# Prove or disprove: If $a_n$ is Cauchy, is $\lfloor a_n \rfloor$ Cauchy?

So all the posts explaining this were above my understanding or had conflicting answers, namely these two answers to the same posts where:

(1) implies if $$a_n$$ converges to $$l$$, then $$\lfloor a_n \rfloor$$ converges to $$\lfloor l \rfloor$$ (link to said answer)

(2) implies if $$a_n$$ converges to $$l$$ where (CASE 1) if $$l\in\mathbb{Q}$$, then $$\lfloor a_n \rfloor$$ converges to $$\lfloor l \rfloor$$ but (CASE 2) if $$l\in\mathbb{Z}$$, then the sequence can "flit" above and below $$l$$ and following $$\lfloor l \rfloor$$ it would flip between $$l$$ and $$l-1$$ and if therefore nonconvergent and not Cauchy. (or at least that seems to be the implication for me) (link to said answer)

Now I'm still quite new to real analysis but (2) does make sense to why it would be the case. If I could get some sort of confirmation to be sure I'm not being led astray that would be great. And if there are any functions that fit this, I would love to see them.

• Which posts seem conflicting to you? Also, the second post makes no reference to the rational numbers ($\mathbb{Q}$). Do you mean the real numbers ($\mathbb{R}$) (i.e., the rational and ir-rational numbers). Both answers state that different things happen when $l$ is exactly an integer and when it isn't. This should be fairly easy to see if you observe that $a_n = 1-1/n$ approaches $1$ but has a floor of $\lfloor a_n \rfloor = 0$.
– Jam
Commented Oct 9, 2022 at 19:29

It is not even true that a continuous function maps Cauchy sequences to Cauchy sequences, and $$x\mapsto\lfloor x\rfloor$$ is not even continuous!
Let $$a_n:=1+\frac{(-1)^n}{n}$$ for $$n\in\Bbb N$$. $$a_n\to1$$ is clear, but: $$\lfloor a_n\rfloor=\begin{cases}0&n\text{ is odd}\\1&n\text{ is even}\end{cases}$$So $$\lfloor a_n\rfloor$$ is not Cauchy.
Note that $$\Bbb Z\subseteq\Bbb Q$$ so your second paragraph doesn't really make sense. Did you mean: If $$\ell\in\Bbb R\setminus\Bbb Q$$, then $$\lfloor a_n\rfloor\to\ell$$? So long as $$\ell$$ is not an integer, then $$\lfloor a_n\rfloor\to\ell$$ is guaranteed.
• Thank you for the example! And yeah I should've written $l\in\mathbb{Q}-\mathbb{Z}$ or something along those lines indeed. Commented Oct 9, 2022 at 19:34