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Function Mapping to Open Intervals

  1. Jul 21, 2007 #1
    Hello, I'm working on some questions and I need some further explanation;

    First I must Consider the open interval (0,1), and let S be the set of point in the open unit square; thats is, S={(x,y):0<x,y<1}.

    Question (a) says Find a 1-1 function that maps (0,1) into, but not necessarily onto, S. (this is easy)

    So my first thought is the function [tex](\frac{1}{a},\frac{1}{b}) : a,b\in\mathbf{N}[/tex]. I think this function is into, but not all the members of the set S are mapped to, thus not being onto. Is this right?

    The next question i had a problem with: (b) Use the fact that every real number has a decimal expansion to produce a 1-1 function that maps S into (0,1). Discuss whether the formulated function is onto. (keep in mind that any terminating decimal expansion such as .235 represents the same real number .2349999... .)

    Heres my thought for (b), i define a set [tex]A_n , n\in\mathbf{N}[/tex] where n represents the number of decimal places. So n=1 would be one decimal place, n=2 would be two decimal places, and so on. With only one decimal place, [tex]A_1[/tex] contains every combination of (x,y) such that 0<x,y<1 for instance (.9,.3) or (.1,.5). [tex]A_2[/tex] contains every combination of (x,y) such that 0<x,y<1 to two decimal places, and so on.

    So if i arrange all the [tex]A_n[/tex] like this:

    [tex]A_1(1)[/tex] [tex]A_1(2)[/tex] [tex]A_1(3)[/tex] [tex]A_1(4)[/tex] ...
    [tex]A_2(1)[/tex] [tex]A_2(2)[/tex] [tex]A_2(3)[/tex] ...
    [tex]A_3(1)[/tex] [tex]A_3(2)[/tex] ...
    [tex]A_4(1)[/tex] ...

    and i arrange [tex]\mathbf{N}[/tex] like this:

    1 3 6 10 ...
    2 5 9 ...
    4 8 ...
    7 ...

    Then S maps into [tex]\mathbf{N}[/tex] since i represented every decimal expansion in the array. Is this right? But I'm not sure how the reminating decimal affects this, since, for instance, .9 which belongs to [tex]A_1[/tex] is also .8999... .

    The puzzling thing about this question is that at the end of the question it says this: The Schroder-Bernstein Theorem discussed in 1.4.13 to follow can now be applied to conclude that (0,1)~S.

    Why would i apply something that follows to this question?
    Last edited: Jul 21, 2007
  2. jcsd
  3. Jul 21, 2007 #2


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    I don't understand how
    [tex](\frac{1}{a},\frac{1}{b}) : a,b\in\mathbf{N}[/tex]
    is a function, let alone how it maps (0,1) into anything.

    For your second question, why exactly are you trying to map S into N? They're asking you to map S into (0,1).

    Finally, the Shroeder-Berstein theorem tells you that if you have two sets X and Y, a 1-1 function that maps X into Y, and a 1-1 function that maps Y into X, then there exists a bijection (i.e. a function that is both 1-1 and onto) from X onto Y. So, because this question asks you to find a 1-1 function from (0,1) into S, and a 1-1 function from S into (0,1), then you can apply the theorem and say that they (0,1)~S, assuming this notation means there's a bijection between them.
  4. Jul 21, 2007 #3
    For the first part, I have to find a function thats maps (0,1) into the set S. I thought that a function was just a rule for selecting object, the objects we want to select in this case are x,y such that 0<x,y<1. so if i can find a way of selecting different values for x,y so that 0<x,y<1 then i can use that method as the function to map the open interval values to the set S. so if i select x,y values by using choosing any natural numbers a,b, and plug them into (1/a,1/b), then I'll get values that satisfy the requirement for values belonging to S which is 0<x,y<1. so 1/a,1/b acts as the into function. thats my reasoning, still wrong?

    the second part is just wrong so i have to work on that...
  5. Jul 21, 2007 #4


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    I still don't see what your function is. For example, what is [itex]1/\pi[/itex] mapped to?
  6. Jul 21, 2007 #5
    but the question says a function that maps into not necessarily onto, doesn't thats mean that not all S must be mapped? or i could just as easily pick any real numbers a,b > 0 which would cover pi, but since it said not necessarily onto i thought might as well just pick a and b from the natural numbers. if i was to specify that a and b are real numbers > 0, would the function still be incorrect?
    Last edited: Jul 21, 2007
  7. Jul 21, 2007 #6


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    That's my problem: What is your function?!

    Let's call it f. What is f(1/2)? 1/2 is certainly in (0,1).
  8. Jul 21, 2007 #7
    well, thats where it gets a little tricky... i have an example in my book where it defines what would be in the first few [tex]A_n[/tex]'s, for example it says [tex]A_1 = {\frac{0}{1}[/tex],[tex]A_2 = {\frac{1}{1},\frac{-1}{1}[/tex], [tex]A_3 = {\frac{1}{2},\frac{-1}{2},\frac{2}{1},\frac{1}{2}[/tex], and so on, the key is that the denominator and the numeration add to n in A_n. It then goes on to map N to the members of the various A_n sets. It then says this: 'Admittedly, writing an explicit formula for this correspondence would be an awkward task, and attempting to do so is not the best use of time.' So i thought this may be a similar case, but it looks like this approach was incorrect...

    so how do i approach this problem?
    Last edited: Jul 21, 2007
  9. Jul 21, 2007 #8
    As stated in the problem, this is actually a very easy and almost trivial problem. I guess the best hint to give is to think about what set you are trying to map into what other set, I mean in this case you want to try to map (0,1) into (0,1)X(0,1) (or S), you want a map that is one-to-one, but not necessarily onto.

    First can you think of a map from (0,1) to (0,1) itself? Next can you somehow use this to find a map from (0,1) to S?
  10. Jul 21, 2007 #9
    well, for any set A a function map to itself would be f(a)=a where a belongs to A. So are you saying the function I'm looking for is f(a,b)=(a,b) where a,b belong to R?
    Last edited: Jul 22, 2007
  11. Jul 22, 2007 #10


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    You seem to be missing what the question is asking.

    You want to construct a function from (0,1) to S, i.e. your function f takes in values x in (0,1), and spits out values f(x) in S. So when you write f(a,b)=(a,b) where a,b belong to R, then you're definitely not answering the question at all. What you're doing is writing down a function from R2 to R2. Your function should look like f(x) = (y1, y2), where x is in (0,1), i.e. 0<x<1, and (y1, y2) is in S, i.e. 0<y1<1 and 0<y2<1.

    An example of such a function is f(x) = (x/2, x^2). Here, for example, f(1/2)=(1/4, 1/4). Another example is f(x) = (1/3, x/7) (and f(1/2) = (1/3, 2/7) in this case).

    Do you understand what you have to do now?
    Last edited: Jul 22, 2007
  12. Jul 22, 2007 #11
    i think i understand, thanks.
    Last edited: Jul 22, 2007
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