Prove this lemma
Lemma: The closure of $\mathbb{N}$ is $\mathbb N$. The closure of $\mathbb Z$ is $\mathbb Z$. The closure of $\mathbb Q$ is $\mathbb R$, and the closure of $\mathbb R$ is $\mathbb R$. The closure of the empty set is the empty set.
My Attempt
1) The closure of $\mathbb N$ is $\mathbb N$: Let $x\in \bar{N}$ such that $\epsilon >0$, then there exists $n\in N$ such that $|n-x| \leq \epsilon$. Let $\epsilon = 1/4$. Then $|n-x|<1/4$ <=> $x-1/4 \leq n \leq x+1/4$.
Not sure where to go from here
Best Answer
Easier method:
If $\mathbb N$ has no limit points than $\mathbb N$ is by definition vacuously closed and by definition $\overline {\mathbb N} = \mathbb N \cup \text{limit Points}(\mathbb N) = \mathbb N$.
Let $x$ by a real limit point of $\mathbb N$, if any exist. There exists an integer $n$ such that $n \le x < n+1$. If $n = x$ then there is no other integer in the interval $(n-1, n + 1)$ other than $n = x$ so $x$ is not a limit point. So $x$, if it exists, is not an integer and $n < x < n+1$. Let $\epsilon = \min(x -n, n+1 - x)$. Then $(x - \epsilon, x + \epsilon) \subseteq (n, n+1)$ and this interval contains no natural numbers. So $x$ is NOT a limit point of $\mathbb N$.
So $\mathbb N$ has no limit points. So $\mathbb N$ is closed.
Same argument of $\mathbb Z$.
To prove $\overline {\mathbb Q} = \mathbb R$. Let $x \in \mathbb R$. Let $\{q_n\}$ be a cauchy sequency of rational numbers converging to $x$. For every $\epsilon$ there exists and $q_i$ s.t. $d(x,q_i) < \epsilon$ so $x$ is a limit point of $\mathbb Q$.
To prove $\overline {\mathbb R} = \mathbb R$ is... obvious. Let $x \in \mathbb R$, $\epsilon > 0$ and $y = x - \epsilon/2$, then $d(x,y) = \epsilon/2 < \epsilon$. So $x$ is a limit point.
Closure of the empty set is the empty set.... well, as the empty set has no elements, no elements exists in any neighborhood of any point so no point is a limit point of the empty set. The empty set has not limit points.