Why the subset $\{1/n:n \in N^*\} \subseteq \mathbb{R}$ is discrete, but $\{0\} \cup \{1/n: n \in \mathbb{N}^*\}$ is not?
In "Hermite’s Constant and Lattice Algorithms" by Phong Q. Nguyen, a discrete subset of $\mathbb{R}^n$ is defined as follows.
A subset $D$ of $\mathbb{R}^n$ is discrete when it has no limit point. That is, for all $x \in D$, there exists $\rho > 0$ such that $B(x,p) \cap D = \{x\}$.
Here $B(x,r)$ is defined as the open ball of radius $r$ centered at $x$.
The paper then gives an example:
The set $\{1/n:n \in \mathbb{N}^*\} \subseteq \mathbb{R}$ satisfies the definition, but the set $\{0\} \cup \{1/n: n \in \mathbb{N}^*\}$ does not.
I fail to see why this is the case. Which balls can we choose in the set $\{1/n: n \in \mathbb{N}^*\}$, and why do these not work once we add the zero?
If you consider $D = \{\frac{1}{n}, n \in \mathbb{N}^* \}$. Let $x \in \{\frac{1}{n}, n \in \mathbb{N}^* \}$. For every point $\frac{1}{n}$ consider the ball $B(\frac{1}{n},\frac{1}{n} - \frac{1}{n+1})$. Then, $B(\frac{1}{n},\frac{1}{n} - \frac{1}{n+1}) \cap D = \{ \frac{1}{n} \}$. That corresponds to the definition. (Well, actually, I didn't prove that the intersection between this ball and this point is actually $\frac{1}{n}$. If you want to do so, you can notice that the sequence $\{ \frac{1}{n}\}_{n \in \mathbb{N}^*}$ is increasing. Therefore, you just have to prove that the two adjacent points are far enough)
However, for $\{\frac{1}{n}, n \in \mathbb{N}^* \} \cup \{ 0 \}$, consider the point $x = 0$. Does there exist any $\rho$ such that $B(0,\rho) = \{ 0 \}$ ? Let's assume, by absurd, that such a $\rho$ exists. Consider the number $n = \lceil \rho \rceil +1$ . Then, $n > \rho$. Then, $\frac{1}{n} < \rho$. Then, $\frac{1}{n} \in B(0,\rho)$. Then, you have an absurdity.