Completion of metric spaces
Shiu-Tang Li
Finished: April 10, 2013
Last updated: November 2, 2013
Theorem. Let (M, d) be a metric space. Then, there exists a metric
˜ and an isometry φ : M → M̃ so that φ(M ) = M̃ .
space (M̃ , d)
Proof.
Step 1. First we define an equivalence relation ∼ on all the Cauchy
sequences in M by {an }n ∼ {bn }n whenever limn→∞ d(an , bn ) = 0, and the
equivalent class is denoted by [{an }n ].
Let M̃ be the collection of all such equivalent classes [{an }n ], and let the
metric d˜ on M̃ be defined as
˜
d([{a
n }n ], [{bn }n ]) = lim d(an , bn ).
n→∞
To see d˜ is well-defined, we pick two representatives {an }n , {bn }n from
[{an }n ], [{bn }n ], respectively. Since both {an }n , {bn }n are Cauchy sequences,
{d(an , bn )}n is also a Cauchy sequence in R (by the triangular inequality)
and hence the limit of {d(an , bn )}n exists. It’s not hard to see from the
triangular inequality that the limit remains the same if we choose different
representatives. Besides, d˜ satisfies the triangular inequality, since we have
˜
˜
d([{a
n }n ], [{bn }n ]) + d([{bn }n ], [{cn }n ])
= lim d(an , bn ) + lim d(bn , cn )
n→∞
n→∞
= lim d(an , bn ) + d(bn , cn )
n→∞
≥ lim d(an , cn )
n→∞
˜
=d([{a
n }n ], [{cn }n ]).
Step 2. We now define φ : M → M̃ by φ(a) = [{a, a, · · · }], and it
is an isometry. To see φ(M ) = φ(M ) ∪ φ(M )0 = M̃ , we first prove that
1
M̃ ⊂ φ(M ) ∪ φ(M )0 . For any a := [{an }n ] ∈ M̃ , if [{an }n ] = [{b, b, · · · }]
for some b ∈ M , then a ∈ φ(M ) . If a ∈
/ φ(M ), then we define bn :=
[{an , an , an , · · · }] for all n ∈ N, and bn 6= a for all n ∈ N. It follows that
˜ bn ) = limj→∞ d(aj , an ). Since {an }n is a Cauchy sequence, we have
d(a,
˜ bn ) = 0. As a result, a ∈ φ(M )0 .
limn→∞ d(a,
Step 3. To see φ(M ) ⊂ M̃ , it suffices for us to check M̃ is a closed space.
For any Cauchy sequence {xn }n ∈ M̃ , where each xn := [{an,j }j ], we define
a function ψ : N → N by ψ(n) := min{N ∈ N : d(an,N , an,m ) < 2−n for all
m ≥ N }.
We claim that {an,ψ(n) }n is a Cauchy sequence in M . For every > 0,
˜ n , xm ) =
we may find some N = N () so that for any n, m > N , we have d(x
limj→∞ d(an,j , am,j ) < 6 , and we select K = K(n, m) > max(n, m) so that
d(an,K , am,K ) < 3 . Now, for every n, m > max(N, log2 3 ), we have
d(an,ψ(n) , am,ψ(m) ) ≤ d(an,ψ(n) , an,K ) + d(an,K , am,K ) + d(am,ψ(m) , am,K )
1
1
< n + + m < ,
2
3 2
which proves our claim.
Define x := [{aj,ψ(j) }j ] ∈ M̃ . We have
˜ n , x) = lim d(an,j , aj,ψ(j) )
d(x
j→∞
≤ lim d(an,j , an,ψ(n) ) + d(an,ψ(n) , aj,ψ(j) )
j→∞
= lim d(an,j , an,ψ(n) ) + lim d(an,ψ(n) , aj,ψ(j) )
j→∞
−n
≤2
j→∞
+ lim d(an,ψ(n) , aj,ψ(j) ).
j→∞
Since {aj,ψ(j) }j is a Cauchy sequence in M , limn→∞ limj→∞ d(an,ψ(n) , aj,ψ(j) ) =
˜ n , x) = 0, and we’ve proved that M̃ is a closed space.
0. Thus, limn→∞ d(x
2