Are the following sets open subsets of $\mathbb{R}$ I need to determine whether $[2,4]$ and $\mathbb{R}-\mathbb{Q}$ are open subsets of $\Bbb R$. 
For $[2,4]$: 
I know for a subset to be open then $\forall x\in V$ $\exists (a,b)$ s.t. $x\in (a,b)\subseteq V$
I'm thinking that there's no open interval (a,b) for which $2\in (a,b) \subseteq V$ 
Am I thinking about this wrong? I tried asking my professor but he told me to look at the practice problems but I don't even understand the practice problems. I'm having trouble determining whether a set is open or not. 
 A: You’ve answered the first one correctly: $[2,4]$ is not open, because it does not contain an open interval around $2$. (It also doesn’t contain an open interval around $4$.) You can think about $\Bbb R\setminus\Bbb Q$, the set of irrational numbers, the same way. $\sqrt2\in\Bbb R\setminus\Bbb Q$; does $\Bbb R\setminus\Bbb Q$ contain an open interval around $\sqrt2$? For that matter, does $\Bbb R\setminus\Bbb Q$ contain any non-empty open interval at all?
A: Another way to prove $\Bbb R - \Bbb Q$ is not open is this: Let $\Bbb I$ be the set of irrational numbers, and let $A=\Bbb R - \Bbb Q$. A set is open if and only if its complement is closed. Since $A^c=\Bbb R - \Bbb I$, we have that $\overline{A^c}=\overline{\Bbb R - \Bbb I}=\Bbb R \neq A^c$, i.e, $A^c$ is not equal to its closure, which means that $A^c$ is not closed, and hence, $A$ is not open.
A: Alternative answer: (just for fun!)
A set is open iff its complement is closed, and a set is closed iff it contains all its limit points.
The complement of $[2,4] \subset \mathbb{R}$ is $(-\infty, 2) \cup (4, \infty)$. But this set does not contain all its limit points. For example, it contains the elements in the sequence $1, 1.9, 1.99, 1.999, \ldots$ which converges to $2$, i.e., an element that is not in the set.
The complement of the rationals is the irrationals. Again, this set does not contain all its limit points. For example, it contains the elements in the sequence $\sqrt{2}/1, \sqrt{2}/2, \sqrt{2}/3, \ldots$ which converges to $0$, i.e., an element that is not irrational.
Neither your first nor second set has a closed complement, hence neither your first nor second set is open. QED
