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The Real Numbers
Here we show one way to explicitly construct the real numbers R. First we need a definition.
Definitions/Notation: A sequence of rational numbers is a funtion f : N → Q. Rather than
write f and f (n) for the sequence and its values, we typically write {an } and an . A sequence {an }
is called a Cauchy sequence if, for all > 0 (which we may take to be a rational epsilon for now),
there is an N ∈ N such that |an − am | < for all n, m ≥ N . For any rational number c, we let “c”
(with the quotes) denote the sequence {an } defined by an = c for all n ∈ N. In other words, “c”
is the sequence consisting entirely of the number c. This is clearly a Cauchy sequence. We say a
sequence {an } converges to the (rational) number c, and write {an } → c, if for all > 0, there is
an N ∈ N such that |an − c| < for all n ≥ N .
Lemma: If {an } and {bn } are Cauchy sequences, then so are {an + bn } and {an bn }. All
Cauchy sequences are bounded.
Proof: Let > 0. By definition, there are N1 , N2 ∈ N such that |an − am | < /2 for all
n, m ≥ N1 and |bn − bm | < /2 for all n, m ≥ N2 . Let N = max{N1 , N2 }. Then for all n, m ≥ N
we have
|(an + bn ) − (am + bm )| = |(an − am ) + (bn − bm )|
≤ |an − am | + |bn − bm |
< /2 + /2
= .
This shows that {an + bn } is a Cauchy sequence.
Next, get N3 , N4 ∈ N such that |an − am | < 1 for all n, m ≥ N3 and |bn − bm | < 1 for all
n, m ≥ N . Set B1 = maxn≤N3 {|an | + 1} and B2 = maxn≤N4 {|bn | + 1}. One readily verifies that
|an | < B1 and |bn | < B2 for all n ∈ N (this shows that all Cauchy sequences are bounded). Note
that B1 , B2 > 0 by construction. Now there are N5 , N6 ∈ N such that |an − am | < /(2B2 ) for
all n, m ≥ N5 and |bn − bm | < /(2B1 ) for all n, m ≥ N6 . Let N 0 = max{N5 , N6 }. Then for all
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n, m ≥ N 0 we have
|an bn − am bm | = |an (bn − bm ) + bm (an − am )|
≤ |an (bn − bm )| + |bm (an − am )|
= |an | · |bn − bm | + |bm | · |an − am |
< B1 · /(2B1 ) + B2 · /(2B2 )
= .
This shows that {an bn } is a Cauchy sequence.
Definition: We say two Cauchy sequences {an } and {bn } of rational numbers are equivalent,
and write {an } ∼ {bn }, if the sequence {an − bn } → 0.
Lemma: This is an equivalence relation.
Proof: For any sequence {an } of rational numbers, {an − an } = “0” → 0, so that {an } ∼ {an }
for any Cauchy sequence {an }.
Suppose {an } ∼ {bn } and let > 0. Then for some N ∈ N, |an − bn | < for all n ≥ N . Thus
|bn − an | < for n ≥ N and {bn − an } → 0. In other words, {bn } ∼ {an }.
Suppose {an } ∼ {bn } and {bn } ∼ {cn } and let > 0. Then for some N ∈ N, |an − bn | < /2
for all n ≥ N and for some M ∈ N, |bn − cn | < /2 for all n ≥ M . This implies that
|an − cn | = |an − bn + bn − cn | ≤ |an − bn | + |bn − cn | <
+ =
2 2
for all n ≥ max{N, M }. Thus {an − cn } → 0 and {an } ∼ {cn }.
Definition: The real numbers R is the set of equivalence classes of Cauchy sequences of
rational numbers. We’ll write [{an }] to denote the equivalence class which contains the sequence
{an }. We view Q as a subset of R by identifying the rational number c with [“c”].
Definition: Define addition and multiplication of real numbers by
[{an }] + [{bn }] = [{an + bn }]
and
[{an }] · [{bn }] = [{an · bn }].
Lemma: These operations are well-defined, i.e., depend only on the equivalence classes and
not the particular elements of the equivalence classes used.
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Proof: Suppose {an } ∼ {a0n } and {bn } ∼ {b0n } are all Cauchy sequences. By the lemma above,
both {an + bn } and {an bn } are Cauchy sequences.
Let > 0. Then for some N1 , M1 ∈ N, |an − a0n | < /2 for all n ≥ N1 and |bn − b0n | < /2 for
all n ≥ M1 . This implies that
|(an + bn ) − (a0n + b0n )| = |(an − a0n ) + (bn − b0n )| ≤ |an − a0n | + |bn − b0n | <
+ =
2 2
for all n ≥ max{N1 , M1 }. Thus, {an + bn } ∼ {a0n + b0n } and addition is well-defined.
By a previous lemma, all four sequences {an }, {a0n }, {bn } and {b0n } are bounded. Thus, there
is a B > 0 such that |an |, |a0n |, |bn |, |b0n | ≤ B for all n ∈ N. For some N2 , M2 ∈ N we have
|an − a0n | <
2B
for all n ≥ N2 and |bn − b0n | <
2B
for all n ≥ M2 . This implies that
2|an bn − a0n b0n | = |(an − a0n )(bn + b0n ) + (an + a0n )(bn − b0n )|
≤ |(an − a0n )(bn + b0n )| + |(an + a0n )(bn − b0n )|
= |an − a0n | · |bn + b0n | + |an + a0n | · |bn − b0n |
≤ |an − a0n | · (|bn | + |b0n |) + (|an | + |a0n |) · |bn − b0n |
<
· (2B) +
· (2B)
2B
2B
= 2
for all n ≥ max{N2 , M2 }. Thus, |an bn − a0n b0n | < for all n ≥ max{N2 , M2 } and {an bn } ∼ {a0n b0n }.
This shows that multiplication is well-defined.
Theorem 1: The real numbers are a field.
Proof: Since the rational numbers are a field,
([{an }] + [{bn }]) + [{cn }] = [{(an + bn ) + cn }] = [{an + (bn + cn )}] = [{an }] + ([{bn }] + [{cn }])
for any Cauchy sequences {an }, {bn } and {cn }, and similarly for multiplication. Also,
[{an }] + [{bn }] = [{an + bn }] = [{bn + an }] = [{bn }] + [{an }],
and similarly for multiplication. Next,
([{an }]+[{bn }])·[{cn }] = [{(an +bn )·cn }] = [{(an ·cn )+(bn ·cn )}] = ([{an }]·[{cn }])+([{bn }]·[{cn }]).
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Since 0 is the additive identity element of Q,
[“0”] + [{an }] = [{0 + an }] = [{an }]
for any [{an }] ∈ R, so that [“0”] is an additive identity for R. Clearly
[{an }] + [{−an }] = [{an + (−an )}] = [“0”]
for any [{an }] ∈ R, so that [{−an }] is an additive inverse for [{an }].
Since 1 is the multiplicative identity element of Q
[{an }] · [“1”] = [{an · 1}] = [{an }]
for all [{an }] ∈ R. Clearly “0” 6∼ “1”. Suppose that [{an }] 6= [“0”], i.e., {an } 6∼ “0”. This means
that for some > 0, there are infinitely many n ∈ N such that |an | ≥ . Since {an } is a Cauchy
sequence, there is an N ∈ N such that |an − am | < /2 for all n, m ≥ N . Since there must be an
n0 ≥ N such that |an0 | ≥ , then
|am | ≥ |an0 | − |an0 − am | > −
=
2
2
for all m ≥ N . In particular, am 6= 0 for m ≥ N . Further, it is not difficult to see that the sequence
{bn } defined by
bn =
aN
if n ≤ N ,
an
if n ≥ N
is a Cauchy sequence equivalent to {an }. Thus, we may assume without loss of generality that
an 6= 0 for all n ∈ N. Now since the rational numbers are a field,
−1
[{an }] · [{a−1
n }] = [{an · an }] = [“1”],
so that [{an }] has a multiplicative inverse.
Definition: For [{an }] ∈ R, [{an }] = [{|an |}].
Lemma: This is well-defined, i.e., [{an }] ∈ R and depends only on the equivalence class of
{an }.
Proof: Let > 0. Then for some N ∈ N we have |an − am | < for all n, m ≥ N . Since
|an | − |am | ≤ |an − am |, this implies that |an | − |am | < for all n, m ≥ N and {|an |} is a Cauchy
sequence.
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Suppose {a0n } ∼ {an } and let > 0. Then for some N ∈ N, |a0n − an | < for all n ≥ N . As
above, this implies that |a0n | − |an | < for all n ≥ N , so that {|a0n |} ∼ {|an |}.
Definition: We say a real number [{an }] is greater than zero (or positive), and write [{an }] >
0, if there is an N ∈ N and an > 0 such that an ≥ for all n ≥ N .
Lemma: This is well-defined.
Proof: Suppose {a0n } and {an } are equivalent Cauchy sequences. Suppose further that there
is some N ∈ N and an > 0 such that an ≥ for all n ≥ N . There is an M ∈ N such that
|a0n − an | < /2 for all n ≥ M . This implies that |a0n | ≥ |an | − |a0n − an | > −
2
=
2
for all
n ≥ max{N, M }.
Definition: We say a real number [{an }] is greater than a real number [{bn }], and write
[{an }] > [{bn }], if [{an }] − [{bn }] > 0.
Theorem 2: For any [{an }], [{bn }] ∈ R, the following three properties hold:
a) |[{an }] ≥ [“0”], with equality if and only if [{an }] = [“0”];
b) [{an }] · [{bn }] = |[{an }]| · |[{bn }]|;
c) [{an }] + [{bn }] ≤ [{an }] + [{bn }].
In other words, | · | is an absolute value on R.
Proof: Starting with the first property, we have |an | ≥ 0 for all n. Suppose that {an } 6∼ “0”.
Then there must be an > 0 such that |an | ≥ for infinitely many n ∈ N. Let N ∈ N be
such that |an − am | < /2 for all n, m ≥ N . Since there is an n0 ≥ N with |an0 | ≥ , we have
|am | ≥ |an0 | − |an0 − am | > − /2 = /2 for all m ≥ N . Thus, |[{an }]| = [{|an |}] > [“0”].
For the second property, using |an bn | = |an | · |bn | for all n, we have
[{an }] · [{bn }] = |[{an bn }]| = [{|an bn |}] = [{|an |}] · [{|bn |}] = |[{an }]| · |[{bn }]|.
Now for the third property - the “triangle inequality.” Suppose first that there is some > 0
and some N ∈ N such that |an + bn | + ≤ |an | + |bn | for all n ≥ N . Then by the definitions,
[{an }] + [{bn }] = [{|an + bn |}] < [{|an | + |bn |}] = [{an }] + [{bn }].
So suppose this is not the case and let > 0. Then there are infinitely many n ∈ N such that
|an + bn | + /2 > |an | + |bn |. Also, there is an N ∈ N such that
|an | − |am |, |bn | − |bm |, |an + bn | − |am + bm | < /6
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for all n, m ≥ N . Choose an n ≥ N such that |an + bn | + /2 > |an | + |bn |. Then for all m ≥ N we
have
+ |bn | + − |an + bn | +
6
6
6
= |an | + |bn | − |an + bn | +
2
|am | + |bm | − |am + bm | < |an | +
< .
By the triangle inequality for rational numbers, |am | + |bm | ≥ |am + bm |. Thus,
|am | + |bm | − |am + bm | < for all m ≥ N and {|an + bn |} ∼ {|an | + |bn |}. By the definitions, this means that
[{an }] + [{bn }] = [{an }] + [{bn }].
Lemma: Suppose [{an }] and [{bn }] are two unequal real numbers. Then there is a rational
number c such that [{an }] − [“c”] < [{an }] − [{bn }].
Proof: Since {an } 6∼ {bn }, {an −bn } 6∼ “0”. By Theorem 2, [{an }]−[{bn }] = |[{an −bn }]| > 0
so that there is an > 0 and an N ∈ N such that |an − bn | ≥ for all n ≥ N . Also, for some M ∈ N
we have |an − am | < /2 for all n, m ≥ M . Let c = aN +M . Then for all n ≥ N + M , |an − c| < /2
and |an − bn | ≥ . In particular, |an − bn | − |an − c| ≥ /2 for all n ≥ N + M . By the definitions,
this means that [{an }] − [“c”] < [{an }] − [{bn }].
Theorem 3: Every Cauchy sequence of real numbers converges, i.e., the real numbers are a
topologically complete field.
[Technically speaking, the in the definition of Cauchy sequence of real numbers is allowed
to be real. However, it suffices to restrict to the case where is a positive rational number by the
lemma above.]
Proof: Let {rn } be a Cauchy sequence of real numbers. If there is an r ∈ R and an N ∈ N
such that rn = r for all n ≥ N we are done, since in this case {rn } → r. So suppose this is not the
case. For each n ∈ N let n0 ∈ N be least such that n0 > n and rn 6= rn0 . For each n ∈ N choose (via
the lemma above) an an ∈ Q such that |rn − [”an ”]| < |rn − rn0 |.
Let > 0. Then there is an N ∈ N such that |rn − rm | < /3 for all n, m ≥ N . By the triangle
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inequality, we have
|an − am | = |[“an ”] − [“am ”]| = |[“an ”] − rn + rn − rm + rm − [“am ”]|
≤ |[“an ”] − rn | + |rn − rm | + |rm − [“am ”]|
< |rn − rn0 | + |rn − rm | + |rm − rm0 |
<
+ +
3 3 3
=
for all n, m ≥ N . This shows that {an } is a Cauchy sequence (of rational numbers), i.e., [{an }] ∈ R.
Let > 0 again. Then there are is an N ∈ N such that |rn − rm |, |an − am | < /3 for all
n, m ≥ N . In particular, |rN − [“aN ”]| < /3 and also |[“aN ”] − [{am }]| < /3. Using the triangle
inequality once more,
|rn − [{am }]| = rn − rN + rN − [“aN ”] + [“aN ”] − [{am }]
≤ |rn − rN | + |rN − [“aN ”]| + |[“aN ”] − [{am }]|
<
+ +
3 3 3
=
for all n ≥ N . Thus, {rn } → [{am }] ∈ R.
Lemma: The sum and product of two positive real numbers is positive.
Proof: Suppose [{an }] and [{bn }] are positive real numbers. Then by definition there are
N1 , N2 ∈ N and 1 , 2 > 0 such that an ≥ 1 for all n ≥ N1 and bn ≥ 2 for all n ≥ N2 . This
implies that an + bn ≥ 1 + 2 > 0 and an bn ≥ 1 2 > 0 for all n ≥ max{N1 , N2 }. In other words,
[{an }] + [{bn }] = [{an + bn }] > 0 and [{an }] · [{bn }] = [{an bn }] > 0.
Theorem 4: The real numbers are totally ordered by <.
Proof: Suppose x < y. By definition, this means y − x > 0. Since (y + z) − (x + z) = y − x,
we have y + z > x + z. Suppose x < y and y < z. Then z − x = (z − y) + (y − x) is positive, being
the sum of two positive real numbers. Suppose x < y and z > 0. Then z(y − x) is positive, so that
zy > zx.
Suppose x 6< y and y 6< x, and write x = [{an }], y = [{bn }]. Let > 0. There are infinitely
many n ∈ N such that an − bn < /3 and infinitely many n ∈ N such that bn − an < /3. Also,
there are N, M ∈ N such that |an − am | < /6 for all n, m ≥ N and |bn − bm | < /6 for all
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n, m ≥ M . Choose an n0 , m0 ≥ max{N, M } such that an0 − bn0 < /3 and bm0 − am0 < /3. Then
bn0 − an0 < bm0 + /6 − am0 + /6 < 2/3. For any n ≥ max{N, M } we have
|an − bn | = |an − an0 + an0 − bn0 + bn0 − bn |
≤ |an − an0 | + |an0 − bn0 | + |bn0 − bn |
<
2
+
+
6
3
6
= .
Thus {an − bn } → 0 and [{an }] = [{bn }]. This shows that either x < y or y < x or x = y.
Finally, suppose x > y and y < x. Then [“0”] = (x − y) + (y − x) is positive since it is the
sum of two positive numbers. Similarly, if x = y and either x > y or y > x, then [“0”] is positive,
which is clearly not the case, so that at most one of x > y, y > x and x = y holds.
Theorem 5: Suppose S ⊂ R, S 6= ∅, and there is a B ∈ R with x ≤ B for all x ∈ S. (We
say B is an upper bound for S.) Then there is an upper bound b ∈ R for S such that for all c < b
there is an x ∈ S with x > c. (We say b is a least upper bound for S.)
Proof: Chose an x0 ∈ S and let m0 ∈ N be largest such that B − m0 is an upper bound for
S. Then m0 exists since B − 0 is an upper bound for S and B − n is not an upper bound for S
whenever n > B − x0 .
We claim that there is a sequence B1 , B2 , . . . of upper bounds for S such that Bn − 2−n is not
an upper bound for S and 0 ≤ Bn−1 − Bn ≤ 2−n for all n ∈ N. We prove this claim by induction.
Let m1 ∈ N be largest such that B0 − m1 2−1 is an upper bound for S. Then m1 is either 0 or 1
since B0 − 0 is an upper bound for S and B0 − 1 = B0 − 2 · 2−1 is not. Let B1 = B0 − m1 2−1 . Then
B1 is an upper bound for S and 0 ≤ m1 2−1 = B0 − B1 ≤ 2−1 . Also, B1 − 2−1 = B0 − (m1 + 1)2−1
in not an upper bound for S by the definition of m1 .
Now suppose B1 , . . . , Bn have been chosen which satisfy the requirements above. Let mn+1
be the largest integer such that Bn − mn+1 2−n−1 is an upper bound for S. By the induction
hypothesis, Bn is an upper bound for S and Bn − 2−n is not, so that mn+1 is either 0 or 1. Let
Bn+1 = Bn −mn+1 2−n−1 . Then Bn+1 is an upper bound for S and 0 ≤ mn+1 2−n−1 = Bn −Bn+1 ≤
2−n−1 . Finally, Bn+1 −2−n−1 = Bn −(mn+1 +1)2−n−1 is not an upper bound for S by the definition
of mn+1 .
We claim that this is a Cauchy sequence. Indeed, let > 0 and suppose n, m ∈ N with n > m.
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Then
Bn − Bm = (Bn − Bn−1 ) + · · · + (Bm+1 − Bm )
≤ 2−n + · · · + 2−m−1
= 2−m−1 (1 + · · · + 2m+1−n )
= 2−m−1
1 − 2m−n
1 − 2−1
= 2−m (1 − 2m−n )
< 2−m .
Thus, if n, m ≥ N where 2−N ≤ , then |Bn − Bm | < .
Let b ∈ R be the limit of this Cauchy sequence. Let x ∈ S. Suppose x > b and let = x − b.
For some n ∈ N, |Bn − b| < . But this implies that x − b > |Bn − b|, so that x > Bn . This
contradicts the fact that all Bn ’s are upper bounds for S. Thus, b is an upper bound for S.
Finally, suppose c < b is an upper bound for S. Then b − c ≥ 2−n0 for some n0 ∈ N, which
implies that b − 2−n0 is an upper bound for S. Choose N ∈ N such that |Bn − b| < 2−n0 −1 for
all n ≥ N . We then have Bn − 2−n0 −1 > b − 2−n0 for all n ≥ N . But Bn − 2−n is not an upper
bound for any n, so that Bn − 2−n0 −1 ≤ Bn − 2−n is not an upper bound for any n > n0 . This
contradition shows that c ≥ b, so that b is a least upper bound for S.
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