Nested sequence of half open intervals with non-empty intersection Consider the sets $B_{2^n}^k = [\frac{k}{2^n},\frac{k+1}{2^n})$ with $n \in \mathbb{N}$ and $k \in \mathbb{Z}$.
Now we pick a sequence $(k_n)_{n \in \mathbb{N}}$ such that we get a nested sequence $B_{2^1}^{k_1} \supset B_{2^2}^{k_2} \supset \dots$. 
I suspect that the intersection $\underset{{n \in \mathbb{N}}}{\cap}B_{2^n}^{k_n} = \{x \}$, i.e. is a singleton. 
As this sequence resembles the binary representation of a real number. For your intuition of this sequence: at each step we half the needle and pick either the right or the left part.
Problem is that I don't know how to prove it, as we cannot use the nested interval theorem (which only works for nested compact sets).
 A: The intersection doesn't need to be a singleton. for example, taking $k_n=-1$ for all $n\in\mathbb N$, we see that:
$$B_{2^n}^{k_n} = \left[\frac{k_n}{2^n}, \frac{k_n+1}{2^n}\right) = \left[-\frac{1}{2^n}, 0\right)\supset\left[-\frac{1}{2^{n+1}}, 0\right) = \left[\frac{k_n}{2^n}, \frac{k_n+1}{2^n}\right)=B_{2^{n+1}}^{k_{n+1}}$$
so your condition is met, however the intersection 
$$\bigcap_{n=1}^\infty B_{2^n}^{k_n} = \bigcap_{n=1}^\infty \left[-\frac{1}{2^n}, 0\right)=\emptyset$$
is empty.

What you can prove is that the intersection contains at most one element. The idea of the proof is that if it contains two distinct elements, you can reach a contradiction since these two elements $a,b$ must be elements of an interval with an arbitrarily small length, e.g. with a length of $\frac{a+b}{2}$, which is impossible.
A: This is not necessarily true, by the same argument as usual counterexamples to the nested interval theorem without closedness:
If you take $k := 2^n-1$, you get an empty intersection (intervals approach 1 from below, but it is not contained), corresponding to the binary representation $0.111... = 1$.
