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.

  • $\begingroup$ Nothings wrong with you intuition. But hone your intuition. $a_n \to c$ means that "eventually" the $a_n$s gets "very close" to $c$. And as $c$ is between two integers so the $a_n$s getting "very close" will mean being within these two integers as well. And that would mean the floor values will "eventually" all be constant; the lower of the integers. See my answer for a technical formal write-up. $\endgroup$ – fleablood Oct 6 '19 at 17:00
  • $\begingroup$ Oh... $b_n = [c]$ for all $n > $ some eventual value. So That means for all $n,m > $ the eventual value $|b_n -b_m| = 0$ and that is less than any positive $\epsilon$ so this is Cauchy. We can say that $b_n$ is constant almost everywhere or constant for all but a finite number of values or that $b_n$ is "eventually constant". Those sequences where $b_n =k$ for all $n > $ some $N$ are very easy to show to be Cauchy. $\endgroup$ – fleablood Oct 6 '19 at 17:08

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 $\lfloor l\rfloor$.


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 $k$.


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