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Show that the set of points on a line is equinumerous

  1. Jan 31, 2005 #1
    Show that the set of points on a line is equinumerous with the set of points in space.

    If you consider R^3, then sum(x,y,z) for all ordered triples will get mapped to some real number. But what function can I make so that I can take all real numbers so that they will map to some ordered triple? All I need is 1 bijective function or two total functions ( 1 going from R^3-->R and 1 going from R--->R^3) to show that the two sets are equinumerous right?
  2. jcsd
  3. Feb 1, 2005 #2
    I don't know of any handy formula you can use, but how about just describing a mapping like this:

    Points on a line between 0 and .999999... can be represented as: [itex]0.x_1x_2x_3x_4x_5x_6x_7x_8x_9...[/itex]

    and map to points in space with coordinates (0,0,0) to (.999...,.999...,.999...): [itex](.x_1x_4x_7...,x_2x_5x_8...,x_3x_6x_9...)[/itex]

    and similarly for points 1.000000... to 1.999999... and so on
  4. Feb 1, 2005 #3
    I'm not sure of a formal proof, but I've seen an idea which says that given a real number x, form the ordered triplet (u,v,w) where u is made up of every 3rd digit of x starting with the first, v every 3rd starting with the second, and w the third.

    For example, map 123456.789101112... -> (14.711..., 25.801..., 36.912...). This does seem to show that given any point p in R3 there is some number in R which maps to p.
  5. Feb 1, 2005 #4
    Just had a thought... I'm not sure exactly how you'd go about getting negative values for two of the coordinates. If x<0 you could arbitrarily choose u,v or w to be negative, but the other two would always be positive. Of course, since {(u,v,w) : v,w>=0} will have the same cardinality as {(u,v,w)} = R3 then this may not be a problem.
  6. Feb 1, 2005 #5

    matt grime

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    Firstly since card(R^3)=>card(R), as one is a subset of the other, it suffices only to show that there is an injection from R^3 to R.

    Your original idea will not work, indeed there can be no such nice function, since that would be contintuous, and for various reasons that isn't going to happen. You need to look at the ideas given above for sending representations of triples of real numbers to representations of real numbers.
  7. Feb 1, 2005 #6
    Thanks for all the help. I understand it very clearly now. This math logic class is going to be good, Godel's proof here we come!
  8. Apr 9, 2005 #7

    Sorry to bring this up again, but this problem came up again in a different class and i was pointed out something I never thought of before. If we use this method then say for the real number 0.999999999999999..........

    would get mapped to (0.99999999........, 0.9999999999999999...........)

    However 0.9999......... is actually equal to 1. But if we take 1.0000000 then that would get mapped to (1.00000000000, 0.000000000000) which is completely different than what one would get with 0.9999....... even though .99999......=1.000000000 thus the function wouldn't be 1-1 right?

    Is there a way around this problem?
    Last edited: Apr 9, 2005
  9. Apr 9, 2005 #8


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    Well, obviously, you're not going to evaluate the function at 0.9999... and at 1.000... you'll just pick one (yes, they represent the same number, but they are different decimal expansions). And that means you have to stipulate which decimal expansion will be chosen in the case where there are multiple possibilities (and by multiple possibilites, I mean 2, and this occurs if and only if the number has a terminating decimal expansion, and thus can also be expressed as a decimal with repeating 9's at the end).

    By the way, the problem you mention wouldn't prevent it from being 1-1, it would prevent it from being a function altogether (since it would map 1 to (1,1,1) and (1,0,0)). Once you stipulate that you'll only use the terminating decimal expansion when given the choice (or the repeating 9 expansion, either one, it doesn't matter), it becomes a function, and obviously, a 1-1 function.
    Last edited: Apr 9, 2005
  10. Apr 9, 2005 #9
    That simple enough, thanks AKG
  11. Apr 9, 2005 #10
    I just thought of another problem with this though. Say we stipulate that we will only use the decimal expansions such as .999999......... instead of 1.000000..... and 1.999.... instead of 2 etc. so .99999....... maps to (.999999......, .99999999)
    What Real number ,however, maps to the point in the plane (1.0000000....,0.000000) then if I am trying to create a bijective function from the reals to the pts. in the Real plane?
  12. Apr 9, 2005 #11

    matt grime

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    Why don'#t you stipluate othe restrictions about the representations of the coordinates of the plane
  13. Apr 9, 2005 #12


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    gravenewworld, try:

  14. Apr 11, 2005 #13
    duh, I'm such an idiot for not thinking of that. I have been pulling too many all nighters recently
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