floor of a Cauchy sequence is Cauchy?

Question

I am required to show that if $(a_n)$ is a Cauchy sequence then so is $(\operatorname{floor}(a_n))$. Any hints on how we may go about showing this please do not provide complete solutions.

Answer 1

Consider this particular example:

$$1+(-1)^n \frac1n$$

The sequence converges to $1$, hence it is Cauchy.

Try to examine the floor function.

Answer 2

$\lfloor a_n\rfloor - \lfloor a_m\rfloor \in \mathbb Z$, so for any $0 < \epsilon < 1$ the only way to have $|\lfloor a_n\rfloor - \lfloor a_m\rfloor| < \epsilon$ is to have $|\lfloor a_n\rfloor - \lfloor a_m\rfloor|=0$ which implies both $a_n$ and $a_m$ are between the same two integers. (i.e. there exists a $k \in \mathbb Z$ so that $k \le a_n < k+1$ and $k \le a_m < k+1$).

So for $\{\lfloor a_n\rfloor\}$ to be cauchy is for $\{a_n\}$ to "reach a point" where all the terms are between two integers. (To be more formal, there must be an $M \in \mathbb N$ and $k \in \mathbb Z$ where $n > M \implies k \le a_n < k+1$).

Must that be true for all Cauchy sequences? A cauchy sequence must converge to a value/limit but the individual terms can be on either side of the value/limit. If the limit is an integer you can have the individual terms "get close" it the integer but not always be on one side.

A simple example is $a_n=(-1)^n*\frac 1n$. Clearly Cauchy and $a_n \to 0$. But for even $n$, $a_n > 0$ and $\lfloor a_n \rfloor = 0$ but for odd $n$, $a_n < 0$ and $\lfloor a_n \rfloor = -1$.

So the statement is not true.

However if $a_n \to L$ and $L \not \in \mathbb Z$ then statement is true.

Which should be intuitively obvious. If $a_n$ converges to a point that is between two integers there "comes a point" are all the terms are between the two integers.

Formally:

If $L \not \in \mathbb Z$ then $\rfloor L \lfloor < L < \rfloor L \lfloor + 1$ and if $D = \min(L - \rfloor L \lfloor,\rfloor L \lfloor + 1 - L)$ then, as $\{a_n\}$ is cauchy, for any $\epsilon > 0$ we can find an $M$ where for any $n > M$ we have $|a_n - L | < {\min(D, \epsilon)} \le D$.

So if $a_n \ge L$ we have $\rfloor L \lfloor < L \le a_n$ and $a_n-L < \rfloor L \lfloor + 1 - L$ so $a_n < \rfloor L \lfloor + 1$.

And if $a_n < L$ then we have $L - a_n < L - \rfloor L \lfloor$ so $\rfloor L \lfloor< a_n <L < \rfloor L \lfloor + 1$.

Either way $\rfloor L \lfloor< a_n < \rfloor L \lfloor + 1$ so $\rfloor a_n \lfloor = \rfloor L \lfloor$ and $\rfloor a_n \lfloor\to \rfloor L \lfloor$