MAT102 Lecture 6

• If a and b are real numbers $a<b$
  • $(a,b)=\{x\in \mathbb{R}:a<x<b\}$
  • $[a,b]=\{x\in \mathbb{R}:a\le x\le b\}$
• 2
  • a
    • Show that $(0,1)$ is open.
    • We need to prove that there is always a number $r$ such that $(x-r,x+r)\subseteq (0,1)$.
    • $\mathrm{\forall }x\in S,\mathrm{\exists }r>0:(x-r,x+r)\subseteq S$
    • Let's set up some numbers:
      • $n\in \mathbb{R},n<r$
      • $n>0$
      • $n=\frac{r}{2}$
      • This means that there is always a number less than $r$, which means there is always room to expand leftwards towards the $(\underset{―}{0},1)$ part of the set.

    ---

    • We want to prove that the set $(0,1)$ is open.
    • Let's set up two variables. $x$ is in the original set $(0,1)$. $y$ is in a set which is inside $(0,1)$.
    • $x\in (0,1)$
    • $y\in (x-r,x+r)$
    • $r=min(x,1-x)$
      • $r>0$
    • $(x-r,x+r)\subseteq (0,1)$
    • Because of r being positive, this will result in:
      • $x-r<x<x+r$
      • $0\le x-r<x<x+r\le 1$
      • So $x-r\ge 0$ and $x+r\le 1$ always. Meaning the set $y$ is enclosed by the open interval $x$.
    • Example:
      • $x=0.3$
      • $r=min(0.3,0.7)$
      • $r=0.3$
      • $(x-0.3,x+0.3)$ should be in $(0,1)$
      • $(0.3-0.3,0.6)$
      • $(0,0.6)$
      • This is contained by $(0,1)$

    ---

    • You can also use two seperate cases.
    • $\{\begin{array}{ll}0<x\le \frac{1}{2},& r=x\\ \frac{1}{2}<x<1,& r=1-x\end{array}$
    • Let $x\in (0,1)$
    • If case 1 $r=x$.
      • Need to show $(x-r,x+r)\subseteq (0,1)$ = $(0,2x)$
      • Need to show $2x\le 1$.
      • From the start, we know $x\le \frac{1}{2}$. $2x\le 1$.
    • Case 2 $r=1-x$
      • $(x-r,x+r)=(1-2x,1)$
      • Need to show $1-2x\ge 0$
      • If $x\ge \frac{1}{2}$
      • $1-2x\ge 0$