(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!
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??
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?
The metric on $\displaystyle \overline{X}$ 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 $\displaystyle \overline{X}$.