Thread: Show that X bar (X closure) is a completion of X.

(X,dx), a metric space, is embedded in (Y,dy), another metric space. (Y,dy) is complete and (X,dx) has dx as the induced metric. Show that X closure (Xbar) is a completion of X.

Thanks a lot!

Do you know what you have to show?

I have to show that there is a subspace W of X bar that is dense in X bar and W is isometric to X.

If dx is the induced metric of dy on X, why won't you try W=X?

So if I let W=X, then "clearly" W and X are isometric. So all that remains to show is that X (or W) is everywhere dense in X bar. So, if I let a sequence in X be x_n --> x in X bar. Then d (x_n,x) < epsilon. So we can make the distance between them as small as we want. So X is dense in X bar??

Yes, that's true (and that's why I don't understand exactly the point of the exercise).

When are we using the fact that Y is complete?

I feel like using your suggestion makes the proof almost trivial!

To see that is complete (because each closed subset of a complete metric space is complete for the induced metric).

Yep! I get you. Thank you soooooo much!

The question was from a past exam. You've been very helpful.

Oh just btw, you know how in the general proof for completion, we use the metric on the completion as lim n--> inf of d (x_n, y_n). Here we can just use the same metric right? How do we justify the use of the same metric for X bar and X?

Since X has the induced metric of Y, you can give to the same metric.

Because X closure would be the smallest subset containing X, so if X is a subspace of Y, then either X closure is in Y or is Y. So, X bar will also have the same induced metric?

Anyway, thanks a lot!

The metric on is the same as the induced metric on X when you take two elements in X. By continuity, we can check that it's the case for all the elements of .