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# 2.7: The Infinities. Upper and Lower Limits of Sequences

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## The Infinities

As we have seen, a set $$A \neq \emptyset$$ in E^{1} has a lub (\mathrm{glb}) if A is bounded above (respectively, below), but not otherwise.

In order to avoid this inconvenient restriction, we now add to $$E^{1}$$ two new objects of arbitrary nature, and call them "minus infinity" $$(-\infty)$$ and "plus infinity" $$(+\infty)$$, with the convention that $$-\infty<+\infty$$ and $$-\infty<x<+\infty$$ for all $$x \in E^{1}$$.
It is readily seen that with this convention, the laws of transitivity and trichotomy (Axioms 7 and 8) remain valid.

The set consisting of all reals and the two infinities is called the extended real number system. We denote it by $$E^{*}$$ and call its elements extended real numbers. The ordinary reals are also called finite numbers, while $$\pm \infty$$ are the only two infinite elements of $$E^{*}$$. (Caution: They are not real numbers.)

At this stage we do not define any operations involving $$\pm \infty$$. (This will be done later. However, the notions of upper and lower bound, maximum, minimum, supremum, and infimum are defined in $$E^{*}$$ exactly as in $$E^{1} .$$ In particular,
$-\infty=\min E^{*} \text{ and } +\infty=\max E^{*}$

Thus in $$E^{*}$$ all sets are bounded.

It follows that in $$E^{*}$$ every set $$A \neq \emptyset$$ has a lub and a glb. For if $$A$$ has none in $$E^{1},$$ it still has the upper bound $$+\infty$$ in $$E^{*},$$ which in this case is the unique (hence also the least) upper bound; thus sup $$A=+\infty .$$ I Similarly, inf $$A=-\infty$$ if there is no other lower bound.? As is readily seen, all properties of lub and glb stated in §§8-9 remain valid in $$E^{*}$$ (with the same proof). The only exception is Theorem 2$$\left(\mathrm{ii}^{\prime}\right)$$ in the case $$q=+\infty$$ (respectively, $$p=-\infty ) \sin \mathrm{ce}+\infty-\varepsilon$$ and $$-\infty+\varepsilon$$ make no sense. Part (ii) of Theorem 2 is valid.

We can now define intervals in $$E^{*}$$ exactly as in $$E^{1}$$ §§8-9, Example (3), allowing also infinite values of $$a, b, x .$$ For example,

\begin{aligned} (-\infty, a) &=\left\{x \in E^{*} |-\infty<x<a\right\}=\left\{x \in E^{1} | x<a\right\} \\ (a,+\infty) &=\left\{x \in E^{1} | a<x\right\} \\ (-\infty,+\infty) &=\left\{x \in E^{*} |-\infty<x<+\infty\right\}=E^{1} \\ [-\infty,+\infty] &=\left\{x \in E^{*} |-\infty \leq x \leq+\infty\right\} ; \text { etc. } \end{aligned}

Intervals with finite endpoints are said to be finite; all other intervals are called infinite. The infinite intervals

$(-\infty, a),(-\infty, a],(a,+\infty),[a,+\infty), \quad a \in E^{1}$

are actually subsets of $$E^{1},$$ as is $$(-\infty,+\infty) .$$ Thus we shall speak of infinite intervals in $$E^{1}$$ as well.

## Upper and Lower Limits

In Chapter 1, §§1-3 we already mentioned that a real number $$p$$ is called the limit of a sequence $$\left\{ x_{n} \right\} \subseteq E^{1} \left( p=\lim x_{n} \right)$$ iff

$(\forall \varepsilon > 0)( \exists k)( \forall n>k) \quad \left| x_{n} - p \right| < \varepsilon , \text{ i.e., } p - \varepsilon < x_{n} < p - \varepsilon$

where $$\varepsilon \in E^{1}$$ and $$n, k \in N$$.

This may be stated as follows:

For sufficiently large $$n(n>k), x_{n}$$ becomes and stays as close to $$p$$ as we like ("$$\varepsilon$$-close").

We also define (in $$E^{1}$$ and $$E^{*}$$)

\begin{align} \lim _{n \rightarrow \infty} x_{n} &= +\infty \Longleftrightarrow\left( \forall a \in E^{1} \right)( \exists k)( \forall n>k) \quad x_{n}>a \text{ and} \\ \lim _{n \rightarrow \infty} x_{n} &= -\infty \Longleftrightarrow\left( \forall b \in E^{1} \right)( \exists k)( \forall n>k) \quad x_{n}<b.\end{align}

Note that $$(2)$$ and $$(3)$$ make sense in $$E^{1},$$ too, since the symbols $$\pm \infty$$ do not occur on the right side of the formulas. Formula $$(2)$$ means that $$x_{n}$$ becomes arbitrarily larger than any $$a \in E^{1}$$ given in advance) for sufficiently large $$n(n>k) .$$ The interpretation of $$(3)$$ is analogous. A more general and unified approach will now be developed for $$E^{*}$$ (allowing infinite terms $$x_{n},$$ too).

Let $$\left\{x_{n}\right\}$$ be any sequence in $$E^{*} .$$ For each $$n,$$ let $$A_{n}$$ be the set of all terms from $$x_{n}$$ onward, i.e.,

$\left\{x_{n}, x_{n+1}, \ldots\right\}$

For example,

$A_{1}=\left\{x_{1}, x_{2}, \dots\right\}, A_{2}=\left\{x_{2}, x_{3}, \ldots\right\}, \text{ etc.}$

The $$A_{n}$$ form a contracting sequence (see Chapter 1, §8) since

$A_{1} \supseteq A_{2} \supseteq \cdots.$

Now, for each $$n,$$ let

$p_{n}=\inf A_{n} \text{ and } q_{n}=\sup A_{n}$

also denoted

$p_{n}=\inf _{k \geq n} x_{k} \text{ and } q_{n}=\sup _{k \geq n} x_{k}.$

(These infima and suprema always exist in $$E^{*},$$ as noted above.) Since $$A_{n} \supseteq A_{n+1},$$ Corollary 2 of §§8-9 yields

$\inf A_{n} \leq \inf A_{n+1} \leq \sup A_{n+1} \leq \sup A_{n}$

Thus

$p_{1} \leq p_{2} \leq \cdots \leq p_{n} \leq p_{n+1} \leq \cdots \leq q_{n+1} \leq q_{n} \leq \cdots \leq q_{2} \leq q_{1}$

and so $$\left\{p_{n}\right\} \uparrow,$$ while $$\left\{q_{n}\right\} \downarrow$$ in $$E^{*} .$$ We also see that each $$q_{m}$$ is an upper bound of all $$p_{n}$$ and hence

$q_{m} \geq \sup _{n} p_{n}\left( = \operatorname{lub} \text{ of all } p_{n} \right).$

This, in turn, shows that this sup (call it $$\underline{L} )$$ is a lower bound of all $$q_{m},$$ and so

$\underline{L} \leq \inf _{m} q_{m}.$

We put

$\inf _{m} q_{m}=\overline{L}.$

Definition

For each sequence $$\left\{x_{n}\right\} \subseteq E^{*},$$ we define its upper limit $$\overline{L}$$ and its lower limit $$\underline{L},$$ denoted

$\overline{L}=\overline{\lim } x_{n}=\limsup _{n \rightarrow \infty} x_{n} \text{ and } \underline{L}=\lim x_{n}=\liminf _{n \rightarrow \infty} x_{n}$

as follows.

We put $$(\forall n)$$

$q_{n}=\sup _{k \geq n} x_{k} \text{ and } p_{n}=\inf _{k \geq n} x_{k},$

as before. Then we set

$\overline{L}=\overline{\lim } x_{n}=\inf _{n} q_{n}\text{ and } \underline{L}=\underline{\lim} x_{n}=\sup _{n} p_{n}, \text{ all in } E^{*}.$

Here and below, inf $$_{n} q_{n}$$ is the inf of all $$q_{n},$$ and $$\sup _{n} p_{n}$$ is the sup of all $$p_{n}$$.

Corollary $$\PageIndex{1}$$

For any sequence in $$E^{*},$$

$\inf _{n} x_{n} \leq \underline{\lim} x_{n} \leq \overline{\lim } x_{n} \leq \sup _{n} x_{n}.$

For, as we noted above,

$\underline{L}=\sup _{n} p_{n} \leq \inf _{m} q_{m}=\overline{L}.$

Also,

$\underline{L} \geq p_{n}=\inf A_{n} \geq \inf A_{1}=\inf _{n} x_{n}\text{ and }$

$\overline{L} \leq q_{n}=\sup A_{n} \leq \sup A_{1}=\sup _{n} x_{n},$

with $$A_{n}$$ as above.

Example $$\PageIndex{1}$$

(a) Let

$x_{n}=\frac{1}{n}.$

Here

$q_{1}=\sup \left\{1, \frac{1}{2}, \ldots, \frac{1}{n}, \ldots\right\}=1, q_{2}=\frac{1}{2}, q_{n}=\frac{1}{n}.$

Hence

$\overline{L}=\inf _{n} q_{n}=\inf \left\{1, \frac{1}{2}, \ldots, \frac{1}{n}, \ldots\right\}=0,$

as easily follows by Theorem 2 in §§8-9 and the Archimedean property. (Verify!) Also,

$p_{1}=\inf _{k \geq 1} \frac{1}{k}=0, p_{2}=\inf _{k \geq 2} \frac{1}{k}=0, \ldots, p_{n}=\inf _{k \geq n} \frac{1}{k}=0.$

since all $$p_{n}$$ are $$0,$$ so is $$\overline{L}=\sup _{n} p_{n} .$$ Thus here $$\underline{L}=\overline{L}=0.$$

(b) Consider the sequence

$1,-1,2,-\frac{1}{2}, \ldots, n,-\frac{1}{n}, \ldots$

Here

$p_{1}=-1=p_{2}, p_{3}=-\frac{1}{2}=p_{4}, \ldots ; p_{2 n-1}=-\frac{1}{n}=p_{2 n}.$

Thus

$\underline{\lim} _{n} x_{n}=\sup _{n} p_{n}=\sup \left\{-1,-\frac{1}{2}, \ldots,-\frac{1}{n}, \ldots\right\}=0.$

On the other hand, $$q_{n}=+\infty$$ for all $$n .$$ (Why?) Thus

$\overline{\lim } x_{n}=\inf _{n} q_{n}=+\infty.$

Theorem $$\PageIndex{1}$$

(i) If $$x_{n} \geq b$$ for infinitely many $$n,$$ then

$\overline{\lim } x_{n} \geq b \quad\text{ as well }.$

(ii) If $$x_{n} \leq a$$ for all but finitely many $$n,$$ then

$\overline{\lim } x_{n} \leq a \quad\text{ as well }.$

Similarly for lower limits (with all inequalities reversed).

Proof

(i) If $$x_{n} \geq b$$ for infinitely many $$n,$$ then such $$n$$ must occur in each set

$A_{m}=\left\{x_{m}, x_{m+1}, \ldots\right\}.$

Hence

$(\forall m) \quad q_{m}=\sup A_{m} \geq b;$

so $$\overline{L}=\inf _{m} q_{m} \geq b,$$ by Corollary 1 of §§8-9.

(ii) If $$x_{n} \leq a$$ except finitely many $$n,$$ let $$n_{0}$$ be the last of these "exceptional" values of $$n .$$

Then for $$n>n_{0}, x_{n} \leq a,$$ i.e., the set

$A_{n}=\left\{x_{n}, x_{n+1}, \ldots\right\}$

is bounded above by $$a ;$$ so

$\left(\forall n>n_{0}\right) \quad q_{n}=\sup A_{n} \leq a.$

Hence certainly $$\overline{L}=\inf _{n} q_{n} \leq a. \square$$

corollary $$\PageIndex{2}$$

(i) If $$\overline{\lim } x_{n}>a,$$ then $$x_{n}>a$$ for infinitely many $$n.$$

(ii) If $$\overline{\lim } x_{n}<b,$$ then $$x_{n}<b$$ for all but finitely many $$n.$$

Similarly for lower limits (with all inequalities reversed).

Proof

Assume the opposite and find a contradiction to Theorem 1. $$\square$$

To unify our definitions, we now introduce some useful notions.

By a neighborhood of $$p,$$ briefly $$G_{p},$$ we mean, for $$p \in E^{1},$$ any interval of the form

$(p-\varepsilon, p+\varepsilon), \quad \varepsilon>0.$

If $$p=+\infty($$ respectively $$, p=-\infty), G_{p}$$ is an infinite interval of the form

$(a,+\infty] \text{ (respectively, } [-\infty, b)),\text{ with } a, b \in E^{1}.$

We can now combine formulas (1)-(3) into one equivalent definition.

Definition

An element $$p \in E^{*}$$ (finite or not) is called the limit of a sequence $$\left\{x_{n}\right\}$$ in $$E^{*}$$ iff each $$G_{p}$$ (no matter how small it is) contains all but finitely many $$x_{n},$$ i.e. all $$x_{n}$$ from some $$x_{k}$$ onward. In symbols,

$\left(\forall G_{p}\right)(\exists k)(\forall n>k) \quad x_{n} \in G_{p}.$

We shall use the notation

$p=\lim x_{n}\text{ or } \lim _{n \rightarrow \infty} x_{n}.$

Indeed, if $$p \in E^{1},$$ then $$x_{n} \in G_{p}$$ means

$p-\varepsilon<x_{n}<p+\varepsilon,$

as in (1). If, however, $$p=\pm \infty,$$ it means

$x_{n}>a\left(\text { respectively, } x_{n}<b\right),$

as in (2) and (3).

Theorem $$\PageIndex{2}$$

We have $$q=\overline{\lim } x_{n}$$ in $$E^{*}$$ iff

(i) each neighborhood $$G_{q}$$ contains $$x_{n}$$ for infinitely many $$n,$$ and

(ii') if $$q<b,$$ then $$x_{n} \geq b$$ for at most finitely many $$n .$$

Proof

If $$q=\overline{\lim } x_{n},$$ Corollary 2 yields (ii')

It also shows that any interval $$(a, b),$$ with $$a<q<b,$$ contains infinitely many $$x_{n}$$ (for there are infinitely many $$x_{n}>a,$$ and only finitely many $$x_{n} \geq b$$ by $$\left(\mathrm{ii}^{\prime}\right) ).$$

Now if $$q \in E^{1},$$

$G_{q}=(q-\varepsilon, q+\varepsilon)$

is such an interval, so we obtain $$\left(i^{\prime}\right) .$$ The cases $$q=\pm \infty$$ are analogous; we leave them to the reader.

Conversely, assume $$\left(\mathrm{i}^{\prime}\right)$$ and $$\left(\mathrm{ii}^{\prime}\right).$$

Seeking a contradiction, let $$q<\overline{L} ;$$ say,

$q<b<\overline{\lim } x_{n}.$

Then Corollary 2$$(\mathrm{i})$$ yields $$x_{n}>b$$ for infinitely many $$n,$$ contrary to our assumption $$\left(\mathrm{i} i^{\prime}\right) .$$

Similarly, $$q>\overline{\lim } x_{n}$$ would contradict $$\left(\mathrm{i}^{\prime}\right).$$

Thus necessarily $$q=\overline{\lim } x_{n} . \square$$

Theorem $$\PageIndex{3}$$

We have $$q=\lim x_{n}$$ in $$E^{*}$$ iff

$\underline{\lim} x_{n}=\overline{\lim } x_{n}=q.$

Proof

Suppose

$\underline{\lim} x_{n}=\overline{\lim } x_{n}=q.$

If $$q \in E^{1},$$ then every $$G_{q}$$ is an interval $$(a, b), a<q<b$$; therefore, Corollary 2(ii) and its analogue for $$\underline{\lim} x_{n}$$ imply (with $$q$$ treated as both $$\overline{\lim} x_{n}$$ and $$\underline{\lim} x_{n}$$ that

$a<x_{n}<b\text{ for all but finitely many n}.$

Thus by Definition $$2, q=\lim x_{n},$$ as claimed.

Conversely, if so, then any $$G_{q}$$ (no matter how small) contains all but finitely many $$x_{n} .$$ Hence so does any interval $$(a, b)$$ with $$a<q<b,$$ for it contains some small $$G_{q} .$$

Now, exactly as in the proof of Theorem $$2,$$ one excludes

$q \neq \underline{\lim} x_{n}\text{ and } q \neq \overline{\lim } x_{n}.$

This settles the case $$q \in E^{1} .$$ The cases $$q=\pm \infty$$ are quite analogous. $$\square$$

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