If $\{x_n\}$ is a sequence in $\mathbb{N}$ and $x_n \rightarrow x$, prove there exists $N$ such that $x_n = x$ for $n \geq N$

Question

Since $x_n \rightarrow x$, we know that for all $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $n \geq N$, $|x_n - x| < \epsilon$.

We want to show that for some $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $n \geq N$, $x_n = x$, but that would imply that $x_n - x = 0$, and $0 \not\in \mathbb{N}$, so I don't know what $\epsilon > 0$ to choose to make $|x_n - x| = 0$, since for all $\epsilon > 0$, we have $|x_n - x| < \epsilon$, which means that it can never happen that $|x_n - x| = 0$ since $\epsilon > 0$.

Answer 1

Choose $\epsilon = \dfrac{1}{3} \to \exists N_0 \in \mathbb{N}$ such that if $m, n > N_0 \to |x_n-x_m| < \dfrac{1}{3} $ since its a Cauchy sequence. But $x_n, x_m \in \mathbb{N} \to |x_m-x_n| \in \mathbb{N}\cup \{0\} \to |x_m-x_n| = 0 \to x_m = x_n$. This shows the sequence is a constant after $N_0$, and we can call this constant value $y$. It is easy to show that $y$ is the limit of the sequence $\{x_n\}$. And since $x$ is the limit of $\{x_n\}$, we have $y = x$. Thus $x_n = x, \forall n \geq N_0$.

Answer 2

The fact that $0 \not\in \mathbb{N}$ (if you use that convention) is not relevant.

What you need to do is choose an $\varepsilon > 0$ such that $|x_n - x| < \varepsilon$ implies that $x_n = x$. In general you can't do this, but in this situation you can. As a hint for which $\varepsilon$ to choose, what is the smallest $|x_n - x|$ can be without being zero? What would happen if you chose $\varepsilon$ to be smaller than this value?