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Thread: bijective maps

  1. #1
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    bijective maps

    Let n \in \mathbb{Z^+}, X is nonempty.
    (i) Find a bijective map f: X^\omega \times X^\omega \rightarrow X^\omega
    (ii) If A \subset B find an injective map A^\omega)^n \rightarrow B^\omega" alt="gA^\omega)^n \rightarrow B^\omega" />

    The cartesian product A_1 \times A_2 \times ... is the set of all \omega-tuples of elements of X is denoted by X^\omega.

    I've tried many examples but none works... Can I get some help with these two please?
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  2. #2
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    (i) define f for example like this: f(\left<g,h\right>)=u where u(n) = \begin{cases} g(m) & \mbox{if } n=2m \\ h(m) & \mbox{if } n=2m+1 \end{cases}

    (ii) if A is empty then g = \emptyset (the empty mapping) is injective. if A is nonempty, you can define g similarly as in (i):

    g(\left<h_0, \ldots, h_{n-1}\right>) = v where v(i) = \begin{cases} h_0(m) & \mbox{if } i=nm \\ h_1(m) & \mbox{if } i=nm+1 \\ \cdots \\ h_{n-1}(m) & \mbox{if } i=nm+n-1\end{cases}
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  3. #3
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    Quote Originally Posted by dori1123 View Post
    Let n \in \mathbb{Z^+}, X is nonempty.
    (i) Find a bijective map f: X^\omega \times X^\omega \rightarrow X^\omega
    This makes no sense. Did you mean X\subset Z^+ rather than n \in Z^+?

    (ii) If A \subset B find an injective map A^\omega)^n \rightarrow B^\omega" alt="gA^\omega)^n \rightarrow B^\omega" />

    The cartesian product A_1 \times A_2 \times ... is the set of all \omega-tuples of elements of X is denoted by X^\omega.

    I've tried many examples but none works... Can I get some help with these two please?
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  4. #4
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    Quote Originally Posted by HallsofIvy View Post
    This makes no sense. Did you mean X\subset Z^+ rather than n \in Z^+?
    Hi,
    what exactly doesn't make sense? condition n \in \mathbb{Z}^+ applies to part (ii).
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  5. #5
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    it is n \in \mathbb{Z^+}, it's a problem from the book Topology by Munkres
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  6. #6
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    When you write f(\left<g,h\right>) = ..., is this \left<g,h\right> the cyclic group generated by g and h?
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  7. #7
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    no, it is an ordered pair.
    f: X^\omega \times X^\omega \rightarrow X^\omega means that f takes a pair of functions from \omega to X and returns another function from \omega to X.
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  8. #8
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    Thanks I just want to make sure because I've been using the symbol < > for cyclic groups..

    Can you also help me with one more problem?
    with the same conditions given,
    find a bijective map f: X^n \times X^\omega \rightarrow X^\omega
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  9. #9
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    if you understand previous two solutions, this one should be no problem for you..
    define f(\left<g,h\right>)=u where u(i) = \begin{cases} g(i) & \mbox{if } i<n \\ h(i-n) & \mbox{if } i \ge n \end{cases}

    can you verify it is well-defined, injective and surjective?
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