text: # Just a few problems I've been working on for too long... (Bijections, sup/inf)
  
  • Sep 13th 2009, 01:24 PM
  thaopanda
  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...
  
  - Nicole
  • Sep 13th 2009, 05:36 PM
  Taluivren
  Hi
  1) $\displaystyle S = (S\smallsetminus T) \cup (S\cap T)\$, $\displaystyle T = (T\smallsetminus S) \cup (S\cap T)\$. Note that both unions are in fact disjoint unions.
  Because $\displaystyle |S\smallsetminus T| = |T\smallsetminus S|$, there's a bijection $\displaystyle g: (S\smallsetminus T) \rightarrow (T\smallsetminus S)$. Now we're ready to construct a bijection $\displaystyle f:S\rightarrow T$ like this:
  
  $\displaystyle f(x) = \begin{cases} g(x) & \mbox{if } x\in (S\smallsetminus T) \\ x & \mbox{if } x\in S\cap T \end{cases}$
  Verify it is really a bijection!
  
  2) $\displaystyle |S|=|S_1|$ means there's a bijection $\displaystyle g:S\rightarrow S_1$
  $\displaystyle |T|=|T_1|$ means there's a bijection $\displaystyle h:T\rightarrow T_1$
  Define $\displaystyle f:S\times T \rightarrow S_1\times T_1$ as follows: $\displaystyle f\left((a,b)\right) = (g(a),h(b))$
  Verify it is a bijection.
  
  3) Conditions say that $\displaystyle \sup A$ and $\displaystyle \inf B$ exist.
  Is $\displaystyle \sup A - \inf B$ an upper bound of $\displaystyle A-B$ ? Yes: let $\displaystyle z \in A-B$, so $\displaystyle z = a-b$ for some $\displaystyle a\in A$, $\displaystyle b\in B$.
  But $\displaystyle a<\sup A$ and $\displaystyle b>\inf B$, so $\displaystyle z=a-b <\sup A - \inf B$.
  
  Is $\displaystyle \sup A - \inf B$ the least upper bound of $\displaystyle A-B$ ? Suppose it is not, then there exist $\displaystyle \varepsilon >0$ such that $\displaystyle \sup A - \inf B -\varepsilon$ is an upper bound of $\displaystyle A-B$.
  Because $\displaystyle \sup A$ is a supremum, there must exist some $\displaystyle a\in A$ such that $\displaystyle a>\sup A-\varepsilon/2$.
  Because $\displaystyle \inf B$ is an infimum, there must exist some $\displaystyle b\in B$ such that $\displaystyle b<\inf B+\varepsilon/2$.
  We get $\displaystyle z=a-b>\sup A-\varepsilon/2 - (\inf B+\varepsilon/2) = \sup A - \inf B -\varepsilon$ which contradicts the assumption that $\displaystyle \sup A - \inf B -\varepsilon$ is an upper bound of $\displaystyle A-B$.
  We conclude that $\displaystyle \sup A - \inf B = \sup(A-B)$.
  
  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 $\displaystyle A=\{\frac{1}{1},\frac{1}{2},\frac{1}{3},\frac{1}{4 },\frac{1}{5},\ldots\} \subseteq [0,1]$ and set $\displaystyle B=\{\frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5 },\ldots\} \subseteq [0,1)$. You can construct bijection $\displaystyle g:A\rightarrow B$ by setting $\displaystyle g\left(\frac{1}{i}\right)=\frac{1}{i+1}$ (verify it is a bijection).
  
  Next, we see that $\displaystyle [0,1]\smallsetminus A = [0,1)\smallsetminus B$, so the identity on $\displaystyle [0,1]\smallsetminus A$ is a bijection from $\displaystyle [0,1]\smallsetminus A$ to $\displaystyle [0,1)\smallsetminus B$.
  
  Now we're ready to construct a bijection $\displaystyle f:[0,1]\rightarrow [0,1)$ as follows:
  
  $\displaystyle f(x) = \begin{cases} g(x) & \mbox{if } x \in A \\ x & \mbox{if } x\in [0,1]\smallsetminus A \end{cases}$