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Homework Help: Prove (0,1)=[0,1]

  1. Dec 2, 2008 #1
    1. The problem statement, all variables and given/known data
    prove (0,1)=[0,1]
    Define f;(0,1)-->R as follows.
    For n[tex]\in[/tex]natural numbers, n[tex]\geq[/tex]2, f(1/n)=1/(n-1) and for all other x[tex]\in[/tex](0,1), f(x)=x.


    2. Relevant equations



    3. The attempt at a solution
    I know I need to start proving f is a 1-1 function from (0,1) into (0,1]. The problem si getting started.
    I know for 1-1 f(y)=f(z) implies y=z.
     
  2. jcsd
  3. Dec 2, 2008 #2
    Prove (0,1) = [0,1] ?
    Or perhaps disprove? Remember, intervals are just sets. To prove 2 sets are equal, you must prove each set is a subset of the other.

    (0,1) is a subset of [0,1], but [0,1] is not a subsets of (0,1) since 0 and 1 are not in (0,1). So It is not true that (0,1)=[0,1]

    As for your second question. You want to prove F is 1-1? A function F:A->B is 1-1 if for all a and a' in A, f(a) = f(a') implies a = a'

    So suppose f(a) = f(a') ... now this is all you have to work with, evaluate what f(a) and f(a') is.
     
  4. Dec 2, 2008 #3
    Well sorry it's to prove [0,1]=(0,1), but it is possible.
     
  5. Dec 2, 2008 #4
    It seems what you really want to prove is |(0,1)| = |[0,1]|, where | | represents cardinality.
     
  6. Dec 2, 2008 #5
    The intervals we're talking about are sets of real numbers and it doesn't matter what order you put the equality. Either way, [0,1] is not a subset of (0,1). Why?

    If [0,1] were equal to (0,1) then [0,1] would be a subset of (0,1). That means that every number in [0,1] would be inside (0,1) but that is clearly not the case since the boundary points of [0,1] are not included in (0,1).

    Please show me proof that [0,1] = (0,1)
     
  7. Dec 2, 2008 #6
    Ah that makes more sense mutton.
     
  8. Dec 2, 2008 #7
    It's a real analysis proof
     
  9. Dec 3, 2008 #8

    HallsofIvy

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    Science Advisor

    The first thing you need to do is read the problem carefully!

    You have twice now told us that the problem is to prove that [0,1]= (0,1).

    You can't- it's not true.

    You also said "I know I need to start proving f is a 1-1 function from (0,1) into (0,1]", which would NOT prove that [0,1]= (0,1). It would prove that the cardinality of (0, 1) is the same as the cardinality of [0,1]. That is completely different from saying "(0,1)= [0,1]".

    Try mapping the irrational numbers to themselves, then look at the rationals only. Since they are countable, they can be put into a sequence- and then "shifted".
     
  10. Dec 3, 2008 #9
    Oops, i guess it is (0,1)=[0,1].
    Ok, then evaluate for a and a'.
    f(1/a)=1/(a-1))
    f(1/a')=1/(a'-1)
    1/(a-1)=1/(a'-1)
     
  11. Dec 3, 2008 #10
    No, it's not. Get your notation right.
     
  12. Dec 3, 2008 #11
    yes it is. I just checked my book and my book says to prove (0,1)=[0,1].
     
  13. Dec 3, 2008 #12
    Definitely not true. (0,1) doesn't include 0. [0,1] does.
     
  14. Dec 3, 2008 #13
    This is frustrating haha so this will be my last post on this thread :) Using standard interval notation (0,1) is not equal to [0,1]. I think you should do a little more research into set theory and what an interval actually is (It's a set) and then you might understand what you are trying to tell us. Either you are not including enough information, or there is a typo in your book. There is no convincing us otherwise.
     
  15. Dec 3, 2008 #14
    Ok, sorry it's frustrating me too because I've always been told that (0,1) doesn't include 0 and 1, but [0,1] does, but now I have to prove these 2 are equal. It's hard to grasp.
     
  16. Dec 3, 2008 #15
    Haha k I guess it wont be my last post :) You've been told correctly. (0,1) does not include 0 and 1 but [0,1] does. Both of those intervals are sets. Sets are just collections of numbers. Are the sets {1,2} and {1,2,3} equal? What about the sets {1} and {2}? Of course not! By the same reasoning [0,1] is not equal to (0,1) because (0,1) doesn't have 0 and 1.

    [Edit]

    Well more generally, a set is a collection objects but in our case we'll be using real numbers.
     
  17. Dec 3, 2008 #16
    You are not proving that they are equal. You are proving that they have the same number of elements.

    Just check that your function is a bijection between the two sets. (Read HallsofIvy's last paragraph.)
     
  18. Dec 3, 2008 #17
    Oh geez, I should have been a little more careful with my last post. But yes, listen to mutton and HallsofIvy.
     
  19. Dec 7, 2008 #18
    I did a little more...


    Define f(0,1)-->R as follows
    For n element of natural numbers, n>=2,f(1/n)=1/n-1 and for all other x elemts of(0,1), f(x)=x.

    I have some steps to follow, proving that f is a 1-1 function from (0,1) into (0,1] and proving that f is a function from (0,1) onto (0,1] are the first 2 steps.
    I know how to prove 1-1.
    Let a=1/n and b=1/m
    f(a)=1/n-1
    f(b)=1/m-1
    1/n-1=1/m-1
    m-1/n-1=1
    m-1=n-1
    m=n and thus 1-1.
    I see that but I'm unsure how to prove that it goes to (0,1] because of the 1. I know 1/2-->1, but I assume this doesn't really prove it.

    I'm unsure about proving onto.
    So, we use im f.
    y=1/n-1
    y(n-1)=1
    n-1=1/y
    n=1/y +1
    Let a= x and b=z
    n=1/x+1 and n=1/z+1
    1/x+1=1/z+1
    1/x=1/z
    1=x/z
    z=x, thus onto.
    My problem is proving for the specific intervals.
     
  20. Dec 7, 2008 #19

    statdad

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    Homework Helper

    It has to be to show the sets have the same cardinality. Notice that

    [tex]
    [0,1] = (0,1) \cup \{0,1\}
    [/tex]

    and that [tex] (0,1) [/tex] and [tex] \{0,1\}[/tex] are disjoint. You should know the cardinality of both [tex] [0,1] [/tex] and [tex] (0,1) [/tex], so what can you conclude?
     
  21. Dec 7, 2008 #20
    We have done nothing with cardinality.
     
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