Surjectivity of an Isometry given the metric space is complete.

In summary, the conversation discusses a problem involving a complete metric space and its natural isometry. The poster is trying to show that the isometry is bijective and is asking for guidance on their proof. Another user provides tips and suggests looking at a specific exercise in a textbook. The poster then presents their rough proof and thanks the other user for their help. In the end, it is concluded that the isometry is indeed bijective.
  • #1
arturo_026
18
0
Hello, the following is a post that was in progress and I am continuing it here after I received a message saying that most of the members had moved from mathhelpforum here.

Me:
I have a problem where I am 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

TaylorM0192:

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. Me:

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 don't know how to get here

.
.

Thank you again
 
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  • #2
for your help TaylorM0192. I think i understand better now, so I will try to complete my proof. Since X is complete, then for an arbitrary [t_n] element of X*, there exists an [s_n] element of X such that d([s_n], [t_n]) < e. This implies that lim d([s_n], [t_n]) = 0. Since g is an isometry, then lim d([g(s_n)], [t_n]) = 0. Thus g(s_n) = t_n. Therefore, F : X -> X* is surjective and thus bijective.
 
  • #3
for your help. I will definitely take a look at exercise #3.24(e) in Rudin's text. I think I may be on the right track with my proof, but I will definitely need to work out the details and make sure it is solid. If I still have trouble after reading the article, I will let you know. Thanks again!
 

FAQ: Surjectivity of an Isometry given the metric space is complete.

What is surjectivity in the context of an isometry?

Surjectivity in the context of an isometry refers to the property of the isometry mapping every point in the metric space to a unique point in its image. In other words, every point in the image of the isometry has a corresponding point in the original metric space.

How is surjectivity related to completeness of a metric space?

Surjectivity is related to completeness of a metric space because for an isometry to be surjective, it must also be bijective (one-to-one and onto). In a complete metric space, every Cauchy sequence converges to a point in the space, and this ensures that the isometry can map every point in the space to its image.

What does it mean for an isometry to be complete?

An isometry is complete if it preserves the completeness of the metric space. This means that every Cauchy sequence in the original metric space also converges in the image under the isometry.

Can an isometry be surjective if the metric space is not complete?

No, an isometry cannot be surjective if the metric space is not complete. Incomplete metric spaces do not have the property of completeness, meaning that there are Cauchy sequences that do not converge. Therefore, the isometry would not be able to map every point in the space to its image, and therefore would not be surjective.

How can surjectivity of an isometry be proven in a complete metric space?

To prove surjectivity of an isometry in a complete metric space, it must first be shown that the isometry is bijective, meaning it is both one-to-one and onto. Then, it must be shown that the isometry preserves the completeness of the metric space, meaning that every Cauchy sequence in the original space also converges in its image. This would prove that the isometry is surjective in the complete metric space.

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