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Liminf, Limsup and the Bolzano-Weierstrass Theorem¶

Read Section 2.3. Check your understanding against the following questions:

What is $\limsup$ (resp. $\liminf$) of a sequence?
What is the Bolzano-Weierstrass theorem?
What means by $\lim x_n = \infty$, $\lim x_n = -\infty$?

The Bounded Case¶

Suppose $(x_n)$ is a bounded sequence, i.e. there exists a $B$ such that $|x_n| < B$ for all $n$. Then

$$S_k := \{x_n \colon n \ge k\} \qquad (k=1,2,3...)$$

form a decreasing chain (i.e. $S_{k} \supseteq S_{k+1}$) of nonempty bounded sets.

Therefore, both $a_k:=\sup S_k$ and $b_k:=\inf S_k$ ($k \in \Nn$) exist and

$$b_1 \le b_2 \le \cdots \le b_k \le b_{k+1} \le \cdots \le a_{k+1} \le a_k \cdots \le a_2 \le a_1.$$

The sequence $(a_k)$ is decreasing and bounded below by $b_1$ (in fact by any of the $b_k$'s) so

$$\lim a_k = \inf \{a_k \colon k \in \Nn\} = \inf_k \sup S_k$$

exists. Denote this limit by $\limsup x_n$.

Likewise, $(b_k)$ is an increasing sequence bounded above by $a_1$ (in fact any of the $a_k$'s) so

$$\lim b_k = \sup \{b_k \colon k \in \Nn\} = \sup_k \inf S_k.$$

exists. Denote this limit by $\liminf x_n$.

These limits are called the limit superior and limit inferior of $(x_n)$. Clearly, $\liminf x_n \le \limsup x_n$. Moreover, for each $k$,

$$b_k \le x_k \le a_k.$$

Therefore, if $\liminf x_n = \limsup x_n$, then $(x_n)$ converges to this common value by the squeeze lemma.

Checkpoints: What are the $(a_k)$ and $(b_k)$ are of the following sequences and what are their $\limsup$ and $\liminf$?

$$ ((-1)^n),\quad (1/n),\quad ((-1)^n/n)$$

Proposition 1. (2.3.2, 2.3.5) For any bounded sequence of real numbers $(x_n)$. $\limsup x_n$ and $\liminf x_n$ exist with $\liminf x_n \le \limsup x_n$. Moreover,

$(x_n)$ is convergent if, and only if, $\limsup x_n = \liminf x_n$ and the limit of $x_n$ is the their common value.

Proof. The only thing remains to be proved is that if $(x_n)$ is convergent then $\limsup x_n = \liminf x_n$.

Suppose $\lim x_n = L$. If we can show that there are subsequences $(x_{n_k})$ and $(x_{m_k})$ of $(x_n)$ that converges to $\limsup x_n$ and $\liminf x_n$, respectively, then since any subsequence of a convergent sequence converges to the limit of the sequence, so we must have

$$ \limsup x_n = L = \liminf x_n. $$

Therefore, all we need is to establish the following theorem.

Theorem 2. For any bounded sequence $(x_n)$, there is a subsequence that converges to $\limsup x_n$ (resp $\liminf x_n$).

Proof. We prove the statement about $\limsup x_n$ and leave the statement about $\liminf x_n$ as an exercise.

For a bounded sequence $(x_n)$, by the definition of $\limsup$, we have

$$a_n := \sup\{x_m \colon m \ge n\} \to \limsup x_n.$$

Let $n_1 = 1$ and suppose for some $k \ge 1$, $n_k \in \Nn$ has be chosen. By the definitiion of $a_{n_k + 1}$, there exists some natural number $n_{k+1} \ge n_k + 1 > n_k$ such that

$$ a_{n_k + 1} - \frac{1}{k} < x_{n_{k+1}} \le a_{n_k+1}. \tag{*} $$

Since $n_{k+1} > n_k$, we have constructed, by induction, a strictly increasing sequence of natural number $(n_k)$ such that ($*$) holds for each $k$. As a subsequence of $(a_n)$, $(a_{n_k +1})$ converges to $\limsup x_n$ as well. It follows from ($*$) and the squeeze lemma that $(x_{n_k})$ is a subsequence of $(x_n)$ that converges to $\limsup x_n$.

We have just proved a very fundamental result real in Real Analysis.

Bolzano-Weierstrass Theorem. Every bounded sequence of real numbers has a convergent subsequence.

Exercise. Give a sequence of real numbers that has no convergent subsequence.

Proposition 3. If $(x_{n_k})$ is a subsequence of $(x_n)$, then

$$ \liminf_n x_n \le \liminf_k x_{n_k} \le \limsup_k x_{n_k} \le \limsup_n x_n. $$

Consequently, $\limsup x_n$ (resp. $\liminf x_n$) is the largest (resp. the smallest) limit of any convergent subsequence of $(x_n)$.

Proof. For any $m \in \Nn$, it is clear that

$$ S'_{m} := \{x_{n_k} \colon k \ge m\} \subseteq S_{n_m}: = \{x_{n} \colon n \ge n_m\}. $$

So, $a'_m:=\sup S'_{m} \le a_{n_m} :=\sup S_{n_m}.$ Since $(a_{n_m})$ is a subsequence of $(a_n)$ which converges to $\limsup x_n$, therefore,

$$ \limsup_k x_{n_k} := \lim_m a'_m \le \lim_m a_{n_m} = \lim_n a_n = \limsup x_n. $$

Now if $(x_{n_k})$ converges, then by Proposition 1, $\lim_k x_{n_k} = \limsup_k x_{n_k} = \liminf_k x_{n_k}$.

Hence, $\liminf x_n \le \lim x_{n_k} \le \limsup x_n$.

Facts.

B-W Theorem is logically equivalent to the least-upper-bound property.
B-W Theorem holds for sequences in $\Rr^n$ as well.

The Unbounded Case¶

Read also Section 2.3.4

What if $(x_n)$ is unbounded? What can we say about $\limsup x_n$ and $\liminf x_n$?

First, note if $S_k:=\{x_n \colon n \ge k\}$ is bounded above for some $k$ then the maximum of $\{x_1,x_2,\ldots, x_{k-1},B\}$, where $B$ is an upper bound of $S_k$, will be an upper bound of $\{x_n \colon n \ge 1\}$. Thus, if the sequence $(x_n)$ is not bounded above, then none of the $S_k$ is bounded above. In this case, we simply define $\limsup x_n$ to be $+\infty$. Likewise, if $(x_n)$ is not bounded below, we declare $\liminf x_n$ to be $-\infty$.

Suppose $\limsup x_n = +\infty$, i.e. $(x_n)$ is not bounded above. So there exists some natural number $n_1 (\ge 1)$ such that $x_{n_1} \ge 1$. Suppose for some $k \ge 1$, a strictly increasing sequence of natural numbers

$$n_1 < n_2 < \cdots < n_k$$

has been constructed so that $x_{n_k} \ge k$. Now since $S_{n_k + 1}$ is unbounded, so there exists $n_{k+1} \ge n_k +1 > n_k$ such that $x_{n_{k+1}} \ge k+1$. Thus, by induction we constructed a subsequence $(x_{n_k})$ of $(x_n)$ that diverges to $+\infty$.

Clear, if $(x_n)$ has a subsequence $(x_{n_k})$ that diverges to $+\infty$, the set of terms of $(x_n)$ cannot be bounded above.

So we have just proved.

Proposition 3. The following statements are equivalent:

$(x_n)$ is not bounded above (resp., not bouned below).
There exists a subsequence of $(x_n)$ that diverges to $+\infty$ (resp. diverges to $-\infty$).

Few more examples¶

Discuss the limit inferior and limit supreior of the following sequences $(x_n)$ where

$x_n = (-2)^n$
$x_n = 5(1+n)$
$x_n = (-1/2^n)$
$x_n = (-1)^n\left(1 - \dfrac{1}{n}\right)$.

For (1) $(x_n) = (-2,4,-8,...)$. For each $k$, the set $S_k = \{(-2)^k \colon k \ge n\}$ contains both of positive and negative numbers of arbitrary large absolute values, hence both $\inf S_k = -\infty$ and $\sup S_k = \infty$ for each $k$. Therefore, $\limsup x_n = \infty$ and $\liminf x_n = - \infty$.

Another way of seeing this is that, since there is a subsequence $((-2)^{2k})_k = (4^k)_k$ that diverges to $\infty$, so $\limsup$ of the sequence must be be $\infty$. Likewise since the subsequece $((-2)^{2k-1}) = (-2^{2k-1})$ diverges to $-\infty$, so $\liminf$ of the sequence must be $-\infty$.

For (2), $(x_n) = (10,15,20,...)$. Then $\inf S_k$ and $\sup S_k$ are both $\infty$ for each $k$ and hence so are the liminf and $\limsup$.

For (3), $(x_n) = (-1/2,1/4,-1/8,...)$. Note that $\lim x_n = 0$ hence $\limsup x_n = \liminf x_n = \lim x_n =0$.

For (4), $(x_n) = (0,(1-1/2),-(1-1/3),...)$. This sequence is divergent but bounded, hence both $\liminf$ and $\limsup$ of $(x_n)$ are finite. In this case, $\limsup x_n =1$ and $\liminf x_n = -1$.