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Imbedding of the ratiomals

  1. Jul 13, 2013 #1
    Hi'
    can the rational numbers be imbedded in a countable complete metric space X?


    If D is the set of isolated points of X,then D is dense in X\D is countable complete metric space so it is homeomorphic to Q.Where am i wrong?
     
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  3. Jul 13, 2013 #2

    jgens

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    Let X be a countable discrete space. This is certainly a countable complete metric space, but clearly there is no embedding of the rationals into this space.
     
  4. Jul 13, 2013 #3
    why,beacause the imbedded image must be closed hence topologically complete?Why does it have to be closed?
     
  5. Jul 13, 2013 #4

    micromass

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    The Baire category theorem implies that any countable, complete metric space is discrete. The rationals don't embded in the discrete topology.
     
  6. Jul 13, 2013 #5

    WannabeNewton

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    Assume there is an embedding ##i_{\mathbb{Q}}:\mathbb{Q}\rightarrow X## where ##X## is a discrete space. Then, since the embedding map is a homeomorphism onto its image, ##\{q\} = i_{\mathbb{Q}}^{-1}(i_{\mathbb{Q}}(\{q\}))## must be open in ##\mathbb{Q}## meaning there exists an open subset ##U\subseteq \mathbb{R}## such that ##\{q\} = U\cap \mathbb{Q}## which is clearly a contradiction because ##q## is a limit point of ##\mathbb{Q}##.
     
    Last edited: Jul 13, 2013
  7. Jul 13, 2013 #6

    lavinia

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    The real numbers are the limits of all of the Cauchy sequences of rationals. So any complete metric space that contains the rationals must also contain the real numbers.
     
  8. Jul 13, 2013 #7
    But Cauchy sequence in Q may not be Cauchy in the complete space.Homeomorphism preserves the topology,not the metric.
     
  9. Jul 14, 2013 #8

    lavinia

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    The it is not an embedding
     
  10. Jul 14, 2013 #9

    micromass

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    An embedding is merely continuous. It doesn't need to send Cauchy sequences to Cauchy sequences. You'll need something stronger than continuous (for example, uniform continuity).
     
  11. Jul 14, 2013 #10
    Why is it that continuous functions map convergent sequences to convergent sequences, but not Cauchy sequences to Cauchy sequences? That's somewhat counterintuitive.
     
  12. Jul 14, 2013 #11

    WannabeNewton

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  13. Jul 14, 2013 #12

    micromass

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    WbN's link is pretty neat. But allow me to give a bit more of a higher level answer.

    First, let me look at metric spaces. Take two points ##x## and ##y## in a metric space. We say that they are ##\varepsilon##-close if ##d(x,y)<\varepsilon##. So metric spaces allow us to measure when two points are sufficiently close to eachother.

    Of course, this definition uses the metric in a very crucial way. However, the entire point of a topology is to be able to generalize the above. Indeed, in a topological space, we can take a neighborhood ##U## of a point ##x##. We then say that ##y## is ##U##-close to ##x## if ##y\in U##.

    So a topological space is able to say when a point ##y## is very close to a point ##x##. Indeed, we can say that ##y## is "very close" to a point ##x## if ##y## is ##U##-close to ##x## for "many neighborhoods" of ##x##. This is all just intuitive of course.

    However, this makes things clear that we can describe convergence of sequences in a topological space. Indeed, let ##(x_n)_n## be a sequence. We say that ##x_n\rightarrow x## if for any neighborhood ##U##, there exists a point ##m## such that ##x_n## is ##U##-close to ##x## for ##n\geq m##.

    The crucial thing to notice here is that we compare the sequence elements to a fixed point ##x##. We don't let ##x## vary. The same with continuity. We define continuity at a point ##x##. The crucial point is again that we don't let ##x## vary.

    However, when we look at Cauchy sequences, we see that something different is going on. In the definition of a Cauchy sequence, we see that for all ##p,q>N## we have that ##d(x_p,x_q)## are ##\varepsilon##-close. So here, the ##x_p## and ##x_q## vary. It is not clear (or possible) to write this definition using some notion of ##U##-closeness.

    The problem is this. Given ##4## ponts ##x##, ##y##, ##z## and ##t##. In a metric space, we can easily say that ##x## and ##y## are as close to eachother as ##z## and ##t## are. Indeed, we just compare ##d(x,y)## and ##d(z,t)##.
    In a topological space, we can say that ##y## is ##U##-close to ##x##. But ##U## is a neighborhood of ##x##. It has nothing to do with ##z## and ##t##. So we can't say that ##z## is ##U##-close to ##t##, since ##U## isn't a neighborhood of ##t##. Neither is there any way to "translate" a neighborhood of ##x## to a neighborhood of ##y##. So the intuitive problem here is that neighborhoods of one point don't have anything to do with neighborhoods of another point.

    A space where we make this possible is a uniform space. A uniform space is a set ##X## together with a uniformity. A uniformity is a collection of subsets of ##X\times X## that satisfies some axioms. Take an element of the uniformity ##U##. Now we can say that ##x## and ##y## are ##U##-close if ##(x,y)\in U##. Take our four points from above. We can say that ##(x,y)\in U## and ##(z,t)\in U##. This would measure that ##x## and ##y## are as ##U##-close to eachother a ##z## and ##t##.

    Now we can make sense of the notion of a Cauchy sequence. Indeed, we say that ##(x_n)_n## is Cauchy if for all elements of the uniformity ##U## holds that there exists a ##N## such that for all ##p,q>N## holds that ##x_p## and ##x_q## are ##U##-close.

    Clearly, a metric space induces a uniform space. We take as uniformity the sets ##\{(x,y)~\vert~d(x,y)<\varepsilon\}## (and some combinations of those for technical purposes).

    Every uniform space also induces a topological space. Indeed, given a point ##x## and an element of the uniformity ##U##, we say that ##\{y\in X~\vert~y~\text{is U-close to x}\}## is a neighborhood of ##x##. This determines a topological space.

    To make a long story short. Cauchy sequences are no topological concept. They are a uniform concept.
     
    Last edited: Jul 14, 2013
  14. Jul 14, 2013 #13

    Bacle2

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    Look at the homeomorphism between (-1,1) and ℝ, and what it does to Cauchy sequences.

    Basically, Cauchy sequences are not defined in terms of open sets.

    Consider too, f(x)= 1/(x-√2) defined on the Rationals into the Reals, and consider a sequence that converges to √2 , and its image. Notice it does not extend into a continuous function from ℝ
    to itself. It is a standard result that you need uniform continuity, so that your function takes Cauchy sequences into Cauchy sequences ( and the space into which you extend must be complete).
     
    Last edited: Jul 14, 2013
  15. Jul 14, 2013 #14

    lavinia

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    OK. I thought the metrical relations were meant to be preserved.
     
    Last edited: Jul 14, 2013
  16. Jul 15, 2013 #15

    Bacle2

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    No, an embedding is not merely continuous; it is a homeomorphism onto/into its image.
     
  17. Jul 15, 2013 #16

    micromass

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    I meant "merely continuous" as opposed to "isometric".
     
  18. Jul 18, 2013 #17

    lavinia

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    I found this theorem on Wikipedia. the theorem of Ostrowski.

    https://en.wikipedia.org/wiki/Ostrowski's_theorem#Proof

    It states that the only absolute values on the rationals are the trivial, the Euclidean, and the p-adics.

    If the rationals are embedded in a complete metric space and if they inherit an absolute value from the metric.

    |q| = d(q.0) the distance of q to zero is actually an absolute value

    then their completion under this metric must equal either the real numbers, the p-adic numbers for some prime, p. In both cases these are uncountable fields. The p-adic numbers are topological Cantor sets.

    So if they are embedded in a complete metric space that is countable, the metric can not induce an absolute value.

    There seem to be two possible ways that the embedding space can be countable.

    - all but countably many Cauchy sequences in the rationals map to divergent sequences

    - All but countably many Cauchy sequences converge and they converge to points outside the image of the rationals (because the rationals are embedded).

    The first case cannot happen because the image of every non empty open interval of rationals would have to be unbounded yet the ball of radius 1/2 around the image of 1 is bounded.

    second case?

    An example of a countable space containing the rationals which feels like a candidate would be the quotient of the reals by the irrationals. But I wonder if such a space is metrizable.
     
    Last edited: Jul 18, 2013
  19. Jul 18, 2013 #18

    micromass

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    You mean identifying the irrationals to one point? That's not Hausdorff.
     
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