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?