COMPLETION OF A METRIC SPACE
HOW ANY INCOMPLETE METRIC SPACE CAN BE COMPLETED
REBECCA AND TRACE
Given any incomplete metric space (X,d), ∃(X̄, dX̄ ) a completion, with (X,d) ⊂ (X̄, dX̄ )
where X̄ complete, and X is not closed in X̄. Specifically, the closure of X will, with respect
to dX̄ , be X̄
1. Show equivalence relation
∞
Given any Cauchy sequence {xn }
in X we introduce the formal limit LIMn→∞ xn . we
n=1
say LIMn→∞ xn = LIMn→∞ yn if lim d(xn , yn ) = 0. This forms an equivalence relation
n→∞
on X̄
1.1. Reflexivity.
It can be easily seen shown that LIMn→∞ xn = LIMn→∞ xn since ∀x ∈ X lim d(xn , xn ) =
n→∞
0. So the relation is reflexive.
1.2. Symmetry.
Since d is a metric, then ∀x,y d(x,y) = d(y,x) ⇒ if lim d(xn , yn ) = 0, then lim d(yn , xn ) =
n→∞
n→∞
0. So the relation is symmetric.
1.3. Transitive.
If LIMn→∞ xn = LIMn→∞ yn , and LIMn→∞ yn = LIMn→∞ zn then that means that
lim d(xn , yn ) = 0, and that lim d(yn , zn ) = 0. Then by the triangle equality
n→∞
n→∞
0 ≤ d(xn , zn ) ≤ d(xn , yn ) + d(yn , zn )
0 ≤ lim d(xn , zn ) ≤ lim d(xn , yn ) + lim d(yn , zn )
n→∞
n→∞
n→∞
0 ≤ lim d(xn , zn ) ≤ 0 + 0
n→∞
By the squeeze theorem
lim d(xn , zn ) = 0
n→∞
Since ∀x,y d(x,y) ≥ 0 then we can say that lim d(xn , zn ) = 0. Showing that the relation
n→∞
is transitive.
1
2
REBECCA AND TRACE
¯ is a metric space
2. Show(X̄, d)
Let X̄ be the space of all formal limits of Cauchy sequences in X, with the above relation.
We define a metric dX̄ : X̄ × X̄ → R+ by setting
dX̄ (LIMn→∞ xn , LIMn→∞ yn ) ..= lim d(xn , yn ).
n→∞
2.1. Well defined.
Let x=x’, y= y’, we would like to show that dX̄ (x, y) = dX̄ (x’, y’). It suffices to show that
dX̄ (x, y) = dX̄ (x, y’)
dX̄ (xn , yn ) ≤ dX̄ (xn , yn0 ) + dX̄ (yn0 , yn )
dX̄ (xn , yn0 ) ≤ dX̄ (xn , yn ) + dX̄ (yn , yn0 )
note : dX̄ (yn0 , yn ) = dX̄ (yn , yn0 )
−dX̄ (yn , yn0 ) ≤ dX̄ (xn , yn ) − dX̄ (xn , yn0 ) ≤ dX̄ (yn , yn0 )
lim −dX̄ (yn , yn0 ) ≤ lim dX̄ (xn , yn ) − lim dX̄ (xn , yn0 ) ≤ lim dX̄ (yn , yn0 )
n→∞
n→∞
n→∞
n→∞
0 ≤ lim dX̄ (xn , yn ) − lim dX̄ (xn , yn0 ) ≤ 0
n→∞
n→∞
By the squeeze theorem
lim d (xn , yn )
n→∞ X̄
= lim dX̄ (xn , yn0 )
n→∞
2.2. Gives the structure of a metric space.
Since dX̄ (LIMn→∞ xn , LIMn→∞ yn ) ..= lim d(xn , yn ) then ∀LIMn→∞ xn , LIMn→∞ yn .
n→∞
(1) dX̄ is non negative:
since d is a metric, d(xn , yn ) ≥ 0 for all n > 0, hence
dX̄ (LIMn→∞ xn , LIMn→∞ yn ) = lim d(xn , yn ) ≥ 0
n→∞
(2) Identity holds:
By definition of LIM: dX̄ (LIMn→∞ xn , LIMn→∞ yn ) = lim d(xn , yn ) = 0
n→∞
⇒ LIMn→∞ xn = LIMn→∞ yn
(3) dX̄ is symmetric:
since d is a metric: d(xn , yn ) = d(yn , xn ) for all n > 0, hence
dX̄ (LIMn→∞ xn , LIMn→∞ yn ) = lim d(xn , yn )
n→∞
dX̄ (LIMn→∞ yn , LIMn→∞ xn ) = lim d(yn , xn )
n→∞
lim d(yn , xn ) = lim d(xn , yn )
n→∞
n→∞
⇒ dX̄ (LIMn→∞ xn , LIMn→∞ yn ) = dX̄ (LIMn→∞ yn , LIMn→∞ xn )
COMPLETION OF A METRIC SPACE HOW ANY INCOMPLETE METRIC SPACE CAN BE C
(4) The triangle inequality holds: since d is a metric for all n > 0
d(xn , zn ) ≤ d(xn , yn ) + d(yn , zn ),
hence
lim d(xn , zn ) ≤ lim d(xn , yn ) + lim d(yn , zn )
n→∞
n→∞
n→∞
but since
dX̄ (LIMn→∞ xn , LIMn→∞ yn ) = lim d(xn , yn )
n→∞
dX̄ (LIMn→∞ yn , LIMn→∞ zn ) = lim d(yn , zn )
n→∞
dX̄ (LIMn→∞ xn , LIMn→∞ zn ) = lim d(xn , zn )
n→∞
we conclude that
dX̄ (LIMn→∞ xn , LIMn→∞ zn ) ≤ dX̄ (LIMn→∞ xn , LIMn→∞ yn )+dX̄ (LIMn→∞ yn , LIMn→∞ zn ).
¯
3. Show (X,d) is a subspace of (X̄, d)
Let x ∈ X, with LIMn→∞ x ∈ X̄ we say that x = y ⇐⇒ LIMn→∞ x = LIMn→∞ y.
Let’s assume LIMn→∞ x = LIMn→∞ y then dX̄ (LIMn→∞ x, LIMn→∞ y) ..= lim d(x, y)=
n→∞
0 by definition. Similarly for the other direction. This shows that dX = dX̄ which means
we that dx can be represented by dX̄ |x which shows that X is a subspace of X̄
¯ is complete
4. Show (X̄, d)
Let ϕ :X→ X̄ be defined by ϕ(x) := LIM {x}. Then ϕ is an isometry since for x and y in
X we have,
ϕ(x) := LIM {x} and ϕ(y) := LIM {y}
and dX̄ (ϕ(x), ϕ(y)) = lim d(x, y) = d(x, y).
n→∞
Definition of Density of ϕ(X) ⊆ X̄: ϕ(X) is dense in X̄ if given any x̄ ∈ X̄ and > 0 we
can find a point x ∈ X so that ϕ(x) is -close to x̄. Now we show how to find x in X given
x̄ and .
Given any x̄ = LIM {xn } in X̄ with {xn } Cauchy in X, ∀ε > 0 ∃N > 0, ∀n, m ≥ N
d(xn , xm ) < ε, m = N
⇒ ∀n ≥ N, d(xn , xN ) ≤ ε
⇒ lim d(xn , xN ) ≤ ε
n→∞
Let x = xN ∈ X then
ϕ(x) := LIM {xN } and
dX̄ (x̄, ϕ(x)) = lim d(xn , xN ) ≤ ε.
n→∞
4
REBECCA AND TRACE
This is the first step to completeness: Cauchy sequences in the dense subset ϕ(X) are
convergent in X̄.
If {xk } is Cauchy in X̄ and {xk } ⊂ ϕ(X) = A with A dense in X̄,
xk = ϕ(xk ), with xk ∈ X
Then given ε > 0, ∃N > 0, n, m > N and since ϕ is an isometry
dX̄ (xn , xm ) = dX̄ (ϕ(xn ), ϕ(xm )) = d(xn , xm ) < ε
⇒ {xn }in X is Cauchy for {xk } ⊂ ϕ(X)
The reverse implication also holds.
We just showed that if xk = ϕ(xk ) = LIM {xk } then
is Cauchy in X̄ ⇔ {xk } ⊂ X is Cauchy in X.
{xk }
Let x̄ = LIM {xn } ∈ X̄.
We need to show that the Cauchy sequence xk → x̄ in X̄. So we need to have that for
∀ε > 0 ∃N > 0 with k > N
dx̄ (xk , x̄) = dx̄ (LIM {xk }, LIM {xn }) = lim d(xk , xn ) < ε.
n→∞
The last inequality holds for k, n ≥ N because the sequence {xk } is Cauchy in X.
Given (X̄, dx̄ ) metric space, we see that A=ϕ(X) is a subset of X̄ such that every Cauchy
sequence in A converges in X̄. And since A is dense in X̄ we can show that X̄ is complete,
by Lemma below, finishing the proof.
Lemma: Let (Z, ρ) be a metric space, A a dense subset of Z. Suppose that given any
Cauchy sequence {an } ⊂ A then the sequence is convergent in Z. Then Z is complete.
Proof: Take a Cauchy sequence {zk } ∈ Z, since A is dense then for each zk we can find
a Cauchy sequence {akn } ∈ A such that lim ρ(akn , zk ) = 0. This means that given > 0
n→∞
there is Nk > 0 such that for all n ≥ Nk
ρ(akn , zk ) ≤ /3.
COMPLETION OF A METRIC SPACE HOW ANY INCOMPLETE METRIC SPACE CAN BE C
Increasing if necessary the values of Nk we can ensure that N1 < N2 < N3 < . . . . In
particular, if we choose n = Nk get for all k ≥ 1
ρ(akNk , zk ) ≤ /3.
Claim 1: {an := anNn }n≥1 is Cauchy in A, hence convergent to some z0 ∈ Z by hypothesis:
ρ(an , am ) ≤ ρ(an , zn ) + ρ(zn , zm ) + ρ(zm , am )
ρ(an , am ) ≤ ρ(anNn , zn ) + ρ(zn , zm ) + ρ(am
Nm , zm )
ρ(an , am ) ≤ /3 + /3 + /3,
for the middle term we used the hypothesis that the sequence {zk } is Cauchy. Hence there
is z0 ∈ Z such that
lim an = z0 so then for n ≥ N
n→∞
ρ(an , z0 ) ≤ /2
Finally we show that the Cauchy sequence {zn } converges to that z0 .
Claim 2: d(zn , z0 ) < for n ≤ N
d(zn , z0 ) ≤ d(zn , an ) + d(an , z0 )
d(zn , z0 ) ≤ /2 + /2 ≤ .
Lastly, we let X̄ = Z, dx̄ = ρ and A = A Then we have by the Lemma that X̄ is in fact
complete.
*** A final note to the reader: You will recall that it took us half a semester to show the
completion of the reals. Because we did all of that work, and we already knew that the
reals were complete, we were able to show the completion of a given metric space in much
less time. We were able to define our metric dX̄ : X̄ × X̄ → R+ with
dX̄ (LIMn→∞ xn , LIMn→∞ yn ) ..= lim d(xn , yn ).
n→∞
because we know that the right-hand-side limit exists. Since we have that, {xn }n≥1 and
{yn }n≥1 are Cauchy sequences in X then the sequence of real numbers {d(xn , yn )}n≥1 is a
Cauchy sequence in R and since R is complete the sequence must be convergent to a real
number L = lim d(xn , yn ).***
n→∞