Question about a proof of Cauchy $\implies $ Convergent in $\mathbb R^k$

Question

Suppose $\{x_n\}$ is bounded. Then the range of $\{x_n\}$ is in some $k$-cell $I$ which is compact. Thus $\{x_n\}$ converges in $I$.

We are done if we can prove $\{x_n\}$ is bounded. Let $E_N$ be the range of the subsequence $x_N, x_{N +1}, x_{N + 2}, \ldots.$ Then $\displaystyle{\lim_{N \to \infty} \operatorname{diam} E_N = 0}$. $\color{red}{\text{Thus there's $N \in \mathbb N$ with $d(\operatorname{diam}E_N, 0) < 1$}},$ $ \color{blue}{\text{so the distance between any two points of $E_N$ is less than $1$}}$, so $E_N$ is bounded. Because $\{x_1, x_2, \ldots, x_{N - 1}\}$ is finite, we can bound it by the maximum distance between two of its points. Note, $\{x_1, x_2, \ldots, x_{N - 1}\} \cup E_N$ which is the range of $\{x_n\}$ is a union of two bounded sets and so $\{x_n\}$ is bounded.

My question is about the parts in $\color{red}{\text{red}}$ and $\color{blue}{\text{blue}}$ of the quote above. If $\displaystyle{\lim_{N \to \infty} \operatorname{diam} E_N = 0}$, then for any $\epsilon > 0$, there's some $K \in \mathbb N$ s.t. $N \ge K \implies d(\operatorname{diam} E_N, 0) < \epsilon.$ Since $d(\operatorname{diam} E_N, 0) < \epsilon$ is true for any nonnegative $\epsilon$, in particular we have $d(\operatorname{diam} E_N, 0) < 1 \iff |\operatorname{diam}E_N| < 1$ which means the distance between any two points of $E_N$ is less than $1$. But this is true only for $N \ge K.$ That means all of $E_N, E_{N+1}, E_{N+2},\ldots, E_{K -1}$ have a diameter greater or equal to $1$. Shouldn't we have to have $|\operatorname{diam}E_N| < 1$ true for all $N$ for the $\color{blue}{\text{conclusion in blue}}$ above to hold? Thanks.

Answer 1

In the sentence you've highlighted red and blue, the proof is choosing a particular value of $N$. Before the red part, $N$ could be any natural number, and by the time we reach the blue part, $N$ is to mean a certain number where we do know that $|\mathrm{diam}\,E_N| < 1$.

You are right that the proof would be clearer if it didn't "reuse" $N$ in this way, instead saying

Thus there's $K \in \mathbb{N}$ with $d(\mathrm{diam}\, E_K,0) < 1$, so the distance between any two points of $E_K$ is less than $1$, so $E_K$ is bounded.

and so on, using $K$ instead of $N$ for the remainder of the text.