MAT102 Tutorial W4

• 1
  • a
    • An open set that is not an open interval could be $\mathbb{R}$. Because $\mathrm{\forall }x\in \mathbb{R},\mathrm{\exists }r>0(x-r,x+r)\subseteq \mathbb{R}$.
    • For any $x$ you can create an open interval around it of size $r$, that will still be a real number.
    • Or
      • $(1,2)\cup (3,4)$
  • b
    • A closed set would be $[0,1]$. Because $[0,1{]}^c$ is $(-\mathrm{\infty },0)\cup (0,\mathrm{\infty })$.
  • c
    • Find a set that is open and closed.
    • $\mathbb{N}$ because it's bounded on the lower end being either $0,1$.
    • $\mathbb{N}=[0,\mathrm{\infty })$

    ---

    • $\mathrm{\varnothing }$
    • Vacuously true. Can't prove if it's open or closed, so $\text{false}⟹\text{true or false}$
    • Meaning true.
  • d
    • Negate the def of being open.
    • $\mathrm{\forall }x\in A,\mathrm{\exists }r>0(x-r,x+r)\subseteq A$
    • $\mathrm{\exists }x\in A,\mathrm{\forall }r>0(x-r,x+r)⊈A$
    • $\mathrm{\exists }x\in A,\mathrm{\forall }r>0(x-r,x+r)\subseteq A^c$
    • For a set not to be open. It means we have a value $x$ where no matter what $r$ or $ϵ$ is, it's always outside the set.
  • e
    • A set that's not open and not closed.
    • $\mathbb{R}\setminus \mathbb{Q}$
      • All real numbers excluding rationals
      • So all irrationals
• 2
  • a
    • a) $\mathrm{\forall }x\in A,\mathrm{\forall }y\in B,(x==y)$
    • b) $\mathrm{\exists }x\in A,\mathrm{\forall }y\in B,(x==y)$
    • c) $\mathrm{\exists }x\in A,\mathrm{\exists }y\in B,(x==y)$
      • If a), $A=B$
      • If b), the x in A, is equal to everything in B.
    • Show that $A⟹B$ but the converse is false
      • $\mathrm{\forall }x\in A,\mathrm{\forall }y\in B,(x==y)⟹\mathrm{\exists }x\in A,\mathrm{\forall }y\in B,(x==y)$
      • This is only true if $A=\{1\}\wedge B=\{1\}$. If there are ever more elements to these sets, it won't work.
      • This also means the second part is true, that there is an x value in A where all y values in B are equal.
      • $\mathrm{\exists }x\in A,\mathrm{\forall }y\in B,(x==y)⟹\mathrm{\forall }x\in A,\mathrm{\forall }y\in B,(x==y)$
      • $\{\begin{array}{l}A=\{1,2\}\\ B=\{1,2\}\end{array}$
      • For this set it doesn't work, will it work for singleton sets?
      • $\{\begin{array}{l}A=\{1\}\\ B=\{1\}\end{array}$

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      • $\mathrm{\forall }x\in A,\mathrm{\forall }y\in B,(x==y)$
      • If every element in A is the same as B, then it works.
        • Singleton sets, or when set $A=B$
      • There is one element in A where every Y in B is that element.
      • Converse:
        • $B⟹A$
        • $\mathrm{\exists }x\in A,\mathrm{\forall }y\in B,(x==y)⟹\mathrm{\forall }x\in A,\mathrm{\forall }y\in B,(x==y)$
        • There is an x in A where all y in B are equal.
        • $\{\begin{array}{l}A=\{1,2,3\}\\ B=\{1\}\end{array}$
        • Second part doesn't hold.
    • Show that $B⟹C$'s converse is false
      • b) $\mathrm{\exists }x\in A,\mathrm{\forall }y\in B,(x==y)$
      • c) $\mathrm{\exists }x\in A,\mathrm{\exists }y\in B,(x==y)$
      • Pick some $x$ and $y_0\in B$.
      • Then we know $x=y_0$ so then C holds.
      • $C⟹B$
      • $\{\begin{array}{l}A=\{1,3,5\}\\ B=\{1,0,2\}\end{array}$
      • C is true for $1==1$
      • But that doesn't mean that $1==0,2$
• 3
  • Let $a\in \mathbb{R},b\in \mathbb{Q}$ be anything, and assume $a,b\ne 0$
  • If $ab=\mathbb{R}\setminus \mathbb{Q}$ (i.e $ab$ is irrational). Then $a$ is irrational.
  • A is any real number
  • B is rational
    • $b=\frac{p}{q}$
    • $p,q\ne 0$
    • $p,q\in \mathbb{Z}$
  • if $a\left(\frac{p}{q}\right)\in \mathbb{R}\setminus \mathbb{Q}$ then $a\in \mathbb{R}\setminus \mathbb{Q}$
  • $\frac{ap}{q}$
  • $a\in {\mathbb{Q}}^c\cap \mathbb{R}$
  • $a\in Q^c$
  • $a\in \mathbb{R}$
  • Since we can't exactly do anything with irrationals, only $\mathbb{R}\setminus \mathbb{Q}$, we negate.

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  • Contrapositive
  • If a is rational, then ab is rational
  • Let $a$ be rational,
    • $a=\frac{p}{q},p\in \mathbb{Z},q\in \mathbb{N}$
  • We know $b\in \mathbb{Q}$
    • $b=\frac{m}{n},m\in \mathbb{Z},n\in \mathbb{N}$
  • $ab=\frac{p}{q}\frac{m}{n}$
  • $ab=\frac{pm}{qn}$
  • Since $p,m\in \mathbb{Z}$, then $pm$ is $\mathbb{Z}$.
  • Since $q,n\in \mathbb{N}$, then $qn$ is $\mathbb{N}$.
  • This means that $ab\in \mathbb{Q}$, since the numerator is $\mathbb{Z}$ and the denominator is $\mathbb{N}$
• 4
  • $a,b\in \mathbb{R}$ and $a,b\ne 0$
  • If ab is negative, and a is positive, then b is negative.
  • $P⟹Q\vee R$
  • Slower Way
    • Let $ab<0$. and let $a>0$
    • This means that the result of a positive multiplied with a negative results in a negative.
    • When $ab<0$, either ($a>0$ or $b<0$) or ($a<0$ or $b>0$). Since $a>0$, the only option is that $b<0$.
  • Krishna's Way
    • We proceed with the contrapositive.
    • We need to show that if $a>0$ and $b>0$, then $ab>0$.
    • Let $a>0\wedge b>0$
    • Notice that multiplying any two positive $\mathbb{R}$ numbers is always going to be positive.
    • $a\cdot b>0$