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How can a curve be one dimensional?

  1. Apr 3, 2012 #1
    I've heard of curves being described as one dimensional but I don't understand how anything other than a straight line can be one dimensional as surely once the curve becomes, well, curved it is now in two dimensions?

    attachment.php?attachmentid=45853&stc=1&d=1333480408.png

    I have illustrated this in the above diagram with the curve and straight line shown in red and the dimensions in green.

    Thanks
    A.
     

    Attached Files:

  2. jcsd
  3. Apr 3, 2012 #2

    Office_Shredder

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    The curve is embedded in a 2 dimensional plane, but the curve itself is one dimensional. For example, I can describe all the points using only one parameter in a very natural way (if the graph is y=f(x), then x is the parameter since it uniquely identifies y)

    Really what you want is dimension to be an intrinsic property. If I took your curve and put it in 3 dimensions, is it a 3 dimensional object now? What about the fact that we're in spacetime, is it 4 dimensional? That's a bit silly, so instead we ask what properties does the curve have that is independent of the space that it is embedded in.

    There are several possible definitions of dimension depending on context but the basic idea is that if it has dimension k you can describe every point using k parameters at least locally - by that I mean, for example the curve x2+y2=1 (a circle) you can describe the top half of it by the set of points of the form [itex](x,\sqrt{1-x^2})[/itex] and the bottom half is of the form [itex](x,-\sqrt{1-x^2})[/itex]. So I wasn't able to describe the whole circle using the x-coordinate but I was able to describe each half of it using a single coordinate, which means the whole thing is 1 dimensional (in fact you can describe the whole circle using a single coordinate if you use trigonometric functions, which I encourage you to try if you haven't seen it)
     
  4. Apr 3, 2012 #3

    Bacle2

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    Would you think of the curve as having non-zero area? I don't mean a closed curve

    that may enclose an area; I mean the curve itself.
     
  5. Apr 3, 2012 #4

    HallsofIvy

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    Yes, such a curve is in two dimensions but it only "one dimensional" itself".

    What is your definition of "dimension"? One common one is: a subset of Euclidean space is said to be "n-dimensional" if and only if any point can be identified using n real numbers.

    Choosing one point on the curve, we can identify any point on the curve by its distance (positive in one direction, negative in the other) from the chosen point. Since that is a single number, the curve is one-dimensional.
     
  6. Apr 3, 2012 #5
    Bit of a weird example here so the curve is like curved corridor in a building, each office on the corridor is numbered with respect to where it is positioned along the corridor and not where it is positioned relative to the 4 outer walls of the building.
     
  7. Apr 3, 2012 #6

    Matterwave

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    Perhaps a somewhat more intuitive way of thinking about it is, if you live "in" the curve (i.e. you're not allowed to go outside of it), you can only ever go forwards or backwards, never to the left or right or up or down.
     
  8. Apr 3, 2012 #7

    Bacle2

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    Altho you may want to consider separately the case of space-filling curves, like the Peano curve.
     
  9. Apr 3, 2012 #8

    Bacle2

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    I'm starting to doubt my answer, if we can define a curve as the continuous image of a connected interval. We can use the fact that, e.g., any compact metric space K is the continuous image of the Cantor set. Then take, e.g., K=[0,1]^n (mapping I into R^n), and extend continuously, using Tietze extension ( C is closed in I ; I is normal), then the image of f has non-empty interior in R^n.

    Note/Edit: I'm referring to a more general definition of a curve, as the continuous image of an interval, and not the curves in the attached files.
     
    Last edited: Apr 3, 2012
  10. Apr 4, 2012 #9

    chiro

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    The easiest way to think about this is finding the minimum parametrization of your object.

    Intuitively for any line, no whatever what its deformation or dimensionality for the embedded space is always a line. Similarly, a flat piece of paper no matter how its deformed or what space its embedded in is always a flat piece of paper.

    If you are given a visualization of the object in question, you can get a very good idea of the number of parameters. If you can't visualize the object but are given a deterministic expression, then you have to use other methods.

    If you are given a non-analytic version of a process where you only know the dimensionality of the embedded space and have to not only figure out the parameterization but also some sort of 'useful' definition for the process as a whole (implicit most likely but could be explicit), then you will have to use even different methods again which will be based on a combination of statistical theory and non-statistical mathematics.

    One method for assessing parameterization is assessing density in various orientations for your function and statistical results used in data mining measure a kind of 'variation' density where you get principal components for your data set. This kind of idea could be applied to your analytic setting in the analytic perspective (not the statistical one).
     
  11. Apr 4, 2012 #10

    Matterwave

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    A curve has the same cardinality as an arbitrary n-dimensional hypercube, but that doesn't mean it has the same dimension. A space-filling curve defines a surjection, but not a homeomorphism.
     
  12. Apr 4, 2012 #11

    Bacle2

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    I was referring to a curve as the continuous, not homeomorphic image of an interval. I was also taking back my statement that a curve must have empty interior. As defined, I think the argument is correct, and a curve does not necessarily have an empty interior.
     
  13. Apr 4, 2012 #12

    Bacle2

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    Clearly, the image of an interval would depend on a single parameter. I was referring to the claim/belief that the continuous image of a curve would have no area, i.e., would have empty interior.
     
  14. Apr 4, 2012 #13

    Bacle2

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    Sorry, one last post on this, to try to avoid confusion:

    0)Define a curve as the continuous image of an (connected)interval
    Claim: A curve can have non-empty interior.

    i)We have that every compact metric space is the continuous image of the
    Cantor set C . So we map C subset I:=[0,1] into, say I^n, so there is an f with f(C)=I^n (n>1)

    ii) By Tietze extension thm. ( C is closed in I normal) , we can extend f to f^, defined on the whole of I, continuously ( on each real variable). Then f^ is a continuous map such that f^(I) has non-empty interior in R^n.
     
  15. Apr 4, 2012 #14

    chiro

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    I'll have a look at your theorems later on, but the thing is if you can deform your object that you have in a way so that is linear, then apply the ideas of dimensionality to get not only the proper dimension but the parameterization of your object, then you are done. By finding the appropriate deformation, you have created a linear representation that can be dealt with in the linear context.

    This is of course equivalent to finding the inverse for each part of your object with respect to the different 'branches' that exist. If you can find the inverse for each 'branch' of your object (might be multi-dimensional and multi-parameterized but the idea is the same), then you can find a linear representation which can be parameterized.

    So the case with a line would result in basically an n dimensional system that represents a line with one parameter t where X(t) = At + (1-t)B in the deformed state to the linear space. If this parameterization wasn't only one dimensional, then the deformation wouldn't produce one but however many parameters for the object.

    With the notion of topology, if you know the object is analytic continuous then there should exist a deformation to the linear representation which depends on the inverse within a given branch-volume that corresponds to where the derivatives are (which correspond to the ideas in the inverse function theorem which are calculated by finding zero-jacobians). With this information you can find the branches and thus deform that region to it's linear counterpart.

    The reason I like to think of something in the deformed linear space is because linear spaces, decompositions, and parameterizations of linear systems are well understood both algebraically, algorithmically and also to an extent visually. It's very easy to parameterize something that is deformed to linear space than to try and analyze it in its non-deformed non-linear space with a function like say f(x,y,z,w,t,u,v,a) = blah for any analytic continuous blah.

    You could also probably consider continuous representations that are not analytic over the whole domain and use the same argument based on topological reasoning but I will only speculate on this (intuitively it makes sense at least).
     
  16. Apr 4, 2012 #15

    Bacle2

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    I think the best we can say is that if f is C^1 (maybe Lipschitz) , then, with a bounded derivative, we can preserve the Hausdorff dimension.
     
  17. Apr 4, 2012 #16

    chiro

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    I'll go with that for finding the dimension (which is the OP's question), but in terms of understanding why (and where) dimension goes from say one thing to another it may be useful to see which 'parts' if they are isolated contribute to where the function requires more dimensions as opposed to less (it can happen but not always).

    The only reason I am saying this (for the OP) is that visually you can see where dimensionality changes. I'm imagining some kind of implicit function that creates branches (like a bifurcation) and does this repeatedly which creates more dimensions at different parts.

    Just out of curiosity, what would you say topologically is a way to do the above? In other words in the case of say a bifurcation like object, how would you isolate this kind of phenomenon in terms of the local characteristics (spatially) of the object?
     
  18. Apr 4, 2012 #17

    Bacle2

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    The way I see it, if the derivative is bounded, then the image of the balls in the covering will not expand by much.

    Sorry, it is 4 a.m., here, I'm out for the night; will try to be back in case there are more posts.
     
  19. Apr 4, 2012 #18

    HallsofIvy

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    Then any two-dimensional ball about a point on the curve will necessarily include some points not on the curve- a curve has no (2 dimensional) interior.
     
  20. Apr 4, 2012 #19

    Bacle2

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    Halls of Ivy:

    I did a specific construction, with a specific definition of curve; what is it about my construction that you think is faulty?

    It seems strange-enough that the hypercube ( as a compact metric space; a space with non-empty interior) is the continuous image of the Cantor set. Then what is wrong with the extension?
     
  21. Apr 5, 2012 #20
    I think the OP is simply reflecting on the fact that there is no such thing as intrinsic curvature in one dimension.
     
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