# 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.

• You are correcting in thinking there is no such open interval. Now, can you try to prove it? For the second one, rememeber that given any interval $(a,b)$ there is a rational there. What does this tell you about $\Bbb R\smallsetminus \Bbb Q$? – Pedro Tamaroff Sep 30 '13 at 2:45

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?

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.
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.
• It's the closure of the set. The closure of a set $S$ is denoted by $\overline{S}$, and it's defined as $S \cup S'$, where $S'$ is the set of accumulation points of $S$. There is a theorem that says that a set $S$ is closed if and only if $S = \overline{S}$. – Twink Sep 30 '13 at 3:06