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    Suppose |A|=|B| and |B|=|C| then there are bijections f:A->B and g:B->C and since gof:A->C inherits bijectivity from f and g then |A|=|C|.

    So for some example my teacher gave he said

    consider f:\mathbb R \to  (0,\infty) defined by f(x)=e^x

    This is a bijection so |\mathbb R | =|(0,\infty)| but |(0,1) | =|(0,\infty)| then |\mathbb R | =|(0,1)|


    i get the final conclusion but i'm not understanding where the (0,1) comes from.
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    Did you have a previous example of a bijection mapping (0, 1) to (0, infty)? (It is not at all hard to find one),
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    (Original post by will'o'wisp2)
    Suppose |A|=|B| and |B|=|C| then there are bijections f:A->B and g:B->C and since gof:A->C inherits bijectivity from f and g then |A|=|C|.

    So for some example my teacher gave he said

    consider f:\mathbb R \to  (0,\infty) defined by f(x)=e^x

    This is a bijection so |\mathbb R | =|(0,\infty)| but |(0,1) | =|(0,\infty)| then |\mathbb R | =|(0,1)|


    i get the final conclusion but i'm not understanding where the (0,1) comes from.
    There's nothing in your post to tell us. Perhaps it's something that was discussed in class/lectures.

    We could create a relationshp.

    E.g. Consider g : (0,\infty)\to (0,1) defined by g(x)=\dfrac{1}{1+x}

    This is a bijection, and a "g" we can use in the transitivity.
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    (Original post by DFranklin)
    Did you have a previous example of a bijection mapping (0, 1) to (0, infty)? (It is not at all hard to find one),
    There is a function on the previous sheet of paper which was the function y=\dfrac{1}{1-x}
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    (Original post by ghostwalker)
    There's nothing in your post to tell us. Perhaps it's something that was discussed in class/lectures.

    We could create a relationshp.

    E.g. Consider g : (0,\infty)\to (0,1) defined by g(x)=\dfrac{1}{1+x}

    This is a bijection, and a "g" we can use in the transitivity.
    oh right i see, yes there was a bijective mapping from (0,1)->(0,infinity) for a function \dfrac{1}{1-x} which ii think probabyl helps, thanks a bunch
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    (Original post by will'o'wisp2)
    oh right i see, yes there was a bijective mapping from (0,1)->(0,infinity) for a function \dfrac{1}{1-x} which ii think probabyl helps, thanks a bunch
    Wtih that domain the codomain for that function is (1,\infty), for a bijection.
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    (Original post by ghostwalker)
    Wtih that domain the codomain for that function is (1,\infty), for a bijection.
    sorry my mistake i misread, the function was \dfrac{x}{1-x} my bad
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    (Original post by will'o'wisp2)
    sorry my mistake i misread, the function was \dfrac{x}{1-x} my bad
    Yep, that will do it.
 
 
 

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