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Continuous Mappings

  1. Aug 30, 2005 #1
    I would like someone to check my working please. Here is the question:

    Is the mapping

    [tex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/tex]

    continuous?

    My solution:

    A mapping [itex]f:X\rightarrow Y[/itex] is continuous at a point [itex]x \in X[/itex] if for every [itex]\epsilon > 0[/itex] there exists a [itex]\delta > 0[/itex] such that

    [tex]d(x,y) < \delta \Rightarrow d(f(x),f(y)) < \epsilon[/tex]

    A function is continuous if it is continuous at every point in its domain.

    This is the definition of continuity that I will be using. Note that I made this choice because the domain of the mapping lies in the rationals with respect to the p-adic metric. Further, [itex]\mathbb{Q}_p[/itex] and [itex]\mathbb{R}[/itex] are both metric spaces - ie there is a very good notion of distance between two points in both sets. This idea is crucial to my solution.

    Loosely, the definition above says that all points [itex]f(y) \in Y[/itex] near [itex]f(x) \in Y[/itex] have their preimages [itex]y \in X[/itex] near [itex]x \in X[/itex].

    In the question at hand, the function [itex]f[/itex] maps points in the metric space [itex]\mathbb{Q}_p[/itex] to points in [itex]\mathbb{R}[/itex] - where [itex]\mathbb{R}[/itex] is the metric space w.r.t to usual Euclidean metric.

    For the function to be continuous the following must be satisfied: The point [itex]f(x) \in \mathbb{R}[/itex] near [itex]f(x) \in \mathbb{R}[/itex] must imply that [itex]y \in \mathbb{Q}_p[/itex] is near [itex]x \in \mathbb{Q}_p[/itex].

    Technically speaking, given an epsilon > 0 there must exist a delta > 0 such that [itex]d(x,y) < \delta \Rightarrow d(f(x),f(y)) < \epsilon[/itex].

    But [itex]d(x,y) := |x-y|_p[/itex]. That is, distance between two points with respect to the p-adic metric gets larger and larger as [itex]x[/itex] gets closer and closer to [itex]y[/itex] with respect to the Euclidean metric. So I can not always guarantee the existance of a positive, non-zero [itex]\delta[/itex] for any given [itex]\epsilon[/itex]. Therefore the mapping [itex]f[/itex] is NOT continuous.

    To make this clearer, the distance between two points [itex]x, y \in \mathbb{Q}_p[/itex] actually gets larger as [itex]x[/itex] and [itex]y[/itex] become closer with respect to the Euclidean metric. So the preimages of 'close' points in [itex]\mathbb{R}[/itex] are actually very far apart in [itex]\mathbb{Q}_p[/itex]. This is the exact opposite of what it means for a function to be continuous.

    How does this sound?
     
    Last edited: Aug 30, 2005
  2. jcsd
  3. Aug 30, 2005 #2

    shmoe

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    It looks like [tex]f[/tex] is the inclusion map from the rationals to the reals, but considering the rationals with the p-adic metric. If so writing [tex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/tex] doesn't make sense as it's not a function on all of [tex]\mathbb{Q}_p[/tex].

    To show it's not continuous at a point x, you have to prove that you can find y's arbitrarily close in the Euclidean metric that are arbitrarily far away in the p-adic metric (it's sufficient to find y's farther than some convenient fixed bound, 1 will work here). You've essentially just declared this was true, but offered no proof. Can you supply details?
     
  4. Aug 30, 2005 #3
    Oh dear, I forgot to mention that

    [tex]f(x) = x[/tex].
     
  5. Aug 30, 2005 #4
    May I ask if this is actually a continuous function?
     
  6. Aug 30, 2005 #5
    I actually considered that this was obvious. But you are right, proving this is much more concrete.

    We need to show that given an [itex]\epsilon > 0[/itex] we need to find a [itex]\delta > 0[/itex] so that whenever [itex]d(x,y) < \delta[/itex] then [itex]d(1(x),1(y)) < \epsilon[/itex]. However, since [itex]1(x) = x[/itex] by definition of the function, it follows that [itex]d(1(x),1(y)) = d(x,y) = |x-y|_p[/itex]. Therefore we cannot necessarily find a [itex]\delta[/itex] which is strictly less that [itex]\epsilon[/itex].
     
  7. Aug 30, 2005 #6

    shmoe

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    It's probably best to not use [tex]d(\cdot,\cdot)[/tex] to represent two different metrics, so I'm going to switch to [tex]|\cdot|_p[/tex] for the p-adic norm and [tex]|\cdot|[/tex] for the usual euclidean norm.

    You have the same problem as before. "cannot necessarily find..." isn't a proof. f is continuous at x if for every [tex]\epsilon>0[/tex] we can find a [tex]\delta>0[/tex] such that for all y with [tex]|x-y|_p<\delta[/tex] we have [tex]|x-y|=|f(x)-f(y)|<\epsilon[/tex].

    To prove f is NOT continuous at x we can find an [tex]\epsilon>0[/tex] that has no [tex]\delta>0[/tex] that will satisfy this. Showing it fails for any specific [tex]\epsilon[/tex] will be enough, so let's take [tex]\epsilon=1[/tex]. Now suppose [tex]\delta>0[/tex] satisfies the definition of continuity for [tex]\epsilon=1[/tex]. Can you find a y where [tex]|x-y|_p<\delta[/tex] but we have [tex]|x-y|>1[/tex]?
     
  8. Aug 30, 2005 #7
    Good point Shmoe!

    If I take y = 2 then (2 must be prime)

    [tex]|0-2|_p = \frac{1}{2} < \delta[/tex]

    But

    [tex]|0-2| = 2 > 1[/tex]

    Is this ok?
     
  9. Aug 30, 2005 #8

    Hurkyl

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    Is this even a well-defined function? x is a p-adic number... in what sense are you embedding the p-adics into the reals?


    Anyways, it's very easy to exhibit a sequence that converges in the p-adics and not in the reals. For example, consider the 3-adic sequence:

    2, 22, 222, 2222, 22222, ...
    which converges to the 3-adic number ...222 (= -1)

    But this sequence clearly does not converge in the reals.


    On the other hand...


    I haven't worked it out, but it seems to me that if I define a function from Qp &rarr; R that maps a p-adic number x to its lexicographical reverse as a real number written in radix p, then this function is a continuous surjection. (not a bijection)

    For example, in the 5-adics,

    f(...41302.134) = 431.20314...(base 5)

    This is in line with my intuition that the p-adics and the reals are opposites.
     
    Last edited: Aug 30, 2005
  10. Aug 30, 2005 #9

    shmoe

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    Oxymoron: You've done a very specific case, when x=0, p=2 and delta>1/2. Let's carry on with this, leaving x=0, and p=2.

    Find me a y where [tex]|0-y|_p<1/2[/tex] but [tex]|0-y|>1[/tex].

    Find me a y where [tex]|0-y|_p<1/4[/tex] but [tex]|0-y|>1[/tex].

    Find me a y where [tex]|0-y|_p<\delta[/tex] but [tex]|0-y|>1[/tex]. y will depend on [tex]\delta[/tex]


    I'm also a little concerned with what you wrote, you know that the p-adic norm [tex]|\cdot |_p[/tex] depends on a specific prime p right? That is, [tex]|\cdot |_2[/tex], [tex]|\cdot |_3[/tex], [tex]|\cdot |_5[/tex] etc. all give different norms.
     
  11. Aug 30, 2005 #10
    The question simply asks to determine whether or not [itex]f:\mathbb{Q}_p \rightarrow \mathbb{R}[/itex] with f(x) = x is a continuous function.

    Clever. I haven't seen or heard anything like that!

    Interesting... how sure are you of this claim (along with the function being surjective if it maps p-adic numbers to their real lexicographic reverse). Im gonna look into this a bit more!

    Yep. thats right.

    Ok, I see where you are going now Shmoe.

    I have indeed chosen a very specific case where I can actually picked a certain prime (2 in this case). I could also have chosen some other prime.

    [tex]|0-y|_2 < 1/2 [/tex]

    Take [itex]y = 4[/itex] then

    [tex]|0 - 4|_2 = \frac{1}{2^2} = 2^{-2} < 1/2[/tex]

    But

    [tex]|0-4| = 4 > 1[/tex]

    Similarly for your second equation I could choose y = 5.
     
    Last edited: Aug 31, 2005
  12. Aug 31, 2005 #11

    Hurkyl

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    Then I would have to say no, because it's not a function at all. It would make sense to ask something like:

    Is [itex]f:\mathbb{Q}_p \cap \mathbb{Q} \rightarrow \mathbb{R} : x \rightarrow x[/itex] a continuous function?

    Because here, the domain is a subset of the range. But [itex]\mathbb{Q}_p[/itex] contains horrible things (it's uncountable, just like the reals), and given one, I can't see any "right" way for viewing it as an element of [itex]\mathbb{R}[/itex].


    A an interesting note, the metric closure of the algebraic closure of [itex]\mathbb{Q}_p[/itex] (called [itex]\Omega_p[/itex] is apparently field isomorphic to [itex]\mathbb{C}[/itex], but Wikipedia doesn't say whether there is an isomorphism as a topological field. (It does say that the existance of the field isomorphism requires the axiom of choice, so it's safe to say that there isn't a nice, canonical isomorphism of the two!)


    I'm optimistic. If it's not actually continuous, I'm pretty sure it's "mostly" continuous, in some intuitive sense.

    It should be obvious that the function is surjective, since the p-adics are isomorphic the set of all left-infinite radix p numbers, and the reals are isomorphic the set of all right-infinite radix p numbers (modulo some relations, such as 1 = 0.zzz... where z = p - 1). :tongue2:
     
  13. Aug 31, 2005 #12
    I spoke to my topology professor today and he said that this function is not continuous - but he did not talk too much about it (since it is an assignment problem). But he approved of what Shmoe and I have been discussing. I think it is safe to say that this function is not continuous.
     
  14. Aug 31, 2005 #13
    I have another function I have been looking at - one which I think IS continuous.

    [tex]f:\{a,b\} \rightarrow \mathbb{N}[/tex]

    with [itex]g(a) = 1[/itex], [itex]g(b) = 2[/itex], and [itex]\{a,b\}[/itex] has the topology [itex]\{\oslash, \{a,b\}, \{a\}\}[/itex].


    I am fairly certain that this function is continuous using the following definition of a continuous function:

    For every open subset [itex]U[/itex] of [itex]Y[/itex] the inverse image [itex]f^{-1}(U)[/itex] is an open subset of [itex]X[/itex].


    I want to use that fact the the topology contains only open sets, which are open subsets of [itex]\{a,b\}[/itex]. So in my opinion it is obvious that the sets in [itex]\{a,b\}[/itex] are open, but I want this for any choice of [itex]U[/itex] in [itex]\mathbb{N}[/itex].

    But then I had another thought.

    [itex]g[/itex] maps [itex]a \in \{a,b\}[/itex] to [itex]1 \in \mathbb{N}[/itex], and [itex]g[/itex] also maps [itex]b \in \{a,b\}[/itex] to [itex]2 \in \mathbb{N}[/itex]. Therefore [itex]g^{-1}(1)[/itex] maps to [itex]a[/itex] which is open with respect to the topology there. But [itex]g^{-1}(2)[/itex] maps to [itex]b[/itex] which is not open with respect to the topology. So the function is not continuous.

    Now I am confused...
     
  15. Aug 31, 2005 #14

    shmoe

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    It almost surely says more than this, something about restricting the function to the rationals? This is what I was pointing out about your notation in post #2 and what Hurkyl has been saying about it not being a function.

    I had meant to keep p fixed at 2, so if you're still working on this

    Find me a y where [tex]|0-y|_2<1/4[/tex] but [tex]|0-y|>1[/tex].

    Find me a y where [tex]|0-y|_2<\delta[/tex] but [tex]|0-y|>1[/tex]. y will depend on [tex]\delta[/tex]


    What's confusing you about your second question? It's not the discrete topology on {a,b}, you have to look at the topology to determine if {b} is open or not.
     
  16. Aug 31, 2005 #15
    b is not open in the topology. So the inverse image of 2 is not mapped to an open set. Hence the function is not continuous.
     
  17. Aug 31, 2005 #16
    .

    Well if I choose [itex]y = 8[/itex] then

    [tex]|0-8|_p = |2^3|_p = 1/8 < 1/4[/tex]

    But

    [tex]|0-8| = 8 > 1[/tex]

    So given a [itex]\delta[/itex] I can find a [itex]y[/itex]:

    [tex]y = \frac{2}{\delta}[/tex]

    such that

    [tex]|0-y|_p < \delta[/tex] but [tex]|0-y|> 1[/tex]
     
  18. Aug 31, 2005 #17

    shmoe

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    For your y given a general delta, are there any conditions on delta? Does it work if delta=1/3 for example?
     
  19. Aug 31, 2005 #18
    Perhaps I should have written

    [tex]y = \frac{p}{\delta}[/tex]

    Because I wrote [itex]y = 2/\delta[/itex] since [itex]p = 2[/itex].
     
  20. Aug 31, 2005 #19
    Delta must be expressible as [itex]\delta = 1/p^n[/itex] for some [itex]n\in\mathbb{N}[/itex]
     
    Last edited: Aug 31, 2005
  21. Sep 1, 2005 #20

    matt grime

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    this only makes sense if you assign a topology to N, since f is continuous iff the inverse image of an OPEN set is open. How do youknow that {2} is open in N?

     
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