contradiction that $t_* < s_*$ and let $s_{n_k}\to s_*$. By assumption, $s_{n_k} \le t_{n_k}$ for all $k\ge K_1$ for some $K_1$.

If $t_* = -\infty$, then for some $K_2$ and all $k\ge K_2$, $t_{n_k} < s_* - 1$. This is impossible because $s_{n_k}\to s_*$.

If $t_* > -\infty$, so $t_*\in\mathbb R$, then $s_* - t_* > c > 0$ for some $c$. So for some $K_3$ and all $k\ge K_3$, $s_{n_k} \le t_{n_k} < t_* + c < s_*$, another contradiction because $s_{n_k}\to s_*$.


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