2.7: The Infinities. Upper and Lower Limits of Sequences
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
\[\inf _{m} q_{m}=\overline{L}.\]
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 \(p_{n}\).
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.
(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
\[\overline{\lim } x_{n}=\inf _{n} q_{n}=+\infty.\]
(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 \(\overline{L}=\inf _{n} q_{n} \leq a. \square\)
(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.
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).
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 \(q=\overline{\lim } x_{n} . \square\)
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. \(\square\)