About the distance between two distinct points in a sequence in a non-sequentially compact space

Question

I was looking at a proof about compactness and there is a step I cannot understand. Since the full proof is quite long and I have trouble understanding only a specific point, I have summarized the situation. Here $\mathcal B_\varepsilon (x)$ represent the open disk of radius $\varepsilon$ and center $x$, namely, the set of points that are at distance from $x$ less than $\varepsilon$.

Let $(X, d)$ be a non-sequentially compact metric space and let $\ell \in \mathbb R ^+$ be a fixed quantity. Let $x_1 \in X$ be a generic point, then consider the set $V_1 = \mathcal B_\ell (x_1)$. Then let $x_2 \notin V_1$ and construct the set $V_2 = V_1 \cup \mathcal B_\ell(x_2)$. Proceed recursively this fashion: $$x_n \notin V_{n-1}, \quad V_n = V_{n-1} \cup \mathcal B_\ell (x_n) $$ and construct the sequence of open sets $\{V_n \}_{n \in \mathbb N}$. This also induced a sequence of points: $\{ x_n \}_{n \in \mathbb N}$.

My textbook says that: $$\forall n \neq m \in \mathbb N, \quad d(x_n, x_m) \geq \ell ,$$ but I fail to understand why this is the case. I manage to prove that this is the case for adjacent terms in the sequence, namely that $$ \forall n \in \mathbb N, \quad d(x_n, x_{n+1}) \geq \ell $$ but how does one further generalize the result for all $ n \neq m $?

Answer 1

Assume $n>m$. Then, we have : $$\mathcal B_l(x_m) \subset V_m \subset \ldots \subset V_{n-1}$$

Since $x_n \notin V_{n-1}$, we have $x_n \notin \mathcal B_l(x_m)$ and $d(x_n,x_m)\geq l$.

Answer 2

Notice that $V_{n-1}$ contains the ball of radius $\ell$ around each point in the set $\{x_1,\ldots,x_{n-1}\}$. At the same time, $x_n\not\in V_{n-1}$. You are not simply choosing $x_n$ to not be in $\mathcal{B}_\ell (x_{n-1})$, you are choosing it not to be in the union $\bigcup\limits_{i=1}^{n-1}\mathcal{B}_\ell(x_{i})$.