Not a homework question, came across this exercise in Churchill's complex analysis. I haven't done any analysis before so I'm not sure how to answer it.
We have a closed interval $$a_0 \leq b_0$$ which we subdivide into $a_1 \leq b_1$, which can be either the left hand or right hand half of $a_0 \leq b_0$. This is then subdivided into $a_2 \leq b_2$, which again can be either the left or right hand half of $a_1 \leq b_1$. This is continued ad infinitum.
Prove that there is a point $x_0$ common to all intervals.
How do I do this? I've got the inequalities $$a_0 \leq a_n \leq a_{n+1} \leq b_0 $$ and $$a_0 \leq b_{n+1} \leq b_n \leq b_0 $$ so $a_n$ and $b_n$ are bounded with
$$\lim_{n\rightarrow \infty} a_n \leq b_0 $$ and $$\lim_{n\rightarrow \infty} b_n \geq a_0 $$ but I'm not sure how to show that the limits of $a_n$ and $b_n$ exist and that they both equal the same number. Any help?
Best Answer
Completeness of $\Bbb R$ ensures that every bounded monotone sequence converges, so $\lim\limits_{n\to\infty}a_n$ and $\lim\limits_{n\to\infty}b_n$ certainly exist; call them $a$ and $b$, respectively. To show that $a=b$, note that by construction $$b_{n+1}-a_{n+1}=\frac12(b_n-a_n)$$ for each $n$, so $\lim\limits_{n\to\infty}(b_n-a_n)=0$. And $a_n\le a\le b\le b_n$ for each $n$, so ... ?