Why is $\mathbb N$ a closed set? I know that we can prove that $\mathbb N$ is a closed set through the use of the compliment of $\mathbb N$. Since $\mathbb R$\ $\mathbb N$ is open, $\mathbb N$ must be closed.
However, the question arises: If that is so, then $\mathbb N$ must contain its limit points. Then what are the limit points of $\mathbb N$ ? I believe they are every element of $\mathbb N$: 1,2,3,4,5,6,... are all limit points of $\mathbb N$. But remember the definition of a limit point: x is a limit point of $A$ if $\forall \epsilon \gt 0$, $V_{\epsilon}(x) \cap A =$ points that are different from $x$. 
Now take $\epsilon = 0.5$ Then we see that $V_{\epsilon =0,5}(1) \cap \mathbb N = {1}$. Hence, EVERY points in $\mathbb N$ is an isolated point, and there is no limit point: A contradiction.
So, please point out what is wrong with my thoughts ? I thank you very much for your help. I am just an introductory real analysis learner, so please help me. I thank you.

Note that $\mathbb{N}$ has no limit points: if $n \in \mathbb{N}$ were to be a limit point, then any neighborhood of $n$ should contain some point of $\mathbb{N}$ other than $n$. However, as $\{n\}$ is open in $\mathbb{N}$ (with the induced topology as a subset of $\mathbb{R}$), $\{n\}$ is a neighborhood of $n$ containing no other point than $n$.
So, $\mathbb{N}$ contains all its limit points: there just are none of those.
