Completion
Theorem. Let (M, d) be a metric space. Then there exists a metric space (M∗ , ∆) and a map
ϕ : M → M∗ such that
(1) ϕ is one–to–one. That is, ϕ(x) = ϕ(y) if and only if x = y.
(2) ∆ ϕ(x), ϕ(y) = d(x, y) for all x, y ∈ M.
(3) M∗ is complete.
(4) ϕ(M) is dense in M∗ . This means that each element of M∗ is a limit of elements of
ϕ(M) = ϕ(x) x ∈ M
Equivalently, for each P ∈ M∗ and each ε > 0, there is a p ∈ M with ∆ ϕ(p), P < ε.
(5) If M is complete, then ϕ(M) = M∗ .
M, d
ϕ 1–1
d(p,q)=∆(ϕ(p),ϕ(q))
ϕ(M), ∆ ⊂ M∗ , ∆
dense complete
Remark.
(a) M∗ is called the completion of M.
(b) ϕ gives a one-to–one correspondance between elements of M and elements of ϕ(M). So we
can think of ϕ as giving the new name ϕ(x) to each x ∈ M and we can think of ϕ(M) as being
the same as M but with the names of the elements changed. Thus, we can think of M as being
ϕ(M) ⊂ M∗ . Using this point of view, the above theorem says that any metric space can be
extended to a complete metric space. I.e. can have its holes filled in.
(c) If M = Q, then we can take M∗ = IR, ∆(x, y) = |x − y| and ϕ(x) = x. It is easy to check that
(1)–(4) are obeyed.
Motivation.
The hard part of the proof is to make a guess as to what M∗ should be. That’s what we’ll do
now. A good strategy is to work backwards. Suppose that, somehow, we have found a suitable M∗
with M ⊂ M∗ . If we can find a way to describe each element of M∗ purely in terms of elements
of M, then we can turn around and take that as the definition of M∗ .
Consider, for example M = Q and M∗ = IR. Each element of IR may be written as the
√
limit of a sequence of rational numbers, and each such sequence is Cauchy. For example, 2 is
the limit of the Cauchy sequence 1, 1.4, 1.41, 1.414, 1.4142, · · ·. Thus specifying a real number
is equivalent to specifying a Cauchy sequence of rational numbers. But there is not a one–to–one
correspondance between real numbers and Cauchy sequences of rational numbers, because many
different Cauchy sequences of rational numbers converge to the same real number. For example
1, 1.4, 1.41, 1.414, 1.4142, · · ·
0, 1.3, 1.40, 1.413, 1.4141, · · ·
2, 1.5, 1.42, 1.415, 1.4143, · · ·
c Joel Feldman.
2014. All rights reserved.
October 19, 2014
Completion
1
√
all converge to 2. (For the second sequence, I just subtracted one from the last nonzero decimal
place. For the third sequence, I added one to the last nonzero decimal place.) To get a one–to–one
correspondance, we can identify each real number P with the set of all Cauchy sequences in Q that
converge to P .
Outline of Proof. First define
o
n
M′ =
pn n∈IN pn n∈IN is a Cauchy sequence in M
That is, M′ is the set of all Cauchy sequences in M. Next we define two Cauchy sequences pn
and qn n∈IN in M to be “equivalent”, written pn n∈IN ∼ qn n∈IN , if and only if
n∈IN
lim d(pn , qn ) = 0
n→∞
This definition is rigged so that any two convergent sequences have the same limit if and only if
they are equivalent. Next, if pn n∈IN , we define the “equivalence class of pn n∈IN ” to be the set
o
n
′ qn n∈IN ∈ M qn n∈IN ∼ pn n∈IN
pn n∈IN =
of all Cauchy sequences that are equivalent to pn n∈IN . We shall shortly prove
∈ M′ then
,
q
n
n∈IN
n∈IN
either
pn n∈IN = qn n∈IN
Lemma 1 If pn
or
pn
n∈IN
∩
qn
n∈IN
=∅
If you think of a Cauchy sequence as one person and an equivalence class of Cauchy sequences as
a “family” of related people, then the above Lemma says, that the whole world is divided into a
collection of nonoverlapping families. Next, we define
o
n
M∗ =
pn n∈IN pn n∈IN ∈ M′
as the set of all “families” and prove
Lemma 2 If pn
, qn
n∈IN
n∈IN
∈ M′ then lim d(pn , qn ) exists.
n→∞
Lemma 3 If pn n∈IN and p′n n∈IN are two representatives of P ∈ M∗ (this means that P =
pn n∈IN = p′n n∈IN ) and qn n∈IN and qn′ n∈IN are two representatives of Q ∈ M∗ , then
lim d(pn , qn ) = lim d(p′n , qn′ )
n→∞
n→∞
This allows us to define ∆(P, Q). If P, Q ∈ M∗ and pn
is a representative of Q ∈ M∗ , we define
n∈IN
is a representative of P and qn
∆(P, Q) = lim d(pn , qn )
n∈IN
(1)
n→∞
By Lemma 3, we always get the same value for ∆(P, Q) no matter which representatives for P and
Q are used in (1). Finally, we define ϕ : M → M∗ by
ϕ(p) = p, p, p, · · ·
The conclusions of the Theorem are now proven as a series of Lemmata.
c Joel Feldman.
2014. All rights reserved.
October 19, 2014
Completion
2
Lemma 4 ∆ is a metric on M∗ .
Lemma 5 M∗ is complete
Lemma 6 ∆ ϕ(p), ϕ(q) = d(p, q) for all p, q ∈ M.
Lemma 7 ϕ is one–to–one.
Lemma 8 ϕ(M) is dense in M∗ .
Lemma 9 If M is complete, then ϕ(M) = M∗ .
So now we just have to prove all of the Lemmata.
Lemma 1 If pn
n∈IN
either
, qn
pn
n∈IN
n∈IN
∈ M′ then
=
qn
n∈IN
or
pn
n∈IN
∩
qn
n∈IN
=∅
Proof: Suppose that the conclusion is false. Then pn n∈IN ∩ qn n∈IN must be nonempty so
that there is a Cauchy sequence an n∈IN that is in both pn n∈IN and qn n∈IN . Consequently
(an ) is equivalent to both (pn ) and (qn ) so that
lim d(an , pn ) = 0 and lim d(an , qn ) = 0
n→∞
n→∞
On the other hand pn n∈IN and qn n∈IN must be different, so that there is a Cauchy sequence
(b ) that is in one of them (say pn n∈IN ) and not in the other (say qn n∈IN ). Since (bn ) is in
n pn n∈IN ,
lim d(bn , pn ) = 0
n→∞
By the triangle inequality
d(bn , qn ) ≤ d(bn , pn ) + d(pn , an ) + d(an , qn )
But this forces lim d(bn , qn ) = 0, which contradicts the assumption that (bn ) is not in
n→∞
Lemma 2 If pn
c Joel Feldman.
n∈IN
, qn
n∈IN
qn
n∈IN
.
∈ M′ then lim d(pn , qn ) exists.
2014. All rights reserved.
n→∞
October 19, 2014
Completion
3
Proof:
We first prove
d(pn , qn ) − d(pm , qm ) ≤ d(pn , pm ) + d(qm , qn )
qn
qm
pn
pm
(2)
by combining the two right hand inequalities of
d(pn , qn ) ≤ d(pn , pm ) + d(pm , qm ) + d(qm , qn ) ⇒ d(pn , qn ) − d(pm , qm ) ≤ d(pn , pm ) + d(qm , qn )
d(pm , qm ) ≤ d(pm , pn ) + d(pn , qn ) + d(qn , qm )
⇒ d(pm , qm ) − d(pn , qn ) ≤ d(pn , pm ) + d(qn , qm )
Since pn n∈IN and qn n∈IN are both Cauchy, both d(pn , pm ) and d(qm , qn ) converge to zero as
m, n → ∞. So the same is true for d(pn , qn ) − d(pm , qm ). So d(pn , qn ) n∈IN is a Cauchy sequence
of real numbers. Since IR is complete, this Cauchy sequence converges.
Lemma 3 If pn n∈IN and p′n n∈IN are two representatives of P ∈ M∗ (this means that P =
pn n∈IN = p′n n∈IN ) and qn n∈IN and qn′ n∈IN are two representatives of Q ∈ M∗ , then
lim d(pn , qn ) = lim d(p′n , qn′ )
n→∞
Proof:
n→∞
Replace, in (2), pm and qm by p′n and qn′ respectively. This gives
d(pn , qn ) − d(p′ , q ′ ) ≤ d(pn , p′ ) + d(q ′ , qn )
n
n
n n
, we have that lim d(pn , p′n ) = 0. Since (qn′ ) ∈
n→∞
lim d(qn , qn′ ) = 0. Consequently, lim d(pn , qn ) − d(p′n , qn′ ) = 0.
Since (p′n ) ∈
pn
n∈IN
n→∞
qn
n∈IN
, we have that
n→∞
Lemma 4 ∆ is a metric on M∗ .
Proof:
Let
P =
pn
n∈IN
∈ M∗
Q=
qn
n∈IN
∈ M∗
R=
rn
n∈IN
∈ M∗
(i) Since every d(pn , qn ) ≥ 0, we have ∆(P, Q) = lim d(pn , qn ) ≥ 0. On the other hand, if
n→∞ ∆(P, Q) = 0, then lim d(pn , qn ) = 0 so that pn n∈IN ∼ qn n∈IN and P = Q.
n→∞
(ii) ∆(P, Q) = lim d(pn , qn ) = lim d(qn , pn ) = ∆(Q, P ).
n→∞
n→∞
(iii) ∆(P, Q) = lim d(pn , qn ) ≤ lim
n→∞
c Joel Feldman.
n→∞
2014. All rights reserved.
d(pn , rn ) + d(rn , qn ) = ∆(P, R) + ∆(R, Q).
October 19, 2014
Completion
4
Lemma 5 M∗ is complete
Proof: Let Q(n) ∈ M∗
be a Cauchy sequence. We must prove that it has a limit, Q ∈ M∗ ,
(n) as n → ∞. Each Q(n) is an equivalence class of Cauchy sequences in M. Say Q(n) = qℓ ℓ∈IN .
(n)
We shall guess Q = qn n∈IN by choosing each qn to be a qℓ with ℓ chosen larger and larger as
n increases. Here we go:
(1) (1) (1)
(1)
qℓ ℓ∈IN Cauchy ⇒ ∃ ℓ1 ∈ IN s.t. ℓ ≥ ℓ1 ⇒ d(qℓ , qℓ1 ) < 1. Pick q1 = qℓ1 .
(2)
(2) (2)
(2)
qℓ ℓ∈IN Cauchy ⇒ ∃ ℓ2 > ℓ1 s.t. ℓ ≥ ℓ2 ⇒ d(qℓ , qℓ2 ) < 12 . Pick q2 = qℓ2 .
(3) (3) (3)
(3)
qℓ ℓ∈IN Cauchy ⇒ ∃ ℓ3 > ℓ2 s.t. ℓ ≥ ℓ3 ⇒ d(qℓ , qℓ3 ) < 13 . Pick q3 = qℓ3 .
..
.
(n)
(n)
(n) (n)
qℓ ℓ∈IN Cauchy ⇒ ∃ ℓn > ℓn−1 s.t. ℓ ≥ ℓn ⇒ d(qℓ , qℓn ) < n1 . Pick qn = qℓn .
..
.
(n)
In the example sketched below, the qℓn ’s are circled.
n∈IN
Q(1)
Q(2)
Q(3) · · ·
(2)
(1)
q4
(1)
q3
(1)
q2
(1)
q1
q4
(2)
q3
(2)
q2
(2)
q1
is Cauchy: Let ε > 0. By the triangle inequality
(n) (n) (n) (m) (m) (m) d(qn , qm ) ≤ d qℓn , qℓ
+ d qℓ , qℓ
+ d qℓ , qℓ m
(n) (m) (n) (n) (m) (m) = d qℓ n , qℓ
+ d qℓ , qℓ
− ∆(Q(n) , Q(m) ) + ∆(Q(n) , Q(m) ) + d qℓ , qℓm
Proof that qn
n∈IN
(3)
for any ℓ ∈ IN.
◦ If ℓ ≥ ℓn , the first term is smaller than n1 .
(n) (m) ◦ By definition, ∆(Q(n) , Q(m) ) = lim d qℓ , qℓ
. So there is a natural number Nn,m such
ℓ→∞
that the second term is smaller than 4ε whenever ℓ ≥ Nn,m .
◦ By hypothesis, the sequence (Q(n) ) is Cauchy. So there is a natural number Ñ such that the
third term is smaller than 4ε whenever n, m ≥ Ñ .
1
◦ Finally, if ℓ ≥ ℓm , the last term is smaller than m
.
4
Choose any natural number N ≥ max Ñ , ε . I claim that d(qn , qm ) < ε whenever n, m ≥ N . So
let n, m ≥ N . Now choose ℓ to be any natural number bigger than max Nn,m , ℓn , ℓm . Then the
four terms in (3) are each smaller than 4ε .
c Joel Feldman.
2014. All rights reserved.
October 19, 2014
Completion
5
Proof that Q = lim Q(n) : Let ε > 0. By definition
n→∞
(m)
(n)
(n)
∆(Q, Q(n) ) = lim d(qm , qm
) = lim d qℓm , qm
m→∞
m→∞
By the triangle inequality
(m) (n) (m) (n) (n) (n) d qℓ m , qm
≤ d qℓ m , qℓ n + d qℓ n , qm
(3)
(n) ◦ Since the sequence qn = qℓn is Cauchy, there is a natural number N ′ such that the first
term is smaller than 2ε whenever n, m ≥ N ′ .
◦ By the construction of ℓn , the second term is smaller than n1 whenever m ≥ ℓn .
Choose any natural number N ≥ max N ′ , 2ε . I claim that ∆(Q, Q(n) ) < ε whenever n ≥ N . So
let n ≥ N . Now choose m to be any natural number bigger than max N ′ , ℓn . Then the two
terms in (4) are each smaller than 2ε .
Lemma 6 ∆ ϕ(p), ϕ(q) = d(p, q) for all p, q ∈ M.
Proof:
∆ ϕ(p), ϕ(q) = lim d ϕ(p)n , ϕ(q)n = lim d(p, q) = d(p, q)
n→∞
n→∞
Lemma 7 ϕ is one–to–one.
Proof:
ϕ(p) = ϕ(q) =⇒ ∆ ϕ(p), ϕ(q) = 0 =⇒ d(p, q) = 0 =⇒ p = q.
Lemma 8 ϕ(M) is dense in M∗ .
Proof: Let P = pn n∈IN ∈ M∗ . Then I claim that the sequence ϕ(pm ) m∈IN converges in
M∗ to P . To check this, it suffices to observe that
∆ P, ϕ(pm ) = lim d pn , ϕ(pm )n = lim d(pn , pm )
n→∞
n→∞
Since pn n∈IN is Cauchy, this converges to zero as m → ∞.
Lemma 9 If M is complete, then ϕ(M) = M∗ .
Proof:
Let Q ∈ M∗ . We must find q ∈ M with ϕ(q) = Q. By Lemma 8
∃ qn ∈ M n∈IN s.t. Q = lim ϕ(qn )
n→∞
=⇒ ϕ(qn ) n∈IN is Cauchy in M∗
=⇒ qn n∈IN is Cauchy in M, by Lemma 6
=⇒ q = lim qn exists, since M is complete
n→∞
=⇒ ϕ(q) = lim ϕ(qn ) by Lemma 6
n→∞
=⇒ ϕ(q) = Q
c Joel Feldman.
2014. All rights reserved.
October 19, 2014
Completion
6