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Math Help - Just a few problems I've been working on for too long... (Bijections, sup/inf)

  1. #1
    Member thaopanda's Avatar
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    Just a few problems I've been working on for too long... (Bijections, sup/inf)

    Even after reviewing my notes over and over again and trying to do these problems over and over again, I still can't figure out how to finish them or start them....

    1. Let S, T be sets such that (S \ T) ~ (T \ S). Prove that S ~ T


    2. Let S, S1, T, T1 be sets such that S ~ S1 and T ~ T1. Show that S x T ~ S1 X T1


    3. Let A, B be subsets of R and nonempty
    A is bounded from above and B is bounded from below
    Define A - B as the set { z = x - y : x e A, y e B}
    Show that sup (A - B) = sup A - inf B


    4. Show that [0,1] ~ [0,2)

    Hopefully the problems above are understandable...

    please help...

    - Nicole
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  2. #2
    Member
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    Hi
    1) S = (S\smallsetminus T) \cup (S\cap T)\, T = (T\smallsetminus S) \cup (S\cap T)\. Note that both unions are in fact disjoint unions.
    Because |S\smallsetminus T| =  |T\smallsetminus S|, there's a bijection g: (S\smallsetminus T) \rightarrow (T\smallsetminus S). Now we're ready to construct a bijection f:S\rightarrow T like this:

    f(x) = \begin{cases} g(x) & \mbox{if } x\in (S\smallsetminus T) \\ x & \mbox{if } x\in S\cap T \end{cases}
    Verify it is really a bijection!

    2) |S|=|S_1| means there's a bijection g:S\rightarrow S_1
    |T|=|T_1| means there's a bijection h:T\rightarrow T_1
    Define f:S\times T \rightarrow S_1\times T_1 as follows: f\left((a,b)\right) = (g(a),h(b))
    Verify it is a bijection.

    3) Conditions say that \sup A and \inf B exist.
    Is \sup A - \inf B an upper bound of A-B ? Yes: let z \in A-B, so z = a-b for some a\in A, b\in B.
    But a<\sup A and b>\inf B, so z=a-b <\sup A - \inf B.

    Is \sup A - \inf B the least upper bound of A-B ? Suppose it is not, then there exist \varepsilon >0 such that \sup A - \inf B -\varepsilon is an upper bound of A-B.
    Because \sup A is a supremum, there must exist some a\in A such that a>\sup A-\varepsilon/2.
    Because \inf B is an infimum, there must exist some b\in B such that b<\inf B+\varepsilon/2.
    We get z=a-b>\sup A-\varepsilon/2 - (\inf B+\varepsilon/2) = \sup A - \inf B -\varepsilon which contradicts the assumption that \sup A - \inf B -\varepsilon is an upper bound of A-B.
    We conclude that \sup A - \inf B = \sup(A-B).

    4) Because x->2x is a bijection from [0,1) to [0,2), it will suffice to construct a bijection between [0,1] and [0,1) and composition of the two bijections will be a bijection from [0,1] to [0,2).

    If you know Cantor-Bernstein Theorem, it is easy, because there obviously exist an injective map from [0,1) to [0,1] (the identity map) and an injective map from [0,1] to [0,1) (x -> x/2).

    If you don't know the theorem, consider set A=\{\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4  },\frac{1}{5},\ldots\} \subseteq [0,1] and set B=\{\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5  },\ldots\} \subseteq [0,1). You can construct bijection g:A\rightarrow B by setting g\left(\frac{1}{i}\right)=\frac{1}{i+1} (verify it is a bijection).

    Next, we see that [0,1]\smallsetminus A = [0,1)\smallsetminus B, so the identity on [0,1]\smallsetminus A is a bijection from [0,1]\smallsetminus A to [0,1)\smallsetminus B.

    Now we're ready to construct a bijection f:[0,1]\rightarrow [0,1) as follows:

    f(x) = \begin{cases} g(x) & \mbox{if } x \in A \\ x & \mbox{if } x\in [0,1]\smallsetminus A \end{cases}
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