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Surjectivity of an Isometry given the metric space is complete.


New member
Mar 7, 2012
Hello, the following is a post that was in progress and Im continuing it here after I recieved a message saying that most of the members had moved from mathhelpforum here.

I have a problem where im asked to show that for a complete metric space X, the the natural Isometry F:X --> X* is bijective.
I showed easily that it is injective, now i have to show it is surjective.

X is a metric space, and X* is it's completion. i.e every cauchy sequence contained in X converges to an element in X*. So if X is complete, this means X=X*

I came up with this but i need someone to tell me how wrong or close to right i am:

To prove F is surjective, i claim that for any element of X*, call it t, there is an element b in X. Since X is a complete metric space then X=X*, so t=b
Thus t is the element in X such that F(t)=t.

Any guidence will be apreciated


So X is already complete, and therefore the isometry which defines the completion of X (call it g, thought it's often called "phi") is such that g(X) = X*, since the image of X is dense in X* if X is not complete. If you start from the beginning (i.e. first step in the construction of a complete metric space from a non-complete metric space, like Q to R), you will see the answer to your question follow immediately.

If you're using Rudin's text, the appropriate exercise is #3.24(e). For better or worse, there's probably solutions all over the internet.

If you still have problems after reading that article, just say so.


This is a rough scketch i have for my proof:

Due to g(X) being dense in X* then for an arbitrary [s_n] element of g(X) there exists a [t_n] element of X* such that d([s_n],[t_n]) < e(epsilon) this is the same as saying lim d([s_n],[t_n]) < e.
Now, since X is complete, then t_n converges in X...

From here I want to say that t_n is an element of X and thus conclude that g([s_n])=t_n thus g being surjective. But i dont know how to get here


Thank you again