TITLE: Two proofs regarding open and closed sets
QUESTION [5 upvotes]: [Question]
1) Let $M=\{\frac{1}{n}:n\in\mathbb{N}\}$. How do I show $M$ is not open in $\mathbb{R}$?
2) Let $K=\{\frac{1}{n}:n\in\mathbb{N}\} \cup \{0\}$. How do i show $K$ is closed in $\mathbb{R}$?
[Solution]
1) 
I've taken the point $1 \in M$ and placed an open ball around it such that $B_r(1)$. 
I want to prove that I can find an $x \in B_r(1)$ such that $x \not\in M$ because then, by definition, $M$ is not open.
I understand that it isnt open, by drawing and looking at the interval. But how do i prove it mathematically?
Could i simply write something like this: 
Let $r>0$ and $x=1+\frac{1}{2}r$ then $x \in B_r(1)$ while $x \not\in M$ therefore $B_r(1) \not\subset M$ and therefore $M$ is not open.
I'm not sure how to argue how/why I defined x as such. Is the above an okay proof and how could I proof/argue how x is defined?
2) To show that $K$ is closed, I will show that $K^c$ is open.
I know $K^c$ is an open union of sets but I'm not certain how to make a proper proof. Here is what I've tried
$K$ will never be negative and will be greater than 1, therefore it will never be within $(-∞;0)$ & $(1;∞)$.
Now I'm not certain what to do.

REPLY [3 votes]: This solution is wrong. The second member of the union is not an intersection but an union. In fact we have:
$K^c=(-\infty,0)\cup(\cup_{n=1}^{\infty}(\frac{1}{1+n},\frac{1}{n}))\cup(1,\infty)$.
Since by definition the arbitrary union of open sets is open, we have $K^c$ is open. Therefore $K$ is closed.