Prove the set is not open. Prove that the set $\mathbb{R}-\{1/n|n \in\mathbb{N}\}$ is not open.
OK, so I am having a little trouble. I know that the definition of open set is : iff every point of $A$ is an interior point of $A$. The definition of a closed set: iff its complement, $A^c$ is open.
I have made a number line to list the numbers in  the set to find the interior points. I came up with,
$(-\infty, 1)\cup(1,1/2)\cup(1/2, 1/3)\cup(1/3,1/4).........$
From my thoughts, the set $A=Int(A)$. Am I on the right track?
Further, I am not sure how this set is not open (I guess meaning that it's closed), if all the points of $A$ is in the interior of $A$. 
Pretty sure I am missing something, so any guidance would be appreciated.
 A: HINT: Show that $\{\frac1n\mid n\in\Bbb N\}$ is not closed. Recall that a set $A$ is closed if and only if whenever $x_n\in A$ is a convergent sequence, then $\lim x_n\in A$ as well.
A: Note that:
$\mathbb{R}-\{1/n|n \in\mathbb{N}\} \neq (-\infty, 1)\cup(1,1/2)\cup(1/2, 1/3)\cup(1/3,1/4).........$
$\mathbb{R}-\{1/n|n \in\mathbb{N}\}=(-\infty,0]\cup(1,\infty)\cup(1/2,1),(1/3,1/2),(1/4,1/3),....$
A: First of all, not every set that is not open is closed. In fact, this set is an excellent example.
Now, you are told to prove that it's not open. You gave a good definition of "open" in $\mathbb{R}$ - a set can be called open iff all its points are interior points (that is, points that every open interval around them is in the set).
Your set has a special point in it. Can you locate it? Can you prove it's not an interior point?
A: In this case the problem comes easy. For example $0\in A=\mathbb{R}-\{\frac{1}{n}:n\in\mathbb{N}\}$ and for any $\delta>0$ exist $n$ such that $\frac{1}{n}<\delta$ so it's imposible to find a ball ($B_{\delta}(0)$) around $0$, contained completly in $A$. Therefore we prove the negation of the definition of an open set.
