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Math Help - Cardinal number

  1. #1
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    Set

    I have this example in my text book:
    Prove that (0,1]~(1,3):
    Here is the answer provided:
    We define two bijections:
    f: (0,1]->(0,1) and g: (0,1)->(1,3), with

    the word after x in the third row means "else" in English.

    Can anyone explain how this thing got solved?
    thanks
    Attached Thumbnails Attached Thumbnails Cardinal number-2.jpg  
    Last edited by tashe; January 16th 2008 at 01:48 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by tashe View Post
    I have this example in my text book:
    Prove that (0,1]~(1,3):
    Here is the answer provided:
    We define two bijections:
    f: (0,1]->(0,1) and g: (0,1)->(1,3), with

    the word after x in the third row means "else" in English.

    Can anyone explain how this thing got solved?
    thanks
    somehow, your question isn't clear to me. how what got solved? what are we supposed to do with f(x) and g(x)? state the question as it is in the text

    and what does this have to do with cardinal numbers?
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  3. #3
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    I'm sorry I made a mistake. Its not about cardinal numbers, its probably about sets, but not sure cause I don't have the title written in the book.
    And for the rest, this is the way it is solved.

    As a small primer, there is another example which is solved this way:
    Prove that (0,1)~(1,3) :
    Answer:
    We define a bijection f: (0,1)~(1,3) with f(x)=2x+1
    Thats all i got.
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  4. #4
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    The function g is clearly a bijection g:\left( {0,1} \right) \leftrightarrow \left( {1,3} \right).
    In this problem we must show \left( {0,\left. 1 \right]} \right. \approx \left( {1,3} \right).

    So what the given function f should map 1 to 2 and then ‘shifts’ the other fractions.
    You must show that the function is a bijection.
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  5. #5
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    So if the function can be represented as two bijections than the function itself is a bijection? Is that what you want to point?
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  6. #6
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    Quote Originally Posted by tashe View Post
    So if the function can be represented as two bijections than the function itself is a bijection? Is that what you want to point?
    I misread you question. I assumed it was the way I have done such before.
    Here is what I do.
    0,1] \to (1,3)" alt="f0,1] \to (1,3)" /> f(x) = \left\{ {\begin{array}{lr}    2 & {x = 1}  \\<br />
   {g\left( {\frac{1}{{n + 1}}} \right)} & {x = \frac{1}{n}\,,\,n \ge 2}  \\<br />
   {g(x)} & {\mbox{else}}  \\<br />
\end{array}} \right.

    It follows that f is bijective.
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