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What exactly the completion of a space is

  1. Nov 10, 2005 #1
    I am having a hard time working out the completion of certain spaces. But first I'd like to be able to fully understand what exactly the completion of a space is.

    Here is the way I see it. (It could be wrong, or unclear)

    If you have a normed vector space V [Does the space you want to complete have to be normed?] then there is a Banach space X and a linear isometric isomorphism T of V onto T(V), where T(V) is a dense subspace of X.

    Ok, so you have a normed space V, which does not have to be complete, however it can be. [If V were complete, does this mean we could simply map, via the linear isometric isomorphism T, straight onto X?] Then we have a linear isometric isomorphism T, which maps elements in V to T(V). But T(V) is no ordinary space, it is actually a dense subspace of a, perhaps, larger Banach space X.

    Whats more, the isomorphism which does this, T, and the resulting Banach space X is actually unique. [So for every V there is a unique map T and Banach space X?].

    Now is this Banach space, X the completion of V?


    Also, I don't seem to have any easy, illustrative examples of the completion of spaces. If anyone could be so kind as to provide a example or two, or has any discussion on this topic, or has an illuminating description which might help me understand this better, it would be much appreciated.
     
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  3. Nov 10, 2005 #2

    Hurkyl

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    Here's are some simple examples of a completion of a metric space:

    The completion of Q is R.
    The completion of [0, 1) U (1, 2] is [0, 2]

    A completion is usually with respect to a metric. Any additional structure (such as a vector space structure) is irrelevant to the process of completing the metric space.
     
  4. Nov 10, 2005 #3
    Ok Hurkyl. The completion [itex]\mathbb{Q}[/itex] is [itex]\mathbb{R}[/itex] as you say, but how?

    The way I see it, completing a metric space (as you said, vector structure is irrelevant, but I think metric structure is, after all metric structure gives you the idea of distance so convergence will make sense) to form a new metric space which is itself complete, is kind of like closing a set to form a new closed set.

    Elements in the completion space, ie [itex]\mathbb{R}[/itex] are actually Cauchy sequences, right? (for example, [itex]\pi = \{3,3.1,3.14,3.141,...\}[/itex] is a Cauchy sequence representation of the real number [itex]\pi[/itex]) I suppose more precisely would be that they are an equivalence class of Cauchy sequences that converge to the same thing. So 3.69999999... and 3.7000000... represent the same element, 3.7 in [itex]\mathbb{R}[/itex] even though they are two different Cauchy sequences in [itex]\mathbb{Q}[/itex] (that is, [itex]3.6999... = \{3,3.6,3.69,3.699,...\}[/itex] and likewise for 3.7000....).

    The original elements of [itex]\mathbb{Q}[/itex] embed into [itex]\mathbb{R}[/itex] in the usual way. So if, say, [itex]{1}\in\mathbb{Q}[/itex] then [itex]\{1,1,1,1,...\}\in\mathbb{R}[/itex]

    So for example, the sequence of rational numbers [itex]\{3,3.1,3.14,3.141,...\}\in\mathbb{Q} [/itex] is a Cauchy sequence, as the distance between each successive term in the sequence is monotonically decreasing.
     
    Last edited: Nov 10, 2005
  5. Nov 10, 2005 #4
    I roughly know how to complete the rational numbers to form the reals. Informally, the rational numbers have gaps, i.e. irrational numbers, numbers which cannot be written as a fraction. Completing the rationals gives you a new set of numbers, the reals, which include irrational numbers.

    It seems strange that you can start with a set of rational numbers, have no idea whatsoever what an irrational number is, complete the set, and you have a new set with these new irrational numbers in it. I might ask "Where did they come from?". After all, before the completion was done, irrational numbers did not exist.

    How can completing the rationals to form the reals be applied to the more general setting. My understanding is, completing a metric space requires the formulation of an isometric isomorphism [itex]T\,:\,\mathbb{Q} \rightarrow T(\mathbb{Q})[/itex].

    Just at a glance, the target space [itex]T(\mathbb{Q})[/itex] will be a dense subspace of a Banach space [itex]X[/itex] and this map, together with X is unique.

    So my first question is: What is the isometric isomorphism which does this? And is the Banach space X I speak of actually [itex]X = \mathbb{R}[/itex]?
     
  6. Nov 10, 2005 #5

    AKG

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    My book defines it this way:

    Let X be a metric space. If h : X -> Y is an isometric imbedding of X into a complete metric space Y, then the subspace cl(h(X)) of Y is a complete metric space. It is called the completion of X.

    An isometric imbedding is an isometry, which I believe you understand. An imbedding is an injective continuous function with the property that the function h-1 : h(X) -> X is continuous. The closure of a set h(X), denoted cl(h(X)), in Y is the smallest closed set in Y containing h(X). Equivalently, it is the intersection of all closed sets containing h(X). Equivalently, it is the set of points y in Y such that every open set containing y intersects h(X) in some points other than y (it may intersect h(X) at y, but must also do so at some other points of h(X) as well). A complete metric space is one in which every Cauchy sequence converges.

    Notice that h(X) is dense in cl(h(X)). A is dense in B iff cl(A) = B.

    So we could redo the definition above, defining Y' = cl(h(X)), and h' : X -> Y' by h'(x) = h(x). Then we have a space X, and an isometric imbedding h' such that h'(X) is dense in Y', so this Y' is a completion of X (in this definition, which is more like yours, it is the codomain of the isometric imbedding which is the completion, not just a subspace of the codomain). The map h' and the completion Y' are not unique. But, you can prove that if h'' : X -> Y'' is such that Y'' is another completion of X, then there is an isometric imbedding from Y'' to Y' which is equal to h'h''-1 on h''(X).

    The space you want to complete has to be metric, but with a norm you can define a metric: d(x,y) = ||x-y||. I don't understand your second red question, but we could map V identically onto V. This would satisfy the criteria. The Banach space X is a completion of V, and it is unique up to an isometric imbedding, as mentioned at the end of the previous paragraph.
     
  7. Nov 10, 2005 #6

    AKG

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    Look at what you said in your previous post, about reals being equivalence classes of Cauchy sequences. You don't know what these new numbers are, but you do know that Cauchy sequences are, and you do know that they don't all converge in [itex]\mathbb{Q}[/itex].

    You can supposedly define an isometric embedding from [itex]\mathbb{Q}[/itex] into the set of equivalence classes of Cauchy sequences of [itex]\mathbb{Q}[/itex] (note you don't have to know what irrationals are to work with a set of equivalence classes of Cauchy sequences of rationals) such that the image of [itex]\mathbb{Q}[/itex] is dense in the target space. We then decide to call this target space the reals, and those points in the closure of the image of [itex]\mathbb{Q}[/itex] which are not actually in the image will be what we call irrationals.
    Maybe you can define an equivalence relation on the set of Cauchy sequences of V such that you can find an isometric imbedding from V into the set of equivalence classes such that the closure of the image of V is equal to the set of equivalence classes. Then this set of equivalence classes is a closure of V.
    Yes, X is the reals, and T is the function T(q) = q.
     
  8. Nov 10, 2005 #7

    AKG

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    To check if the reals complete the rationals, you have to first be aware that we're dealing with the standard metrics for these spaces. You also have to know that the reals are complete. You then have to check that T defined by T(q) = q is an isometric imbedding (this is obvious however), and that the rationals are dense in the reals, but this is a pretty standard fact (take any real number and consider any open interval about it; you know that this interval will contain some rational numbers because if its width is e, you can find some N such that 1/N < e, i.e. N = ceiling[1 + 1/e], and you know that for some k, k/N is in this interval; this tells you that every real number is in the closure of the rationals, so the rationals are dense in the reals). So you're done.

    This is a way to check that the reals complete the rationals. If you had some arbitrary set S and wanted to complete it, then you'd need to construct some space from S and show that it completes S. Often the reals are constructed as a completion of the rationals by treating the reals as the set of equivalence classes of Cauchy sequences of rationals. Link.
     
  9. Nov 11, 2005 #8
    Thank you so much for you time AKG.

    This is something which I have never understood, and I think, for me at least, it is worth the attention. "Unique up to an isometric imbedding". What does that mean? I've seen it written a lot. Things like: "___ is ___ up to isomorphism" or something like that.

    The rest of what you've written AKG makes sense. It has solidified my understanding considerably, I suppose I just need to bounce ideas off someone.

    Lets look at a somewhat more involved problem concerning completion now.

    If you have the space of all real-valued polynomials, [itex]X = P(\mathbb{R})[/itex] with the norm

    [tex]\|f\|_{[0,1]} = \sup_{x\in[0,1]}|f(x)|[/tex]

    what is the completion of [itex]X[/itex].


    The first thing I realized was that, even though Hurkyl and AKG said that X has to be metric, a norm defined like this one, takes care of that. Am I right about this?

    Elements of our space X are polynomials. What is the 'distance' between two polynomials? This norm says the 'distance' between two polynomials in the closed interval is merely the largest vertical extent of f in that interval. So we have our metric.

    Now if I remember correctly, every continuous function on [0,1], or any closed interval, can be approximated by an element of [itex]P(\mathbb{R})[/itex], i.e. the Weierstrass Approximation theorem. Im not exactly sure how to prove it, but I'd say, because of this, [itex]P(\mathbb{R})[/itex] is dense in [itex]C([0,1])[/itex].

    Then somehow we can define an isometric imbedding [itex]h\,:\,P(\mathbb{R}) \rightarrow C([0,1])[/itex] and since [itex]C([0,1])[/itex] is complete, then [itex]h(P(\mathbb{R})) \subseteq C([0,1])[/itex], then the completion of [itex]P(\mathbb{R})[/itex] is [itex]C([0,1])[/itex].

    How does this look? I know I left some rigourous stuff out, simply because I don't know how to do it, but I may have the general jist of it.
     
    Last edited: Nov 11, 2005
  10. Nov 11, 2005 #9

    matt grime

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    The problem is that you a priori know what the real numbers are and know how to write a nice representation of a cuachy sequence of rationals. All that happens in the genearl case is that you don't know beforehand what a nice representation will look like.


    No, they 'existed': we have known about sqrt(2) for millennia. What is true is that it took until quite late to formalize the real numbers as an analytic object (one where convergent sequences made sense)
     
  11. Nov 11, 2005 #10

    matt grime

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    Let V be a finite dimensional vector space. Let V* be its dual. V and V* are not equal, but they are isomorphic as vector spaces. There is an important distinction in modern mathematics between genuine equality and isomorphism.

    Consider the rationals purely as a metric space. They can be completed to R, but there is another (equivalent) completion where we send x in Q to -x in R. This is an isometric embedding into a complete metric space. Differrent from the inclusion of x to x, but isometrically ie by considering metrics alone, they are indistinguishable.

    Example 2. The algebraic closure of the reals is the complexes, but it can be done in two ways since there are two choices of the square root of -1 (i and -i) and they are both completely indistinguishable as algebraic closures though they are not equal.
     
  12. Nov 11, 2005 #11

    Hurkyl

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    Say you wanted to construct the completion of Q. There are two classical ways of doing this:

    (1) The space of equivalence classes of Cauchy sequences over Q.
    (2) The space of Dedekind cuts of Q

    There is an evident map from Q into either of these metric spaces that serves as a completion of Q.

    Aside: It is a fairly common practice to name such maps instead by either its domain or range. For instance, while we ought to say that a particular map Q-->R is a completion of Q, we instead say that R is the completion of Q. Another example is that if we have a map of groups f:G-->H, we say that K is the kernel of f, rather than saying a particular map K-->G is the kernel of f. This happens in a lot of contexts, and is probably the fault of the evil set-theory which lets us say that A is a subset of B, when the "right" thing to do is to consider a monic map A-->B! (Yes, I'm being corrupted by category theory. :smile: I was only jesting when I called set theory evil!)


    So, we're unhappy that the notion of completion isn't specific enough for Q to only have one completion. :frown:

    But, we know that these two completions are isomorphic. (There are several senses in which this sentence can be interpreted) So, they are the same, for all practical purposes (FAPP).

    So, we might wonder if all completions of Q are the same FAPP... and we'd be right!

    We know that the completion of a metric space X is unique up to isomorphism -- any two completions of X are isomorphic!

    So while the completion of a metric space is not literally unique, it is unique for all practical purposes.
     
  13. Nov 11, 2005 #12

    AKG

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    "Unique up to an isometric imbedding" in this context means exactly what I previously wrote:

    The map h' and the completion Y' are not unique. But, you can prove that if h'' : X -> Y'' is such that Y'' is another completion of X, then there is an isometric imbedding from Y'' to Y' which is equal to h'h''-1 on h''(X).

    I don't know if "_______ is unique up to _________" has a formal meaning, or if it can always be interpreted unambiguously, but it normally means that there are a bunch of answers to a question that are technically distinct, but if you ignore certain things that are irrelevant to the context, they are the same. So with groups, the groups [itex]2\mathbb{Z}[/itex] and [itex]\mathbb{Z}[/itex] are technically distinct, but we are only concerned with their algebraic structure as groups, so if we ignore everything else about them, they are the same. And saying that two groups are the same if you ignore whatever differences they have that aren't in terms of their group structures is identical to saying that they're isomorphic. A permutation has a unique representation as a product of disjoint cycles up to reordering of the cycles. So if we look at the permutation (125)(34) and (34)(125) and ignore the order of the disjoint cycles, and look only at what cycles are involved, then they are the same. The antiderivative of a function is unique up to a constant, which means if the only difference between two functions is a constant, and we are ignoring constants, then they are the same.

    In our example, it's slightly less clear. Certainly, simply saying that the completion is unique up to an isometric imbedding doesn't tell you that this imbedding is equal to h'h''-1 on h''(X).
    Check this for yourself. We would want to define our metric d probably by:

    d(f,g) = ||f-g||

    like I mentioned in one of my previous posts. Check that this definition satisfies the conditions to make it a metric. Certainly it takes on positive real values and the metric is zero iff the arguments are both zero. It is also obviously symmetric. It remains to check that the triangle inequality is satisfied, but you can do that (I don't know if it works or not, I assume it would).
    If you look at it this way, I think it even becomes clear that the triangle inequality is satisfied.
    You need to be aware of what metric you're using for C([0,1]). It will be the same, however. You want to show that for every f in C([0,1]) and every e > 0, there is a polynomial p such that d(h(p),f) < e. If we define h by:

    [tex]h(p) = p|_{[0,1]}[/tex]

    (i.e. we're restricting p to [0,1]) then after checking that this is an isometric imbedding (which I think is pretty clear as long as we're sure that if two polynomials are the same on [0,1] then they're the same everywhere - if this weren't true then their difference would be a polynomial that "flatlines" on an entire interval [0,1] but isn't identically 0, and no such polynomial does this) the Weierstrass Approximation theorem does indeed give the desired result. To be "safe," we'd take p such that |f(x) - p|[0,1](x)| < e/2 for all x, that we know that d(f,p) which is the supremum of the differences |f(x) - p|[0,1]| is less than or equal to e/2, and thus strictly less than e, as required. If we just chose p such that |f(x) - p|[0,1](x)| < e for all x, it could turn out that the supremum of these differences was exactly e, and this wouldn't be good enough.
     
  14. Nov 12, 2005 #13
    Perfect, well said Hurkyl. I now understand.

    ...oh, and you too AKG!


    I'll have a read of what you wrote AKG...I think I'll get through it by the end of the week...:wink:
     
  15. Nov 12, 2005 #14
    If I am right, and the completion of [itex]P(\mathbb{R})[/itex] is [itex]C([0,1])[/itex] then I have a slight problem.

    I can see how sequences of real polynomials might tend to curves which are indescribable in terms of polynomials (just like sequences of rational numbers converged to irrational numbers). For example, the sequence of polynomials

    [tex]\{x,x^2,x^3,\dots,x^n\}[/tex]

    I can choose [itex]n[/itex] large enough so that over the closed interval [itex][0,1][/itex] this sequence of polynomials converges to a step function, with the step occurring at [itex]x = 1[/itex].

    But then the completion of [itex]P(\mathbb{R})[/itex] cannot be [itex]C([0,1])[/itex] because the step function is not continuous.

    However, the step function is differentiable, and I think every Cauchy sequence of real polynomials over a closed interval converges to a differentiable function. So perhaps the completion of [itex]P(\mathbb{R})[/itex] is the set of all differentiable functions over [itex][0,1][/itex]?
     
  16. Nov 12, 2005 #15

    shmoe

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    In what sense do you mean "converges" here?
     
  17. Nov 12, 2005 #16
    As n becomes arbitrarily large, the limit of the sequence {x^n} aproaches a step function on [0,1] where the step is at x=1.
     
  18. Nov 12, 2005 #17

    shmoe

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    Let me modify my question:

    Converges in what norm?
     
  19. Nov 12, 2005 #18
    The norm is

    [tex]\|f\|_{[0,1]} = \sup_{x\in[0,1]}|f(x)|[/tex]

    That is, the sequence of polynomials [itex]\{x,x^2,x^3,\dots,x^n\}[/itex] converges to the step function in the above norm. Because the norm of f(x) at x=1 is [itex]\{1,1,1,\dots,1\}[/itex]. At x=0 the norm is [itex]\{0,0,0,\dots,0\}[/itex]. At every other point the norm of f(x) is a monotonically decreasing sequence as the curves get 'sharper' and 'sharper' towards the step function.
     
  20. Nov 12, 2005 #19

    mathwonk

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    as all number theorists know (but not me) there is not just ONE completion of Q, but many. you have to be given the metric. there are also very interesting p - adic metrics which give different completions.

    so R is NOT, "the completion of Q", merely one of them.
     
  21. Nov 12, 2005 #20

    Hurkyl

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    But does it decrease uniformly?

    You're telling us that this sequence of continuous functions converges uniformly to a discontinuous function... there's a problem with that, if you recall. :smile:
     
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