Showing a closure of a set is equal to the set itself, metric spaces

Question

Q:Consider the metric space $(X = [0, ∞), d)$ where d is the metric defined by: $$d(x,y) = |x^2-y^2|$$ Let $A = \mathbb{N} ∪ {0}$. Show that $\bar{A} = A$. Justify your answer.

So my thought process was that if we use the definition of closure is by the set defined by $x\in X$ in $\bar{A}$ iff given any $\epsilon > 0, B_\epsilon(x) \cap A \neq \emptyset$

And the fact that the closure of A, $\bar{A} = A ∪ \{\text{limit points of A}\}$, so just showing that all limit points are inside of A is sufficient I think.

And the way I started was using the definition and then let $\sqrt \epsilon$ be given then defn of limit point $$B_\epsilon^d(x) \backslash \{x\} \cap A \neq \emptyset$$ Specifically, $d(0,\sqrt \epsilon) = \epsilon \backslash \{0\}$

Which with the intersection of $\mathbb{N}$ U $\{0\} \neq \emptyset$.

I don't think it's correct but that's my attempt, if someone could help guide / give the asnwer it would be really helpful thank you!

Answer 1

The closure of any set contains that set. So, in particular, $\overline{\{0\}\cup\Bbb N}\supset\{0\}\cup\Bbb N$. In order to prove that we actually have the equality, I shall take $x\in[0,\infty)\setminus(\{0\}\cup\Bbb N)$ and then prove that $x\notin\overline{\{0\}\cup\Bbb N}$. There is some $n\in\{0\}\cup\Bbb N$ such that $n<x<n+1$. Let$$\varepsilon=\min\{x-n,n+1-x\};$$that is, $\varepsilon$ is the (usual) distance from $x$ to the closest element of $\{0\}\cup\Bbb N$. Then, for each $m\in\{0\}\cup\Bbb N$,\begin{align}d(x,m)&=|x^2-m^2|\\&=|x-m|(x+m)\\&\geqslant|x-m|\\&\geqslant\varepsilon,\end{align}and therefore $x\notin\overline{\{0\}\cup\Bbb N}$.