The limit of the interval endpoints depending on $n$ For $n \in \mathbb{N_0}$ consider the sequence of intervals of the the following from: 
\begin{align}
A_k &:= [ k, k + 1 ), \quad k \in \mathbb{N_0} \\
\frac{A_k}{2} &:= \left[ \frac{k}{2}, \frac{k + 1}{2} \right), \quad k \in \mathbb{N_0} \\
. \\
. \\
. \\
\frac{A_k}{2^n} &:= \left[ \frac{k}{2^n}, \frac{k + 1}{2^n} \right), \quad k \in \mathbb{N_0}
\end{align}
Clearly, for every $n \in \mathbb{N_0}$, the intervals $\frac{A_k}{2^n}$, $k \in \mathbb{N_0}$ are pairwise disjoint and their union covers the the interval $[ 0, \infty )$. Fix some $t \in [ 0, \infty )$. For every $n \in \mathbb{N_0}$ there exists a unique $k_n \in \mathbb{N_0}$ such that $t \in \frac{ A_{k_n} }{ 2^n }$, i.e.
$$
\frac{k_n}{2^n} \leq t < \frac{k_n + 1}{2^n}.
$$
I would like to show that, for example, $\frac{k_n + 1}{2^n} \downarrow t$ as $n \rightarrow \infty$. This seems intuitively clear, since with growing $n$ the endpoints of the interval only get closer. But how can one show this in a rigorous way? It seems that showing that $\frac{ k_n + 1 }{ 2^n }$ is a decreasing sequnce with its infimum being equal to $t$ should suffice.

For the decreasing part, it's enough to show that $\frac{k_n+1}{2^n}$ is also the endpoint of one of the intervals for any $m> n$. Because then it's obvious that $\frac{k_m+1}{2^m}$ is either that or something smaller. And in fact, if $m>n$, you have:
$$\frac{k_n+1}{2^n}=\frac{2^{m-n}(k_n+1)}{2^m}=\frac{\left(2^{m-n}k_n+2^{m-n}-1\right)+1}{2^m}=\frac{k'+1}{2^m}$$
where $k'=2^{m-n}k_n+2^{m-n}-1$.
The convergence part follows from the fact that the length of the intervals goes to $0$ and $\frac{k_n+1}{2^n}-t$ is always smaller than that length.
Hope this helps.
