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Anoter question on vector spaces an rationnal numbers

  1. Dec 20, 2004 #1
    I have some questions concerning the rationnal numbers and the vector spaces.
    Let's take the set of rational number Q with the usual addition and multiplication.

    We can say that (Q,+,.) is a vector space on the Q field. Now, if we add the |x| absolute value, we define have vector space with a norm.

    If we add to this space all the points of the cauchy convergent sequence, do we have a complete space (i.e. banach space)?

    If yes, we can call this set Q* (may there is already an other name ;). Is (Q*,+,.) a field?

  2. jcsd
  3. Dec 20, 2004 #2

    matt grime

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    If you complete Q in its ordinary euclidean norm you jsut get the reals. In fact, that's one of the definitions of the reals.
  4. Dec 20, 2004 #3
    Thanks a lot.
    That is one of my questions and why I am trying to use this model /view to understand some aspects. I am trying to construct the completion of Q with countable series. Thus, how can I get the set of real numbers form this construction if |R is non countable?
    I always have interpreted this result (non exsitence of a bijection between |N and |R) as the existence of real numbers that are not the limit of "rationnal" series (i.e. => Q* =/= |R).

    May you tell me where I am wrong?

  5. Dec 20, 2004 #4

    matt grime

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    Every real number has a decimal expansion, right? Well, the partial sums are rational, and cauchy and the limit is irrational, and arbitrary.

    I've no idea where your interpretation came from, and, as it's not reasonable to me, I can't see how to explain why it's wrong other than to say "it is".

    If I can offer a possible anology, why is the power set of N also uncountable, yet the set of all finite subsets countable?
  6. Dec 20, 2004 #5


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    The nonexistance of a bijection only says a single sequence can't pass through every real number.

    Try counting the number of rational sequences and find the cardinality of that set!
  7. Dec 20, 2004 #6
    Thanks, for the answer. Now I am trying to think a little bit more.
    Ok, I am not an expert of cardinality but I need to understand better the completion of Q, so I prefer to stay with bijection properties bewteen sets.

    Therefore, Can we say, we have a bijection between |Nx|Nx ....|Nx ... (the countable infinite product sequence of integer sets) and |R? (I am trying to re-interpret you statement)

    Thanks in advance,

  8. Dec 20, 2004 #7

    matt grime

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    The completion of Q is R; it is the "best" definition of what R is.

    The completion is an analytic statement, not a set theoretic one.

    I don't know if the countable product of N's has cardinality c (card of R). and it isn't important - clearly it is at least as great as c.
  9. Dec 20, 2004 #8
    Well, I have always missed that point (I am not a specialist – I just want to understand the limit behaviour).
    For any countable finite product of |N, call |N^p, we have a bijection between |N and |N^n (we take the odd and even sets of |N and by induction we can generalise to any finite n).

    May you give me a demo (or a link, or tell me) why for n-->+oO, this bijection become false (for example by induction the remaining odd/even parts of |N to construct the set product bijection becomes empty)?

    Thanks in advance. It helps me in viewing clearer my problem.

  10. Dec 20, 2004 #9

    matt grime

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    Because clearly the infite product of N's has greater cardinality than the infinite product of the two element set {0,1}, which has the cardinality as the power set of N, ie is uncountable.
  11. Dec 20, 2004 #10
    Once again thanks, for these answers. However, I do not have spent to much time learning the cardinality of sets. Therefore, I prefer to stay with the application bijection notion between sets (rather than working with the cardinality) to demonstrate/understand better my intial problem.

    I understand, that you are using a cardinality results (i think {0,1}^n, n-->+oO is uncountable) however, forgive me, but I would like to get the same result without using this "black box" cardinality and properties.

    In fact I would like to understand better the transition between the countable to uncountable sets (based solely on the definition of the non existence of bijection).

    For example, how do you proove that {0,1}^n, n--> +oO is uncountable (if I have correctly understood your statement).
    However, I prefer a demonstration based on the case |Nx|Nx....x|Nx... as it is closer to my initial problem (understanding the completion limit by the cauchy sequences).

  12. Dec 20, 2004 #11
  13. Dec 20, 2004 #12
    Thanks a lot for these links, especially the first one. However, I've quickly checked them :cry: and it is not exactly what I am looking for (anyway I will take more time to check them more carrefully).

    In order to be understood better: I am not looking for the uncountability property of the |R set (for example using the decimal representation, cantor diagonal...), I just want to understand better the limit between the countability and uncountability generated from a countable set (i.e. {0,1}^n, n --> +oO or (|Nx|Nx......)_ntimes, n --> +oO).

  14. Dec 20, 2004 #13

    matt grime

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    Finding bijections is hard, and completely unnecessary if you ask me. If you insist on such things then I suggest you look up the schroeder bernstien theorem that states two sets X and Y have the same cardinality iff there are two injections one from X and one to Y.

    Then {0,1}_{N} which we will use as the notation for the product of the set {0,1} a countably infinte number of times (and can i suggest you stop writing that limit thing?) has the same cardinslity as R is quiet trivial:

    there is an obvious injection: send the string (x_i) of 0s and 1s to the real number

    in base 10

    the reverse injection is slightly harder to right out but isn't that tricky: every number in base 2 has some representation (disallowing strings ending in an infinte number of 1s)


    which we may assume is infinite in both directions by addding zeroes before and after.

    send the x_i to the even places in a string and the y_i to the odd places and yo'uve an injection the other way.

    if you are goign to insist on always thinking in only one basic fashion and not broadening to better results you won't get very far.
  15. Dec 20, 2004 #14
    Thanks for your answer, but, sorry, it is not what I am looking for : (.

    Ok, I admit that my questions are not very clear and it is hard for me to formulate them clearly. However, I think if I could define my question clearly, I guess I would get myself the answer : ). Therefore, I make my excuses and thank you, once again, and all the other participants for the answers and I hope to continue to receive some help.

    I think that your suggestion on using the existence of injections between the 2 concerned sets rather than defining a bijection sounds pretty good (i.e. we are using the tools of cardinality rather than the cardinality theorems I do not want to use): The schroeder bernstien theorem (I knew it with the name “cantor-bernstein”, I am always surprised with the different theorem names/equations we can encounter all around the world!)

    Now, I insist I am trying to see the difference between the countable and uncountable property on a limit point of view (it is a kind of physical point of view if you prefer :) as it can help me to understand/proove better the logical consistence of some physical models. However, I want to keep the mathematic consistency and I understand that it is not the usual mathematical way to do that.
    I accept your demonstrations based on the “number labelling”, but they do not show easily the connection at the “limit” (the behaviour I am trying to see or define if you prefer).

    I also want to start with the properties of |N and Q that are countable and understand how we get (at the limit, see below) a non countable set.

    In order to define a limit, we must have a topology. My hypothesis, that can be wrong, is based on the possibility to define, a priory, a larger set X that contains the limit set (I hope it is not circular and consistent) and to define the topology on it (for example the finest topology consisting of the elements of P(X) or a coarser topology if possible).

    With that definition, I may define, if I am not wrong, the limit of product_ntimes {0,1}= {0,1}_{n} (using your notation where I replace the set |N by n) or the limit of product_ntimes {|N}={|N}_{n} as the element of the set X.
    In the first case, for example, I may define a topology on the set X= union_n_element_of_|N {{0,1}_{n}} (with, may be, additional {} to be consistent). Thus, I may define logically a sequence of sets and a limit.
    The advantage of this complication is only the ability “test”/”see” the evolution of an element of the sequence and the limit itself.
    For example, the case of {|N}_{n} where I can define a simple bijection to |N for any n (using inductive odd/even number subsets). Therefore, I may understand better the limit, with the adequate topology, of this sequence. However, if I want to connect this result to my initial question, It is a little bit more complicated because the |Q Cauchy sequences are only a subset of {|Q}_{|N } (problem to define/identify the subset).

    For the basic set {0,1} and the sequence {0,1}_{n}, it is more complicated because, for any n, I have no bijection between |N and {0,1}_{n} and it is more difficult to see the evolution from a finite (countable) set to an infinite non countable set.

    The ideal should be to find a sequence of infinite countable sets that converge (with the adequate topology) to a non countable set equivalent to the set of all |Q Cauchy sequences or something like that (i.e. a good example to see, how the non countable property comes from the countable property through the limit of a sequence)

  16. Dec 20, 2004 #15

    matt grime

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    consider of sequence of rationals

    [tex]x_n=\sum _{r=1}^n \frac{\delta_r}{10^r}[/tex]

    where delta_r is either one or zero. This is a cauchy sequence of rationals. Moreoever two different choices of the delta+r produce sequences, the difference of which is cauchy adn does not converge to zero. Hence they represent different equivlance classes of cauchy sequences modulo convergence. RIght? Sp it is a subset of the completion of Q

    What is the cardinlity of the set of all possible x_n? It is clearly 2^{aleph-0} or the cardinality of the reals.

    Ok? what part of that is troubling?
  17. Dec 21, 2004 #16
    I am sorry, for this late answer. I was trying to think on explaining better what I am looking for.

    Thanks for this compact example (analogue to the previous one, but in a user friendly form :). It helps me, I think, to remove the un-useful features (for me). I am ok with your demonstration and the previous one (I think I understand them).

    We define the symbol “:e:” for “element of” (a set) relation.

    Let’s call the Cauchy sequences you define above (xn)_n:e:|N (<=> to the limit of the serie) with xn :e: {0;(10^-r)_r:e:|N*}=S. (I have inserted the n=0 <=> translation by one of your serie label r).

    We have a simple bijection between S and |N (thus the same cardinality).
    The advantage of (xn)_n:e:|N, (thanks to you), is the cauchy convergence whatever (xn):e:S sequence.
    Now let’s call (xq_n)_n:e:|N the q th cauchy sequence (q:e:|N).
    We thus have xq_n:e:S for every q => [(xq_n)_nq:e:|N] :e: SxSxS......={S}_{|N}.

    The countable set of all Cauchy sequences of the form (xn)_n:e:|N belongs to {S}_{|N}.

    You have thus demonstrated that this set {S}_{|N} has the cardinality of |R. In addition, because we have a bijection between S and |N, we also can say that |Nx|Nx|N......= {|N}_{|N} has the same cardinality of |R (bijection between {|N}_{|N} and {S}_{|N}

    (at least the {|N}_{|N} form helps me in understanding that the “limit set” {|N}_{|N} is uncountable because |N is completely depleted form the extraction of odd/even subparts at the “limit”– case of {|N}_{2n}.)

    Therefore, I can concentrate on a topology model to study the limit of {|N}_{n}, if what I have written above is correct.

    Has anyone a simple idea to map, for every n (n=0 is the empty set for example), the value “{|N}_{n}” into a simple topological space? (i.e. a one to one correspondence).

    Thanks in advance,

  18. Dec 21, 2004 #17

    matt grime

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    Can I suggest you take a second to learn tex (sticky thread in the physics forum should help) because what you have written takes far too much effort to decipher.

    (xn)_n:e:|N (<=> to the limit of the serie) with xn :e: {0;(10^-r)_r:e:|N*}=S

    makes no sense at all. I mean, what has teh limit got to do with an if and only iff symbol? you appear to claim something is in [tex]\mathbb{N}[/tex], what is is that? what is the semicolon in that last bracket? What are the subscript symbols _ for?

    how about these symbols

    [tex] x_n = ?[/tex]


    [tex] y \in S[/tex]

    S is the elements of some set of something? what the partial sums x_n?

    HOw are you picking the q'th cauchy sequeunce? This presupposes they set of all of such caouchy sequences is countalbe when it clearly isn't.

    You appear to be confused by the notion of the elements in the set of partial sums of the cauchy sequence and the set of all such cauchy sequences.

    I have not said that there is a bijection between [tex]\mathbb{N}[/tex] producted with itself a countably infinte number of times (though I believe there is one).

    What has topology got to do with this? We do not need nor require a topology on the set of all convergent cauchy sequences. This is not a construction of topological spaces. ANd if you introduce topologies on sequences you are going to get some very ugly and at the moment completely useless mathematics.

    Here is what I've shown: the space of ALL cauchy sequences is uncountable. If we take the set of ALL cauchy sequences of rationals and define the equivlance relation by x_n ~ y_n iff x_n -y_n converges to zero, then the set of equivalence classes of these is the set of real numbers. Furthermore I have shown that I can pick an uncountable subset of the set of all these equivalence classes - those with only 0s and 1s in the finite partial sums of some numbers base 10.
    Last edited: Dec 21, 2004
  19. Dec 21, 2004 #18
    You are completely right for tex (I will make the effort ;).

    “(<=> to the limit of the serie)” is only a comment. I should have replaced the mathematical symbol by the word equivalent.

    I am working with the limit of your sequence (the series), not with the partial sum.
    I am just defining a map between the limit of your sequence [itex]sum_\infty=\sum _{r=0}^\infty \frac{\delta_r_+_1}{10^{r+1}}[/itex] and the definition of the sequence [tex] (x_n)_n_\in_\mathbb{N} [/tex] (i.e. the limit of your sequence is a series that is equal to the sum of all the [tex] (x_n)_n_\in_\mathbb{N} [/tex] (i.e. I have defined a one to one correspondence between the limit of your sequence and one peculiar [tex] (x_n)_n_\in_\mathbb{N} [/tex].
    (I should have labelled [tex] (x_n)_n_\in_\mathbb{N} [/tex] with another letter in order not to confuse with your partial sum xn.

    [tex]sum_\infty=\sum _{r=0}^\infty \frac{\delta_r_+_1}{10^{r+1}}\equiv (x_n)_n_\in_\mathbb{N} \ {(with \ the \ sum)}[/tex]

    To obtain the full equivalence, we define the set S form where the [itex] x_n [/itex] elements belong to.
    [tex] S=\{0;\{10^-^r\}_r_\in_\mathbb{N*}\} [/tex] (previously S={0;(10^-r)_r:e:|N*}).

    We have a one to one correspondence between the elements of S and [itex]\mathbb{N} [/itex] (i.e. the label of the elements of S).

    It is what I am trying to show in my previous post using your demonstration example and using the equivalence between the set S and |N.

    We have a simple bijection between S and |N, thus we have the same cardinality.

    The advantage of the [tex] (x_n)_n_\in_\mathbb{N} [/tex] is that the elements x_n belong to S forcing in an explicit way the Cauchy convergence of [itex]sum_\infty=\sum _{r=0}^\infty \frac{\delta_r_+_1}{10^{r+1}}[/itex].

    Therefore, to every [itex] (x_n)_n_\in_\mathbb{N} \ {;} \ x_n \in S [/itex] corresponds a cauchy convergent series ([itex]sum_\infty=\sum _{r=0}^\infty \frac{\delta_r_+_1}{10^{r+1}}[/itex] i.e. your cauchy sequence on the partial sums).

    Now, we can label by q, each sequence [itex] (x_n)_n_\in_\mathbb{N} [/itex] by [itex] (x_q_n)_n_\in_\mathbb{N} [/itex] we have the following result (and I correct what I've said: we do not know, a priori if the label q is countable or not, q is just a label to distinguish 2 different sequences):

    Each countable cauchy convergent series, [itex] (x_q_n)_n_\in_\mathbb{N} [/itex] belongs to the set {S}_{|N}=[itex] \prod_{n \in \mathbb{N}^*}S[/itex] (definition) by construction. We do not know if the set of different labels is countable or not (in fact it is what we want to know).

    You have thus demonstrated that the set {S}_{|N} (i.e. the set of all cauchy convergent sums) has the cardinality of |R. Therefore, because we have a bijection between S and |N, we also can say that the set “|Nx|Nx|N......”= “{|N}_{|N}” (definition) has the same cardinality as {S}_{|N} and |R (i.e. we have a bijection between {|N}_{|N} and {S}_{|N} through the bijection between |N and S).

    You are perfectly right and I agree. Now, I think that I have also demonstrated that we have the same cardinality for the set I have labelled {|N}_{|N}.

    I know that a topology structure will add some complications, but currently, it is one of the tools that help me to “see” the limit from the finite product of |N sets and the infinite product of |N sets (i.e. that has the same cardinality of |R). Recall that I want to try to “see” the transition between the countable and uncountable property and the limit is a powerful tool and if I have such a tool, I can use it to understand better the coherence of some other physics problems.

    Therefore, I have tried to prove that the cardinality of {|N}_{|N} is the same as |R using your demonstrations (and the last one you gave was perfect for that). It allows me to restrict the search of a topology to the case of the {|N}_{|N} set limit which is simpler than using a more compact subset (the other examples you give):

    I can concentrate on a topology model to study the limit of {|N}_{n} (a finite product of |N sets) to study the limit n --> +oO {|N}_{n}={|N}_{|N } (the infinite countable product of |N sets).

    Has anyone a simple idea to map, for every n (n=0 is the empty set for example), the value “{|N}_{n}” (the finite product of |N sets) into a simple topological space?.

    Thanks in advance,

  20. Dec 21, 2004 #19
  21. Dec 22, 2004 #20

    matt grime

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    Let [tex]\mathbb{N}_r[/tex] = N_r be the product of r copies of the natural numbers for any ordinal r.

    Then N_r has many topolgies itself. Discrete, finite, cofinite, the product topology from any topology defined on N such as the topoogy of arithemetic progressions.

    The topology does not tell you why the set of all equivalence classes of cauchy sequences of rationals is uncountable.

    The fact that I've shown you how to define an uncountable number of inequivalent classes mean nothing?

    If you just want an explanation as to why [tex]\mathbb{N}_{\omega}[/tex] is uncountable (the w, omega, just means indexed by a countable infinite set) whilst the finite ones are countable?

    Ok, but you aren't going to like it.

    Firstly, you have to understand what you mean when you say limit. The are two kinds of limit, the one you have in mind gives you the 'finite power set' if we think in terms of {0,1}s instead of N,then this the set of all strings that are eventually 0. This is a countable set. This roughly correpsonds to taking the direct limit, or the coproduct. The one you need to consider behaves as the inverse limit, or product, and it is the set of all strings.

    THis is simply the difference between direct product and coproduct (union)
    Last edited: Dec 22, 2004
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