Infinitely many $x_0$s in a sequence, same thing as $\exists N\in \mathbb{N}$ s.t $\forall n \geq N, x_n=x_0$?

Question

Let $\{x_n\}_{n=0}^{\infty}$ be a sequence in $\mathbb{R}$. Is the following implication correct? If so, why?

$(x_n)$ contains infinitely many $x_0$s $\iff \exists N\in \mathbb{N}$ s.t $\forall n \geq N, x_n=x_0$

What about a sequence of the form $(x_n)=\{x_1,x_0,x_3,x_4,x_0,x_6,x_7,x_8,x_0,x_{10},x_{11},x_{12},x_{13},x_0,...\}$. Here, we would have infinitely many $x_0$s but the number of terms between them increases like the natural numbers without bound (at first it's 1, then 2, then 3, then 4, and so on). So, clearly, no such $N$ can exist.

If we reorder the sequence, so that all the non-$x_0$ terms appear to the left, we're technically changing it, right? So that can't be the answer.

Thanks!

Answer 1

It's not true, and your logic works but you don't even need the distance between the $x_0$'s to increase without bound; you can take $x_n=(-1)^n$, then there are infinitely many terms equal to $1$ but there is no $N$ such that $x_n=1$ for $n\ge N$ because $x_n=-1$ for any odd $n$.