A short proof of completion theorem for metric spaces

The completion theorem for metric spaces states that every metric space can be embedded in a complete metric space and the original space's image is dense in that complete space. In several sources, the proof is given by a classical method based on the space of the equivalence classes of all the Cauchy sequences, denoted by X ^ˆ , of the given metric space X , see [1]. In this proof, it is shown that X ^ˆ is a complete metric space, X can be embedded in X , the image of X is dense in X ^ˆ and this completion is unique up to isometry.

In this paper, we propose another way to prove the completion theorem by not using the embedding. We prove that if a metric space X i s not complete, there exists a complete metric space X ^ˆ including X such that X is dense in XX , i.e. X = X .

The proof

Lemma 1. Let ( X,d ) be a metric space and ( x n ) be a Cauchy sequence that is not convergent in X . We define a new space X = X u { c } with the metric d as follows:

d ( x , У ) = <

' d ⁽ x , y ) , lim d ( ^x _n , У ) , n ^w lim d ( x , x n ) , n ^w

. 0,

if x, У e X, if x = c, y e X,

if x e X , y = c ,

if x = У = ^c ,

where c is an element that does not belong to X . Then, ( X, d ) is a metric space satisfying the properties d ( x, y ) = d ( x, y ) for each x,y e X , lim x_n = c and X = X .

n ^w

Proof. By the reverse triangle inequality |d(xn,y)-d(xm,y)| < d(xn,xm), the sequence {d(xn,y)} is a Cauchy sequence in R because (xn) is a Cauchy sequence in X. Since R is complete, then {d(xn,y)} is a convergent sequence for each y e X. Therefore, the function d is well-defined from X x X to R. Also, it is easy to see that d is a metric on the set X. Now, we show that the sequence (xn) converges to c in the metric space (X, d). Given s > 0, there exists a natural number ne such that d (x„, x„) < — for all n, m > n,. Then, we have nm            ε

Краткие сообщения d ( xn, c) = lim d ( xn, xm )< S < S m ^rc            2

for each n > n . This proves the above assertion and also proves that X = X .

Lemma 2. Let ( X , d ) be a metric space included two complete metric spaces ( Y_x , p ) and ( Y , P ) with the conditions p ( x , x * ) = d ( x , x * ) and p ( x , x * ) = d ( x , x * ) for every x , x * e X . If X is both dense in ( Y , p ) and ( Y , p ) , then these complete metric spaces are isomorphic to each other.

Proof. Let t e Y . Since X = Y , there exists a sequence ( x ) о X such that lim p ( x_n , t ) = 0 .

^{x    7}                        n ^rc    ^{x      7}

Then, (xn) is a Cauchy sequence in Y, and also is a Cauchy sequence in Y. Since (Y, p ) is a complete space, then there exists y in Y such that lim p (xn , У ) = 0 . Let f : Yi ^ Y2 , У = f (t) . We n ^rc now show that the function f is an isometry. First, we prove that it is well-defined. Let (zn) о X be also convergent to the point t in the space (Y, p). Then, similarly, there exists z e Y such that lim p2 ( n ^rc    x

z n , z ) = ⁰ .

p 2 ( У , z ) = p 2 ( ^lim x n ,^lim z n ) = ^lim p 2 ( x n , z n ) = ^{lim d} ( x n , z n ) = ^lim p l ( x n , z n )

\ n ^rc      n ^rc / n ^rc     ^{x        z}     n ^rc ^{x        z}     n ^rc

= p i ( lim x n ,lim z_n ) = p i ( t , t ) = 0 ^ y = z .

\ n >rc       n >rc /

Second, we prove that f is surjective. Let y e Y . Since X = Y, there exists a sequence (xn ) о X such that lim p (xn, y) = 0. Then, (x) is a Cauchy sequence in X and also in Y. Since Y is com-n ^rc plete, then there exists t e Y such that lim p (x, t) = 0 . By the construction of f, y = f (t) . Final-n ^rc ly, we prove that f is an isometry. Let tx, t2 e Y .Then, there exist two sequences (xn), (zn) о X such that lim pi (xn, ti ) = 0, limpi (zn, 12 ) = 0, limp2 (x„, f (ti )) = 0 and limp2 (zn, f (12 )) = 0.

n ^rc    ^x                 n ^rc                      n ^rc     ^{x            z}              n ^rc     ^{x             z}

( lim x n ,lim z_n

\ n ^rc     n ^rc

p 2 ( ^f ( t i ) , ^f ( t 2 ) ) = p 2

) = lim p 2 ( x n , z n ) = ^{lim d} ( ^x n , ^z n ) = ^lim p i ( x n , z n ) ) n ^rc                  n ^rc                n ^rc

= p i ( ^lim x n ,^lim z n ) = p i ( t i , t 2 ) .

\ n ^rc     n ^rc /

This completes the proof.

Theorem 1. Every metric space has a unique completion up to isometry.

Proof. Consider the family Q of metric spaces ( Y,p ) satisfying the following conditions:

• 1)    ρ is a metric on Y ,

• 2)    X о Y ,

• 3)    p ( x,y ) = d ( x,y ) for each x,y e X ,

• 4)    X = Y .

We define a relation on Q which is as follows:

( Y i , p _i ) < ( Y 2 , p ₂ ) ^ Y_i о Y₂ and p ₂ ( x , y ) = p _i ( x , y ) for each x , y e Y_i

It is easy to see that the pair ( Q , < ) is a poset. We take a chain Q * in the poset ( Q , < ) and define

Y ■= и Y ( Y , p ) eQ *

Kaya U.

A Short Proof of Completion Theorem for Metric Spaces

We now define a function p* from Y*x Y* to R as follows. If x,y e Y* , then there exists (Yo, p0 )eQ* such that x, y e Yo because Q* is a chain. Let p ( x, y ) = Po ( x, y ) •

Then, the function p is well-defined by the definition of the relation < . One can easily show that p* is a metric on the set Y*. Let y e Y*, there exists (Y, p)e Q* such that y e Y. Since (Y, p)eQ, then there exists a sequence (xn) in X such that lim p(xn,y) = 0 , i.e. lim p (xn,y) = 0. Consequent n ^да

n ^да

ly, X = Y* . These results show that the metric space ( Y *, p * ) belongs to Q and forms an upper bound of the chain Q * . By Zorn’s lemma, Q has a maximal element and we denote it by ( X,d ) . We now prove that the metric space ( X, d ) is a completion of ( X,d ) . Since ( X, d ) is an element of Q , then we just prove that ( X, d ) is complete. Assume the contrary. If ( X, d ) has a non-convergent Cauchy sequence, then Lemma 1 requires that there exists a metric space ( X* ,d *) such that X* = X u { c *} and ( X , d ) <  ( X *, d *) , where c * is a point not in X . We now show that ( X *, d * ) eQ . Indeed, it is enough to show that c * is an accumulation point of the original set X . Given £ >  0 . By Lemma 1, there exists c e X such that d ‘ ( c, c * ) <  ^£ . Besides, by the relation ( X,d ) eQ , there exists x e X such that d ( x, c ) <  ^£ . Then, d * ( x, c *) <  d * ( x, c ) + d * ( c, c *) = d ( x, c ) + d * ( c, c *) <  ^ + ^ = £ . Thus, we have ( X *, d *) eQ . The last and the relation ( X,d ) < ( X *, d *) contradict the maximality of ( X,d ) . This completes the proof.

The uniqueness up to isometry of the completion is directly obtained by Lemma 2.