# Metric space completion

## Homework Statement

Let X be an incomplete metric space. and Let X' denote its completion. I would like to show that there is Cauchy sequence in X which does not converge in X but does converge in X'. Moreover, I want to show that X contains every element of the sequence except the limit point.

## Homework Equations

No equations as such other than the

Definition: A complete metric space is one in which every Cauchy sequence converges.

Definition: A Cauchy sequence { x_n } is one for which given any epsilon > 0 there exists a natural N such that if m,n > N then d(x_n,x_m) < epsilon.

## The Attempt at a Solution

Well, negating the definition of a metric space we should be able to find a cauchy sequence which does not converge in X.

My next line of argument would be to say that since X is isometrically embedded in X'. Then { x_n } is also a Cauchy sequence in X' and thus converges in X'.

However, I'm not sure if it is always the case that X' \ X will always only contain the limit point. Can it not contain other points in the sequence?

Last edited:

HallsofIvy
Homework Helper

## Homework Statement

Let X be an incomplete metric space. and Let X' denote its completion. I would like to show that there is Cauchy sequence in X which does not converge in X but does converge in X'. Moreover, I want to show that X contains every element of the sequence except the limit point.

## Homework Equations

No equations as such other than the

Definition: A complete metric space is one in which every Cauchy sequence converges.

Definition: A Cauchy sequence { x_n } is one for which given any epsilon > 0 there exists a natural N such that if m,n > N then d(x_n,x_m) < epsilon.

## The Attempt at a Solution

Well, negating the definition of a metric space we should be able to find a cauchy sequence which does not converge in X.
I assume that was a typo and you mean "negating the definition of complete". Yes, since you are given that X is NOT complete, there must exist a Cauchy sequence in X which does not converge.

My next line of argument would be to say that since X is isometrically embedded in X'. Then { x_n } is also a Cauchy sequence in X' and thus converges in X'.

However, I'm not sure if it is always the case that X' \ X will always only contain the limit point. Can it not contain other points in the sequence?
Technically, there exist an isometric ((distance preserving), one-to-one function , f(xn)= x'n is in X'. Can you determine whether x'n is a Cauchy sequence in X'? If so, then since you know that X' is complete, the sequence {x'n} must converg to some point x'. Does there exist x in X such that f(x)= x'? If so what would be its relation to the sequence {xn}?

I assume that was a typo and you mean "negating the definition of complete". Yes, since you are given that X is NOT complete, there must exist a Cauchy sequence in X which does not converge.

Technically, there exist an isometric ((distance preserving), one-to-one function , f(xn)= x'n is in X'. Can you determine whether x'n is a Cauchy sequence in X'?

Yes, we just choose the same natural number N since d(x_n,x_m) = d'(x_n',x_m').

If so, then since you know that X' is complete, the sequence {x'n} must converg to some point x'. Does there exist x in X such that f(x)= x'? If so what would be its relation to the sequence {xn}?

If there existed x in X such that f(x) = x'. Then d'(x', x_m') = d(x, x_m). But lim d(x,x_m) = lim d'(x',x_m') = 0 which would imply that { x_n } converges in X (contradiction).

I'm still struggling to show that f(X) contains every element of the sequence except the limit point. Is this actually true?

HallsofIvy