Re: Complete Normed Space

Re: Complete Normed Space

Although the question is phrased a bit confusingly, I think that there a second issue might have been brought up: suppose you take a metric space and take its completion. How do you know that the completion is complete? In other words, if you take a Cauchy sequence in the completion, it may contain new elements that weren't considered before. Will it necessarily converge?

I think that it should, and if I recall correctly, a diagonalisation argument is involved. Maybe somebody can go over the details? (I don't have the time at the moment.)

Re: Complete Normed Space

Quote:

Originally Posted by

**roninpro** Although the question is phrased a bit confusingly, I think that there a second issue might have been brought up: suppose you take a metric space and take its completion. How do you know that the completion is complete? In other words, if you take a Cauchy sequence in the completion, it may contain new elements that weren't considered before. Will it necessarily converge?

I think that it should, and if I recall correctly, a diagonalisation argument is involved. Maybe somebody can go over the details? (I don't have the time at the moment.)

This is not a question you ask in a forum, but something you look in a book for. If perhaps the OP had a particular question about the proof of this fact, maybe, but we shouldn't give a proof here. A quick google search turns up this. If the OP has further interests I suggest he take a look at the accessible text Simmons, George Finlay. Introduction to Topology and Modern Analysis. New York: McGraw-Hill, 1963. Print.

If the OP has more questions after reading both of these sources, then he should report back with specific questions.

@OP: I am not trying to be harsh, it just isn't feasible for us to give you a lesson on completions. Best of luck.

Re: Complete Normed Space

This is what I'm dealing with:

The Banach space Lp(a,b) is the completion of the normed space which consists of all continuous real valued functions on [a,b] and norm defined by (integral of abs value of x(t)^p)^1/p. The space can be completed by extending the allowable space of functions to include the functions whose norm converges to a discontinuous function. My question is, will the completed space still converge to a member of the space.

Thanks for reference. I will try to hunt down a cheap copy. The book I have (Oden) doesn't answer the question. I will keep looking.

Re: Complete Normed Space

Re: Complete Normed Space

Quote:

Originally Posted by

**Drexel28** Ah! This is a better question! This is why I asked what norm you were using. The canonical norm on

is the infinity norm (BECAUSE it makes it complete), but it is not complete as you mentioned with respect to the

-norm. So, to answer what I think is your question, which I think is asking does a sequence in

converge to an element of

always (since

is complete we know it converges to an element of

, but will this guy live in

, the answer is no. We enlarged our space precisely because this isn't true.

Is this what you were looking for?

I don't think so. If a cauchy sequence in the space X of continuous functions on [a,b]with the Lp(a,b) norm (used as a metric) converges to a discontinuous function, it is not in X. If you add that function to X to make it complete, is the resulting space X* complete?

Adding a finite number of discontinuous functions to X won't influence the convergence of any infinite series of functions so the space is complete.

The question gets messy if all possible cauchy (convergent) sequences could lead to an infinite number of discontinuous functions not in X, which, if added to X, may converge to other discontinuous functions which aren't in X* and the space would still be incomplete.

I really wasn't looking for any kind of proof. The printed statement was that if you added the discontinuous functions the space would be complete. If the author simply said it wasn't obvious, look it up, I would have been satisfied.

I naively think of the analogy of the operation of cuts of rational numbers completing the rational numbers, ie, if you take cuts of the real numbers do you get new numbers? Taylor shows that by taking cuts of the real numbers you only get real numbers and therefore the real numbers are complete. I'm just paraphrasing. I really don't understand it. How do you know that there isn't some operation which would put numbers between real numbers. It was proved there wasn't and i could "follow" it, but couldn't grasp it.

I assume the answer to my original question properly phrased is it is not obvious but look it up in " ," which suits me fine. BTW, I'm trying to get by with a very basic vocabulary which might not be up to the task.

Re: Complete Normed Space

I think I see what you're trying to get at, but the problem is not so easy to grasp: As you noted, is not complete in the norm, we can (only intuitively!) add certain functions that are not in the space to try to fill the gaps 'in-between' continous functions. However there is a problem, the process by which you 'complete' the space is really abstract: The completion is a set of equivalence clasess, can these be given an interpretation as a set of functions? The answer is no, at least in the usual sense, and an easy argument as to why we can't have functions is this: Let then if and differ only in a point we still have and , so at most we would get a pseudometric if we were to just add discontinous functions. On the other hand, if you were to try and add 'manually' discontinous functions, how would you know when it's enough? are all the functions you included integrable? As far as I know these problems can't be bypassed by an elementary technique.

So, in comes Lebesgue with his integral and the spaces which, it can't be stressed enough, is **not** a function space in the sense that is: Elements of are equivalence classes of functions (via the relation of being equal almost everywhere). Now what you can say is that every continous function in is a representative of an element in , in this sense we can consider as a subset of , the notation in this case is not really saying there is an inclusion as sets, but that the first has a copy of itself inside the second.

Now, without handwaving **a lot** of important concepts it's really not efficient to actually prove that is (isometrically isomorphic to) the completion of under the norm before introducing Lebesgue measure, and as far as I know there is no other way to show that the completion of the continous functions are even a set with a reasonable representation.

If you really want to learn about this, I think Rudin's books cover the subject.

Re: Complete Normed Space

Fell asleep last night in darkness trying to make sense of norm and completion. Woke up and saw the light.

Rational numbers (set of objects):

Def: Size (Norm)

Def: Distance between objects (Metric)

Def: Sequence

Def: Convergence (dist->0)

Result:

Sequence converges to a rational number (member of the set, repeating decimal). Repeat with different sequence.

Sequence converges but is not a member of the set (non-repeating decimal). Add it to the set and repeat convergence test..

Complete Set: Convergence always results in a member of the set. WITH RESPECT TO NORM AS DEFINED.

NOTE: You have to define size and distance for the new members of the extended set before you can repeat convergence tests which include new members of the set. One way is with "cuts." Landau does this systematically in "Foundations of Analysis."

That's it. Analysis. The Mother Lode. The Plan. Notre Dame at sun-rise.

Set

Size (Norm)

Distance (Metric)

Sequence

Convergence

Completion

EDIT: It's interesting to see previous post from this perspective. Thanks Jose27