Suppose $G,R\subset\mathbb{R}$ are closed sets and $[c_0, d_0]\subset F\cup G$. Show that there exists $l$ such that $\lim c_n=\lim b_n=l$.

Question

Suppose $G,R\subset\mathbb{R}$ are closed sets and $[c_0, d_0]\subset F\cup G$.

If $\frac{c_0+d_0}{2}\in F$, then define $c_1=\frac{c_0+d_0}{2},d_0=d_1$; otherwise define $d_1=\frac{c_0+d_0}{2}, c_0=c_1.$

Show that there exists $b\in[c_0,d_0]$ such that $c_n, d_n\to b$ as $n\to\infty$.

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I could recursively show that $\{c_n\}\subset F$ and $\{d_n\}\subset G$ and perhaps use Bolzano-Weierstrass. For the rest, I'm really stuck here. How do I even begin to show that their limits are equal for some $b$? What does $\{c_n\}\subset F$ and $\{d_n\}\subset G$ give me?

Answer 1

Hint: You have $d_1-c_1=\frac{d_0-c_0}{2}$, and $c_1\geq c_0$, $d_1\leq d_0$. You can show by induction that $d_n-c_n=\frac{d_0-c_0}{2^n}$, and $c_{n+1}\geq c_n$, $d_{n+1}\leq d_n$.