I am preparing for my exam by doing some exercises and need help with the following task:
Prove that every cauchy-sequence is convergent in $\mathbb{R^n}$ by using the nested intervals theorem.
Hint: Let {$\mathbf{x}_n$} be a cauchy-sequence in $\mathbb{R^n}$. This means there exist a monotonic increasing function {$R_i$} of real numbers with $$|\mathbf{x}_l-\mathbf{x}_m|<\frac{1}{2^{i+1}} \forall l,m\geq R_i.$$ Use the nested intervals theorem with the sets $$M_i=\{\mathbf{x}\in\mathbb{R^n}: |\mathbf{x}-\mathbf{x}_{R_i}|\leq \frac{1}{2^i}\}, i=1,2,3,…$$
Here is the nested intervals theorem:
Let $M_1,M_2,…$ be a sequence of bounded, closed, nonempty subsets of $\mathbb{R^n}$ with the following properties:
- diam($M_j$):=$\displaystyle\sup_{\mathbf{x,y}\in M_j}|\mathbf{x-y}|\to 0 $ for $j \to \infty$
- $M_1\supseteq M_2\supseteq…\supseteq M_n$
Then there exist exactly one point $\mathbf{x}$ with $\displaystyle\bigcap_{j=1}^{\infty}M_j$={$\mathbf{x}$}
Well to be honest, I am pretty overwhelmed with this task. Here are a few thoughts for this task:
Well I guess its pretty obvious, that our sets are nonempty subsets of $\mathbb{R^n}$. But how do we show that these sets are closed and bounded? Also we have to show that $M_1\supset M_2\supset…\supset M_n$ and that $\displaystyle\sup_{x,y\in M_j} |x-y| \to 0$ for $j \to \infty$. But I don't know how to do that. Furthermore I don't see how the nested intervalls theorem could help us. We get exactly one $\mathbf{x}$ such that $\displaystyle\bigcap_{j=1}^{\infty}M_j$={$\mathbf{x}$}. Is this the point, {$\mathbf{x}_n$} is converging to?
Also I am pretty unsure how to start. Maybe it would be helpful to write $|x_l-x_m|$ as $|x_l-x+x-x_m|\leq|x_l-x|+|x-x_m|$?
Well you see, I am pretty lost right now. Thats why I would be thankful for any advice.
Best Answer
First let's examine if the $M_i$ fulfill the conditions needed for the nested interval property:
The $M_i$ can be seen to be bounded by rewriting their defining condition: $|x-x_{R_i}| \le \frac{1}{2^i}$ means that $x \in \overline{B_{\frac{1}{2^i}}(x_{R_i})}$ - that is: $x$ lies in a ball centered around $x_{R_i}$ with radius $\frac{1}{2^i}$ - which is a bounded set. These balls are closed (as I indicated by writing $\overline{B_{\frac{1}{2^i}}(x_{R_i})}$ instead of $B_{\frac{1}{2^i}}(x_{R_i})$) because the boundary is still part of the $M_i$ (which can be seen by the $\le$ sign instead of a $<$ sign). If you want to make this argument rigourous you can consider a sequence in $M_i$ converging in $\mathbb R^n$ and then you show that the limit still lies in $M_i$.
Now the diameter of a ball like $\overline{B_{\frac{1}{2^i}}(x_{R_i})}$ is obviously $2 \cdot \frac{1}{2^i} = \frac{1}{2^{i-1}}$ which will converge to zero for $i \to \infty$.
Now we have to show that $M_i \supset M_{i+1}$ for any $i \in \mathbb N$. Let $y \in M_{i+1}$ be arbitrary. Then $|y-x_{R_{i+1}}| \le \frac{1}{2^{i+1}}$. We want to show that $|y-x_{R_i}| \le \frac{1}{2^i}$ because that would mean $y \in M_i$. For that we use the triangle inequality: $$|y-x_{R_i}| = |y- x_{R_{i+1}}+x_{R_{i+1}}-x_{R_i}| \le |y- x_{R_{i+1}}| + |x_{R_{i+1}}-x_{R_i}| \le \frac{1}{2^{i+1}}+\frac{1}{2^{i+1}} = \frac{1}{2^{i}}$$ where I used that $|x_{R_{i+1}}-x_{R_i}| \le \frac{1}{2^{i+1}}$ which comes from the hint stating $|x_m-x_l| \le \frac{1}{2^{i+1}}$ for all $m,l \ge R_i$. Setting $m = R_{i+1} \ge R_i$ and $l=R_i$ (which are allowed in this inequality) we obtain what I used.
Therefore the $M_i$ satisfy all conditions for the nested interval property.
Let $x= \bigcap_{i \in \mathbb N} M_i$ be the unique point in the intersection of the $M_i$. We claim (as you said) that $x_n \to x$ as $n \to \infty$. For given $\varepsilon >0$ we choose $R_i$ so big that $|x-x_{R_i}| < \varepsilon$ and $\frac{1}{2^{i+1}} < \varepsilon$. This is possible because $\text{diam}(M_i) \to 0$ for $i \to \infty$ so the mid-points $x_{R_i}$ of the balls will come arbitrarily close to $x$ and because $\frac{1}{2^{i+1}}$ converges to zero.
Choose $n \ge R_i$. Then $$|x_n-x| \le |x_n-x_{R_i}| + |x_{R_i}-x| \le \frac{1}{2^{i+1}} + \varepsilon \le \varepsilon + \varepsilon = 2 \varepsilon$$ Therefore $x_n$ converges to $x$ as $n \to \infty$.