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Math Help - Define the bijection

  1. #1
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    Define the bijection

    Define a bijection between (0,1) and [0,1].



    I know that in order to prove this is to use a piecewise function. So this is what I have.
    (This is a piecewise function just could not figure out how to put it on here)


    This is what I did... Please help me if im wrong.

    f(x) = x if  x \neq \frac {1}{2^n} for any  n \epsilon N
    f(x) =  \frac{1}{2^n} if  n \epsilon N\cup {0}
    Last edited by tigergirl; May 9th 2010 at 08:50 PM. Reason: missed spelling
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    Quote Originally Posted by tigergirl View Post
    Define a bijection between (0,1) and [0,1].



    I know that in order to prove this is to use a piecewise function. So this is what I have.
    (This is a piecewise function just could not figure out how to put it on here)


    This is what I did... Please help me if im wrong.

    f(x) = x if  x \neq \frac {1}{2^n} for any  n \epsilon N
    f(x) =  \frac{1}{2^n} if  n \epsilon N\cup {0}
    Are we in Reals, Integers, Rationals....?
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    we r in reals...
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  4. #4
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    Are you trying to show (0,1) has cardinality c?
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    Quote Originally Posted by tigergirl View Post
    Define a bijection between (0,1) and [0,1].



    I know that in order to prove this is to use a piecewise function. So this is what I have.
    (This is a piecewise function just could not figure out how to put it on here)


    This is what I did... Please help me if im wrong.

    f(x) = x if  x \neq \frac {1}{2^n} for any  n \epsilon N
    f(x) =  \frac{1}{2^n} if  n \epsilon N\cup {0}
    It helps if you specify the domain and codomain, as in

    f: \mathbb{R} \rightarrow \mathbb{R}

    I think what you were going for is something like

    f: (0,1) \rightarrow [0,1]

    defined piecewise (I also haven't mastered the curly brace thing to make it look nice):

    f(x) = 0 if x = \frac{1}{2}

    f(x) = 1 if x = \frac{1}{4}

    f(x) = \frac{1}{2^{n-2}} if there exists n > 2, n \in \mathbb{Z} such that x = \frac{1}{2^n}

    f(x) = x if there does not exist n \in \mathbb{Z} such that x = \frac{1}{2^n}

    Now, can you prove that this is a bijection?

    (A few edits.)
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    Quote Originally Posted by undefined View Post
    It helps if you specify the domain and codomain, as in

    f: \mathbb{R} \rightarrow \mathbb{R}

    I think what you were going for is something like

    [tex]f: (0,1) \rightarrow [0,1][/math]<---we have a problem here; namely, 0 \notin (0,1)

    defined piecewise (I also haven't mastered the curly brace thing to make it look nice):

    f(x) = 0 if x = \frac{1}{2}

    f(x) = 1 if x = \frac{1}{4}

    f(x) = \frac{1}{2^{n-2}} if there exists n > 2, n \in \mathbb{Z} such that x = \frac{1}{2^n}

    f(x) = x if there does not exist n \in \mathbb{Z} such that x = \frac{1}{2^n}

    Now, can you prove that this is a bijection?

    (A few edits.)
    Some times a function mapping from \mathbb{R} to \mathbb{R} can be bijective, but it seems to me that it's quite impossible to know what smallest real number x could go into f(x), let alone finding a bijection.

    By definition, every element in the domain of a relation R must have an image in its codomain in order to be a function. In this case, we have no way of insuring every element in the domain to have an image since we don't even have the smallest element in it.
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    MHF Contributor undefined's Avatar
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    Quote Originally Posted by novice View Post
    f: (0,1) \rightarrow [0,1]<---we have a problem here; namely, 0 \notin (0, 1)
    Sorry I don't follow. 0 \notin (0, 1) doesn't pose a problem? I never try to define f(0).

    Quote Originally Posted by novice View Post
    Some times a function mapping from \mathbb{R} to \mathbb{R} can be bijective, but it seems to me that it's quite impossible to know what smallest real number x could go into f(x), let alone finding a bijection.

    By definition, every element in the domain of a relation R must have an image in its codomain in order to be a function. In this case, we have no way of insuring every element in the domain to have an image since we don't even have the smallest element in it.
    I don't follow this either. We don't need to worry about smallest elements, and obviously in (0,1) there is no smallest element. Also, we're not mapping from \mathbb{R} to \mathbb{R}, we are mapping from a subset of \mathbb{R} to a subset of \mathbb{R}.

    Would you mind re-reading my post and seeing if you still think it's problematic?

    Edit: By the way, I wrote f: \mathbb{R} \rightarrow \mathbb{R} as an arbitrary example, but perhaps this was a poor choice of example since it might be interpreted as tying in with dwsmith's first question about whether we are in reals, etc. I should have chosen an example that has no relation to the problem, like f: \{1,2,5,7,9\} \rightarrow \mathbb{C}.
    Last edited by undefined; May 10th 2010 at 08:28 AM.
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    Quote Originally Posted by undefined View Post
    Sorry I don't follow. 0 \notin (0, 1) doesn't pose a problem? I never try to define f(0).



    I don't follow this either. We don't need to worry about smallest elements, and obviously in (0,1) there is no smallest element. Also, we're not mapping from \mathbb{R} to \mathbb{R}, we are mapping from a subset of \mathbb{R} to a subset of \mathbb{R}.

    Would you mind re-reading my post and seeing if you still think it's problematic?

    Edit: By the way, I wrote f: \mathbb{R} \rightarrow \mathbb{R} as an arbitrary example, but perhaps this was a poor choice of example since it might be interpreted as tying in with dwsmith's first question about whether we are in reals, etc. I should have chosen an example that has no relation to the problem, like f: \{1,2,5,7,9\} \rightarrow \mathbb{C}.
    I am really not an expert in functions. You seem to know more than I do. I am very glad to know that we need not be concerned over the smallest element in the domain. I need some time to think about your proposal.
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    Quote Originally Posted by undefined View Post
    Sorry I don't follow. 0 \notin (0, 1) doesn't pose a problem? I never try to define f(0).



    I don't follow this either. We don't need to worry about smallest elements, and obviously in (0,1) there is no smallest element. Also, we're not mapping from \mathbb{R} to \mathbb{R}, we are mapping from a subset of \mathbb{R} to a subset of \mathbb{R}.

    Would you mind re-reading my post and seeing if you still think it's problematic?

    Edit: By the way, I wrote f: \mathbb{R} \rightarrow \mathbb{R} as an arbitrary example, but perhaps this was a poor choice of example since it might be interpreted as tying in with dwsmith's first question about whether we are in reals, etc. I should have chosen an example that has no relation to the problem, like f: \{1,2,5,7,9\} \rightarrow \mathbb{C}.
    Yes, you are right, we don't need to know the smallest element. I see n>2.
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    Quote Originally Posted by tigergirl View Post
    Define a bijection between (0,1) and [0,1].



    I know that in order to prove this is to use a piecewise function. So this is what I have.
    (This is a piecewise function just could not figure out how to put it on here)


    This is what I did... Please help me if im wrong.

    f(x) = x if  x \neq \frac {1}{2^n} for any  n \epsilon N
    f(x) =  \frac{1}{2^n} if  n \epsilon N\cup {0}
    I know this does not answer your question, but it is true in general that if E\subseteq U is infinite then \text{card }E=\text{card }E\cup\{e_1,\cdots,e_n\}

    In fact, if \gamma is an infinite cardinal number and \kappa<\gamma then \gamma+\kappa=\gamma
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