Floor function of a non-integer Cauchy sequence also a Cauchy sequence? Let $a_n$ be a Cauchy sequence such that $a_n$ converges to a non-integer value.
Then if we have a sequence $b_n$ defined as $b_n \leq a_n < b_n+1$, will $b_n$ be a Cauchy sequence?
Intuitively, I think $b_n$ is a Cauchy sequence, since $a_n$ converges to non-integer, sequence like $\frac{(-1)^n}{n}$ can't be a counterexample.
Then, there should be a large $N$ and some integer M so that for all $n\geq N$, $M \leq a_n <M+1$.
However, I am having a hard time showing that $b_n$ is a Cauchy sequence.
I tried to use the fact that $a_n$ is a Cauchy sequence that converges to a non-integer value and the definition of $b_n$.
However, I am having a hard time formally proving it.  
Could you point out anything wrong with my intuition and if my intuition is correct, please give some guidelines of how to formally prove it.
Thanks!
 A: If $\lim_{n\to\infty}a_n=l\in\mathbb R\setminus\mathbb Z$, let $\varepsilon>0$ such that is smaller than the distance from $l$ to the closest integer. If $n\gg1$, $\lvert a_n-l\rvert<\varepsilon$ and therefore $a_n\notin\mathbb Z$. Therefore, for such $n$ we have $\lfloor a_n\rfloor=\lfloor l\rfloor$. So, $\bigl(\lfloor a_n\rfloor\bigr)_{n\in\mathbb N}$ converges to $\lfloor l\rfloor$.
A: Sinc $a_n\to c$ where $c$ is not an integer.  Let $k$ be the integer so that $k < c < k+1$.  Let $\epsilon = \min(c - k, (k+1)-c) > 0$.  Then there an $N$ where $n> N$ implies $|c-a_n|<\epsilon$.
And that implies $k < c-\epsilon < a_n < c+\epsilon < k+1$.  So $b_n= \lfloor a_n \rfloor = k$ and $b_n \to k$.
And this is an "eventually constant" sequence and almost trivially Cauchy.  For $n > N$ then $n,m > N$ will mean $b_n = b_m = k$ and $|b_n - b_m|=0 < \epsilon$ for any $\epsilon > 0$.
But this is only true if $a_n\to c$ where $c$ is not an integer.  If $c=k$ is an integer integer you can always have a cauchy sequence where the $a_n$ flit about above and below the integer $k$ and $b_n= \lfloor a_n \rfloor$ bounces between $k-1$ and $k$.
