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:
  
  $\displaystyle f(x) = \begin{cases} g(x) & \mbox{if } x \in A \\ x & \mbox{if } x\in [0,1]\smallsetminus A \end{cases}$
explanation: The document contains a series of math problems and their corresponding solutions. The problems are related to set theory, bijections, and properties of real numbers. Each problem is clearly stated, and the solutions are detailed, step-by-step explanations.
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distilabel_metadata.content: # 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:
  
  $\displaystyle f(x) = \begin{cases} g(x) & \mbox{if } x \in A \\ x & \mbox{if } x\in [0,1]\smallsetminus A \end{cases}$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a series of math problems and their corresponding solutions. The problems are related to set theory, bijections, and properties of real numbers. Each problem is clearly stated, and the solutions are detailed, step-by-step explanations.
  
  Label: multiple-question-and-answer