Proving for any $\{x_{m_k} \}\in I_n$ that:$$\lim_{n \to \infty}\{x_{m_k} \}=c$$
I have been trying to solve this problem, but i dont know how to write it properly, so i need your help whit writing and adding steps or theorems.
We have an intervalue that is constantly being divided $n$ times, $I_n$, such that
$$|I_n|=\dfrac{b-a}{2^n}$$
So, $$|I_n|=\dfrac{b-a}{2^n} \to 0$$
And: $$I_n \subset ...\subset I_2 \subset I_1 \subset I_0$$
There exist:
$$\{x_{m_0} \}\in I_0$$ $$\{x_{m_1} \}\in I_1$$ $$\{x_{m_2} \}\in I_2$$ $$...$$ $$\{x_{m_n} \}\in I_n$$
And also exist:
$$c=\bigcap_{i=1}^n \subset I_n$$
Such that:
$$|\{x_{m_k} \}-c|\le |I_n|=\dfrac{b-a}{2^n} \to 0$$
So: $$\lim_{n \to \infty}\{x_{m_k} \}=c$$
Since $\frac{b-a}{2^n}\rightarrow 0$, the sequence is Cauchy: given $\epsilon>0$, choose $n_0$ such that $\frac{b-a}{2^n} <\epsilon$. Then if $n,m\le n_0$, we have that $x_n,x_m \in I_{n_0}$, so $|x_n-x_m|\le \frac{b-a}{2^n} <\epsilon$.
Then $x=\lim_{n\to \infty}x_n$ exists and it remains to be shown, that $x\in \bigcap_{n=1}^{\infty} I_n$. This means showing that $x\in I_n$ for all $n\in \mathbb{N}$. So let $n_0$ be arbitrary but fixed. Assume WLOG $I_{n_0}$ is closed (if not, do the following steps with $I_{n_0+1}$). We can then write $I_{n_0}=[a ,b]$ and by assumption, we have $x_n \in I_n \subset I_{n_0}$ for all $n\ge n_0$. It follows $a\le x_n \le b$ for all $n\ge n_0$, so we also have $$a\le \lim_{n\to \infty} x_n=x\le b.$$
This means exactly that $x\in [a,b]=I_{n_0}$.
Edit: We actually need the assumption that the intervals are closed. The conclusion is of course false for $I_n=(0,2^{-n}),x_n=2^{-n+1}$, since then $\lim_{n\to \infty} x_n=0.$ In this case we even have $\bigcap_{n=1}^{\infty} I_n = \emptyset$.