I am preparing for my analysis exam by trying to solve old exam tasks from my professor, since he often uses his tasks from his earlier exams. I need help with this task:
Let $M_0=\{1,2\}$ and $$M_{n+1}=\left\{\frac{1}{2}(a+b):a,b\in M_n\right\}\cup\left\{\frac {1}{2}ab:a,b\in M_n\right\}, n\in\mathbb {N_0}$$ with $n\in\mathbb {N_0}$ (all natural numbers + 0).
a) Determine the from $n \in$ $\mathbb{N_0}$ independent upper and lower bounds for $M_n$.
My idea is to start the task by showing $M_1$ and $M_2$. By inserting the elements of $M_0$ and $M_1$ in $M_{n+1}$ we get for:
- $M_1=\{1,\frac{3}{2},2\}\cup\{\frac{1}{2},1,2\}$
- $M_2=\{\frac{3}{4},1,\frac{5}{4},\frac{3}{2},\frac{7}{4},2\}\cup\{\frac{1}{8},\frac{1}{4},\frac{3}{8},1,\frac{3}{2},\frac {9}{4},2\}$. I can't guarantee that I missed some elements, but nevertheless we can see that $x\leq 2$ and $x\geq 0$ $\forall x\in M_n$. So regardless which n we choose, the upper bounds are all $\geq 2$ and the lower bounds are $\leq 0$. But how do we show that more mathematical. These two examples wouldn't be sufficient to proof that.
b) Show with help of mathematical induction, that $1 \in M_n$ and $(\frac{1}{2})^n\in M_n$ $\forall n\in \mathbb{N_0}$.
$1 \in M_n$: P(0) is true, since $1\in M_0$. We proof P(n+1) under the term that P(n): $1 \in M_n$ is true. Thus we can choose a=b=1 and then $\frac{1}{2}\cdot(a+b)=1$. Since $1 \in M_{n+1}\implies 1\in M_n$
$(\frac{1}{2})^n\in M_n$: P(0) is true, $(\frac{1}{2})^0=1\in M_0$. We proof P(n+1) under the term that P(n): $(\frac{1}{2})^n \in M_n$ is true. Thus we can choose $a=b=(\frac{1}{2})^n$: $$\frac{1}{2}\cdot (ab)=\frac{1}{2}\cdot (\frac{1}{2})^{2n}=(\frac{1}{2})^{2n+1} \in M_{n+1}.$$ But how do we get $(\frac{1}{2})^{n+1}$ $\in M_{n+1}$.
c) Determine the supremum and infimum of $M=\bigcup_{n=0}^{\infty} M_n$. Are these maximum and minimum of $M$?
Well in a) we already discovered the lower ($\leq 0$) and upper bounds ($\geq 2$), which are independent from $n$. Thus these are the lower and upper bounds of $M$. So $2$ is our supremum and $0$ is our infimum. Since $2 \in M$, the supremum of $M$ is maximum of $M$. Since $0 \notin M$, the infimum $0$ is not a minimum of $M$. I know the solution is pretty obvious. But like in a), I want the proof to be more mathematical, since I don't belive that this kind of proof would give any points. Any advice?