Why is a sequence $(x_1, x_2, x_3,...)$ eventually periodic if the set $\{x_1, x_2, x_3,...\}$ is finite?

Question

"A sequence $(x_1, x_2, x_3...)$ is eventually periodic if the set ${x_1, x_2, x_3,...}$ is finite. (Think about it.)" (Vivaldi)

First of all I am not sure whether this is an if and only if statement as it is written quite colloquially, so if anyone knows that that would be really helpful. I am inclined to say it is only one way since the decimal expansion of any rational is eventually periodic yet the set of each of its terms is infinite.

Second of all I am looking for intuitive reasoning before I attempt a mathematical proof. My thoughts are that if ${x_1, x_2, x_3,...}$ is infinite then we can always think of the next number possibly being outwith the period?

Thanks for any input

Answer 1

Ok so the first part is simple, if the function is peridodic then the period is finite, what is more: the number of elements of the sequence is at most the length of the period. So then the set of elements of the sequence is fiite.

As for the second part: I don't think it is true. Take any irrational number as an example. The $x$'s are all digits from 0 to 9 and they are not periodic.

Answer 2

Let $$x_k=\begin{cases}1&\text{if }\sqrt k\in\mathbb N\\0&\text{otherwise.}\end{cases}$$ Then $\{x_1,x_2,\ldots\}=\{0,1\}$ is finite, but the sequence is not eventually periodic.

Answer 3

0,1,0,0,1,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,...