Show that the set $\Bbb N$ of natural numbers is closed Let $(\mathbb{R},d)$ be the metric space where 
$$d(x,y)=|x−y| \; \forall x, y \in \mathbb{R}$$
Show that the set $\mathbb{N}$ of natural numbers is closed.
I tried the complement method but i don't know what to do after i let an element belong to the complement? 
 A: Here are a few proofs:

To prove that it's closed, we will prove that it's complement is open. Consider the sets $S_i=B(i+1/2,1/2)$, where $B(x,r)$ is the open ball of radius $r$ centered at $x$. Each $S_i$ is open. $$(-\infty,0)\cup\left(\cup_{i\in\mathbb{N}} S_i\right)=\mathbb{R}-\mathbb{N}$$ so $\mathbb{R}-\mathbb{N}$ is the countable union of open sets and so it's open. Therefore complement of $\mathbb{R}-\mathbb{N}$, which is $\mathbb{N}$, is closed.

To prove that it's closed we will prove that it's complement is open. Let $x\in\mathbb{R}-\mathbb{N}$. We wish to show that there is an open ball centered at $x$ that contains no integers. Let $[x]$ denote the distance from $x$ to the nearest integer. This is either the fractional part of $x$ or one minus the fractional part of $x$, whichever is smaller. Then the ball $B(x,[x]/2)$ doesn't contain any integers. Thus $\mathbb{R}-\mathbb{N}$ is open and so $\mathbb{N}$ is closed.

To prove it's closed, we'll prove that every convergent sequence of elements in $\mathbb{N}$ had a limit in $\mathbb{N}$. Let $x_i\in\mathbb{N}$. For every $\epsilon>$ there exists an $N$ such that for $n>N$, $|a_n-L|<\epsilon$. Let $\epsilon=1/3$. Then whatever the limit is, there is only one natural number within a distance of $1/3$ of it. Call that integer $k$. Therefore there exists an $N$ such that for all $n>N,a_n=k$ and so the sequence converges to $k$.
This last proof exploits the fact that $\mathbb{N}$ is discrete. Any discrete set is closed by this argument.
A: The "complement method" will work: we need to show that $\mathbb{R} \setminus \mathbb{N}$ is open. ($\setminus$ is the set-difference symbol.)
Let $x \in \mathbb{R} \setminus \mathbb{N}$. We need to find an open interval - that is, an open ball - around $x$, which doesn't contain a natural.
Firstly, if $x < 0$ then I hope you can find an interval containing $x$ and no natural number yourself. So we only need to show it for positive $x$; say $x \in (n, n+1)$.
Can you give an interval containing the point $\frac{1}{2}$? How about the point $\frac{7}{8}$? How about $\frac{1}{1000}$? Does your method generalise to $x \in (0,1)$?
Now that you can find the answer for $x \in (0,1)$, add $n$ to it to obtain the answer for $x\in (n,n+1)$.
