In the construction of *JR as an ultrapower in Chapter 3, each sequence of
points r = (rn : n E N) in JR gives rise to the single point [r] of *JR, which
we also denote by the more informative symbol [rnl· Equality of *JR-points
is given by
[rn] = [sn] iff {n EN: rn = sn} E F.
This description works for other kinds of entities than points. We will now
see that a sequence of subsets of JR determines a single subset of *JR. In the
next chapter we will see that a sequence of functions on JR determines a
single function on *JR.
11.1 Internal Sets
Given a sequence (An : n EN) of subsets An JR, define a subset [An] *JR
by specifying, for each [rn] E *JR,
[r n] E [An] iff { n E N : r n E An} E F.
Of course it must be checked that this is a well-defined notion that does
not depend on how points are named, which means that if [rn] = [snJ, then
{n EN: rn E An} E F iff {n EN: Sn E An} E F.
This is a slight extension of the argument given in Section 3.9.
The subsets of *.IR that are produced by this construction are called internal.
Here are some examples:
' '
126 11. Internal and External Sets
• If (An) is a constant sequence with An = A JR. for all n E N, then
the internal set [An] is just the enlargement *A of A defined in Section
3.9. Hence we may also denote *A as [A].
Thus the enlargement of any subset of JR. is an internal subset of *JR..
In particular, we see that *N, *Z, and *Q and *JR. itself are all internal,
as is any finite subset A JR., since in that case A = *A.
• More generally, any finite set X = {[r], ... , [r]} of hyperreals is
internal, for then X= [An], where Ano= {r, ... , r}.
• If a < b in *JR., then the hyperreal open interval
(a, b)o= {x E *JR.: a< x < b}
is internal. Indeed, if a = [an] and b = [bn], then (a, b) is the inter
nal set defined by the sequence ((an, bn) : n E N) of real intervals
(an, bn) JR.. This follows because
[an] < [rn] < [bn] iff {n EN: an < Tn < bn} E F.
Similarly, the hyperreal intervals (a,b], [a, b), [a,b], {x E *JR.n: a< x}
are internal. Notice that if a is unlimited, then each of these intervals
is disjoint from JR., so none of them can be the enlargement *A of a
set A JR., since *A always includes the (real) members of A.
• If N E *N, then the set
{k E *No: k
:::; N} = {1, 2, ... , N}
is internal. If N = [Nn], then this is the internal set [An], where
An = {k E N : k :::; Nn} = { 1, 2, ... , N n}
(since N E *N, we have {n: NnoE N} E F, so we may as well assume
N n E N for all n E N).
• If N = [Nn] E *N, then the set
{k } { 1 2 N-1 }
: k E *N U {0} and k :::; N = 0, N, N, ... , , 1
N N
is the internal set [An], where
{ 1 2 Nn -1 }
An= 0, Nn, ... , ,1 .
NnoNn
These last two examples illustrate the notion of hyperfinite set, which will
be studied in the next chapter.
11.2 Algebra of Internal Sets 127
11.2 Algebra of Internal Sets
(1) The class of internal sets is closed under the standard finite set operations
n, U, and -, with
[An] n [Bn] [Ann En],
[An] U [Bn] [An U En],
[An]-[Bn] [An-En]·
(2) [An] [En] iff {n EN: An En} E :F.
(3) [An]= [En] iff {n EN: Ano= En} E :F.
(4) [An] =I [Bn] iff {n EN: An#-En} E :F.
Proof.
( 1) Exercise.
(2) If [An] [En], then there is some hyperreal [rn] E [An]o-[BnJ, so by
(1) we have
I= {n EN: rn E Ano-En} E :F.
But if
J = { n E N : An En},
then I Jc, so Jc E :F and hence J ¢:. :F.
Conversely, if J ¢:. :F, then Jc E :F, so choosing rn E An-En for each
n E Jc and rn arbitrary for n E J, the argument reverses to give a
point [rn] E [An] -[En]·
(3) This follows from (2) and closure properties of :F (note that the result
is not a matter of the definition of [An] via :F, since equality of [An]
and [En] is defined independently of :F to mean "having the same
members"n).
( 4) Exercise.
D
Part (3) above is important for what it says about the sequence (An : n E
N) that determines a certain internal set. We can replace this sequence by
another (En : n E N) without changing the resulting internal set, provided
that An =En for :F-almost all n. Thus we are free to alter An arbitrarily
when n is outside a set that belongs to :F. For instance, if [An] is nonempty,
then as 0 = [0], we can assume that Anoi-0 for every n EN (11.2(4)), while
if [An] is a subset of *N, then as *N = [N], we can assume that An N for
every n (11.2(2)). Moreover, we can combine finitely many such conditions,
using the closure of :F under finite intersections. So if [An] is a nonempty
subset of *N, we can assume that 0 i-An N for every n E N.
128 11. Internal and External Sets
Subsets of Internal Sets
The fact that the intersection of two internal sets is internal allows us to
prove now that
if a set A of real numbers is internal, then so is every subset of A.
•
Proof. Let X A. Then *X is internal, so if A is internal, then so is
An *X. But since A !R,
An*X=An*Xn!R=Anx
(cf. Ex. 3.10.5), so X= An X is internal.
This result will be used in Section 11.7 to show that actually the only
internal subsets of lR are the finite ones.
11.3 Internal Least Number Principle and
Induction
A characteristic feature of N is that each of its nonempty subsets has a least
member (indeed this holds for any subset of Z that has a lower bound). The
same is not true, however, for *N: the set *N-N of unlimited hypernaturals
has no least member, for if N is unlimited, then so is N-1 (why?). But
we do have
Theorem 11.3.1 Any nonempty internal subset of *N has a least member.
Proof. Let [An] be a nonempty internal subset of *N. Then by the observations
above we can assume that for each n E N,
and so An has a least member rn. This defines a point [rn] E *IR with
{n E N : r n E An} = N E F,
so [rn] E [An]· Moreover, if [sn] E [An], then
{n E N : Sn E An} E F and {n E N : Sn E An} {n E N : r n :::; Sn}'
leading to the conclusion [rn] :::; [sn] in *JR. Hence [An] indeed has a least
member, namely the hyperreal number [rn] determined by the sequence of
least members of the sets Ann.
Writing "min X" for the least element of a set X, this construction can
be expressed concisely by the equation
min [An]n= [min An]·
11.4 The Overflow Principle 129
Now, the least number principle for N is equivalent to the principle of
induction:
A subset of N that contains 1 and is closed under the successor
function n 1---7 n + 1 must be equal to N.
The corresponding assertion about subsets of *N is not in general true, and
can only be derived for internal sets:
Theorem 11.3.2 (Internal Induction) If X is an internal subset of
*N that contains 1 and is closed under the successor function n 1---7 n + 1,
then X= *N.
Proof Let Y = *No-X. Then Y is internal (11.2(1)), so if it is nonempty,
it has a least element n. Then nn-=!= 1, as 1 E X, sonn-1 E *N. But now
n -1 Y, as n is least in Y, so n-1 E X, and therefore n = (n -1) + 1 is
in X by closure under successor. This contradiction forces us to conclude
that Y = 0, and so X = *N. 0
11.4 The Overflow Principle
The set N cannot be internal, or else by internal induction it would be
equal to *N. Thus if an internal set X contains all members of N, then
since X cannot be equal to N, it must "overflow" into *No-N. Indeed,
we will see that X must contain the initial segment of *N up to some
unlimited hypernatural. In fact, a slightly stronger statement than this can
be demonstrated by assuming only that X contains "almost all" members
ofoN:
Theorem 11.4.1 Let X be an internal subset of *N and kEoN. If n EX
for all n E N with k n, then there is an unlimited K E *N with n E X
for all n E *N with k n K.
Proof If all unlimited hypernaturals are in X, then any unlimited K E *N
will do. Otherwise there are unlimited hypernaturals not in X. If we can
show that there is a least such unlimited number H, then all unlimited
numbers smaller than H will be in X, giving the desired result.
To spell this out: if *N -X has unlimited members, then these must be
greater thannk, and so the set
Y = {n E *N : k < n E *N -X}
is nonempty. But Y is internal, by the algebra of internal sets, since it is
equal to
( *N-{1, ... , k}) n (*No-X).
130 11. Internal and External Sets
Hence Y has a least element H by the internal least number principle.
Then H is a hypernatural that is greater than k but not in X, so it must
be the case that H rt N, because of our hypothesis that all limited n k
are in X. Thus H is unlimited. Then K = H -1 is unlimited and meets
the requirements of the theorem: H is the least hypernatural greater. than
k that is not in X, so every n E *N with k ::; n ::; H -1 does belong to X.
D
Exercise 11.4.2
Show that overflow is equivalent to the following statement:
If an internal subset X of *N contains arbitrarily small unlim
ited members, then it is unbounded in N, i.e., contains arbitrar
ily large limited members.
(Hint: consider *N -X.) D
The overflow principle implies that any cofinite subset of N is external: if
an internal A N were cofinite, then it would contain { n E N : k ::; n}
for some k E N, so by overflow A would contain some unlimited number,
contradicting A N.
11.5 Internal Order-Completeness
The principle of order-completeness, attributed to Dedekind, asserts that
every nonempty subset of ffi. with an upper bound in ffi. must have a least
upper bound in ffi.. The corresponding statement about *ffi. is false. In fact,
ffi. itself is a nonempty subset of *ffi. that is bounded but has no least upper
bound. This is because the upper bounds of ffi. in *ffi. are precisely the positive
unlimited numbers, and there is no least positive unlimited number.
Just as for the least number principle, order-completeness is preserved
in passing from ffi. to *ffi. for internal sets:
Theorem 11.5.1 If a nonempty internal subset of *ffi. is bounded above/
below, then it has a least upper/ greatest lower bound in *ffi..
Proof We treat the case of upper bounds. In effect, the point of the proof
is to show that the least upper bound of a bounded internal set [An] is the
hyperreal number determined by the sequence of least upper bounds of the
An's:
lub[An] = [lubAn]·
More precisely, it is enough to require that F-almost all An's have least
upper bounds to make this work.
11.6 External Sets 131
Suppose then that a nonempty internal set [An] has an upper bound [rn]·
Write An ::::; x to mean that x is an upper bound of An in JR., and put
J = {n EN: Ano::::; rn}·
We want J E F. If not, then Jc E F. But if n E Jc, there exists some
an with rn < an E An. This leads to the conclusion [rn] < [an] E [An],
contradicting the fact that [rn] is an upper bound of [An]·
It follows that J E F. Since [An] =J 0, this then implies
J' = { n E N : 0 =J An ::::; r n} E F.
Now, if n E J', then An is a nonempty subset of JR. bounded above (by rn),
and so by the order-completeness of JR., An has a least upper bound Bn E JR.
Then if [bn] E [An],
{ n E N : bn E An} n J' { n E N : bn ::::; Sn}
'
leading to [bn] ::::; [sn], and showing that [sn] is an upper bound of [An]·
Finally, if [tn] is any other upper bound of [An], then {no: An ::::; tn} E F
by the same argument as for [rn], and
{ n E N : An ::::; tn} n J' { n E N : Sn ::::; tn}
'
so we get [sn] ::::; [tn]. This shows that [sn] is indeed the least upper bound
of [An] in *JR.
D
Exercise 11.5.2
Let X be an internal subset of *JR. Prove the following.
(i) If X has arbitrarily large limited members, then it has a positive
unlimited member.
(ii) If X has only limited members, then there is some real r such that
X is included in the interval [-r, r] in *JR.
(iii) If X has arbitrarily small positive unlimited members, then it has a
positive limited member.
(iv) If X has no limited members, then there is some unlimited b such
that X {x E *JR.o: x < -b ornbo< x}.
11.6 External Sets
A subset of *JR is external if it is not internal. Many of the properties that
are special to the structure of *JR define external sets:
132 11. Internal and External Sets
•
Unlimited Hypernatumls. Since *Nn-N has no least member, the internal
least number principle (Theorem 11.3.1) implies that it cannot
be internal.
• Limited Hypernaturals. If N were internal, then by 11.2(1) so too
would be *Nn-N, which we have just seen to be false.
Alternatively, by the internal induction principle, Theorem 11.3.2, if
N were internal, it would be equal to *N.
• Real Numbers. lR is external, for if it were internal, then so too would
be !Rn *N = N.
Alternatively, as noted at the beginning of Section 11.5, lR is bounded
but has no least upper bound in *JR, so must fail to be internal by the
internal order-completeness property, Theorem 11.5.1.
The fact that N is external will be used in the next section to show
that all infinite subsets of JR. are external.
• Limited Hyperreals. The set IL of limited numbers is external for the
same reason lR is: it is bounded above by all members of *JR., but
has no least upper bound. Since
lL = n{( -b, b) : b is unlimited},
it follows that the intersection of an infinite family of internal sets
can fail to be internal.
Observe that if X is an internal set that includes IL, then X =f. IL, and
so X must contain unlimited members ( cf. (i) and (ii) of Exercise
11.5.2). In fact, by considering lower and upper bounds of *JR -X
and *JR. -X, respectively, we can show that if X is an internal set
with IL X, then [-b, b] X for some unlimited b.
•
Injinitesimals. The set JI = hal(O) of infinitesimals is bounded above
(by any positive real), so if it were internal, it would have a least
upper bound b E *JR. Such a b would have to be positive but less than
every positive real, forcing b 0. But then b < 2b E JI, so b cannot be
an upper bound of ll after all.
By similar reasoning, any halo hal(r) is seen to be an external set, as
are its "left and right halves" {x >r: x r} and {x < r:x r}.
More strongly, this type of reasoning shows that if X is any internal
subset of JI, then the least upper bound and greatest lower bound of
X must be infinitesimal, and so X [-c, c] for some c 0.
11.7 Defining Internal Sets 133
Exercise 11.6.1
Show that the following form external sets:
the positive real numbers JR+,
the integers Z,
the rational numbers Q,
the (positive/negative) unlimited hyperreals,
the appreciable numbers.
11.7 Defining Internal Sets
In proving the internal least number and order-completeness properties,
we reverted once more to ultrafilter calculations, so it is natural to ask
whether such results could be obtained instead by a logical transfer. The
assertion that a nonempty set A of natural numbers has a least element
can be expressed by the .C9'l-sentence
[(:3x E IR)(x E A) A (\lx E A)(x EN)) (::Jx E A)(\t'y E A)(x y).
This sentence transforms to
[(:3x E *IR)(x E *A) A (\lx E *A)(x E *N)] (::Jx E *A)(\t'y E *A)(x y),
which asserts the least number principle for the enlarged set *A in *JR. But
what we saw in Section 11.3 was that this transformed sentence is true
when *A is replaced by any internal set X *N. The same observation
applies to the assertion
(::Jx E IR)(\t'y E A)(y x)
(3x E IR) [(\lyE A)(y x) A (\lz E IR) [(\t'y E A)(y z) x zJ]
,
which expresses the order-completeness property that if A has an upper
bound, then it has a least upper bound. The transform of this last sentence
is also true when *A is replaced by any nonempty internal set X *IR
(Theorem 11.5.1).
Let us write cp(A) to indicate that cp is an .C9'l-sentence containing the
set symbol A. Then *cp(X) denotes the sentence obtained by putting X in
place ofn*A in *cp. The examples just discussed suggest the following transfer
principle:
(t) If cp(A) is true whenever A is taken as an arbitrary subset of IR, then
*cp( X) is true whenever X is taken as an arbitrary internal subset
of *JR.
134 11. Internal and External Sets
To understand what is happening here we need to look more widely at
formulae that may have free variables. Let c.p(x, y, A) be a formula with
free variables x and y as well as the set symbol A; for example the formula
(V z E IR) (X < z < y z E A).
We can replace x and y in *c.p by elements of *R Thus *c.p([rn], [sn], [An])
would be the sentence
(Vz E *IR) ([rn] < z < [sn] Z E [An]).
It can be shown that this sentence is true if and only if
{n ENo: Vz E IR (rn < z < Sn z E An)} E F.
The general situation here is that
(i)
if and only if (ii)
This fact can then be used to derive the transfer principle (t). But it also
leads to a new way of defining internal sets: holding the hyperreal [sn] and
the internal set [An] fixed, and allowing the value of x to range over *IR,
define
X= {bE *IR: *c.p(b, [sn], [An]) is true}.
Correspondingly, for each n E N put
Bn = {r E IR: c.p(r, Sn, An) is true}.
Then the equivalence of (i) and (ii) amounts to saying that for any hyperreal
[rnJ,
[rn] EX iff {n EN: rn E Bn} E F.
But this shows that X is the internal set [Bn] determined by the sequence
of real subsets (Bn : n E N).
Expressing this phenomenon in the most general form available at this
stage, we have the following statement.
11.7.1 Internal Set Definition Principle. Let
be an LfR-formula with free variables xo, ..n. , Xn and set symbols A 1, .o.. , Ak.
Then for any hyperreals c1•, ..o. , Cn and any internal sets X 1, ... , X k,
11.7 Defining Internal Sets 135
is an internal subset of *JR. 0
Note that this statement applies in particular to formulae that have only
number variables and no set symbols. It provides a ready means of demonstrating
that various sets are internal, including the examples from Section
11.1:
• Taking