Elaborate on $A^{c}:=\{p\in\mathbb Q : 0<p<\sqrt{2}\}$ not open and not closed in $\mathbb R$

I know that $\sqrt{2}\not\in\mathbb Q$ and $\sqrt{2}\in\mathbb R$ but it is not obvious to me why $\{p\in\mathbb Q : 0<p<\sqrt{2}\} \subset \mathbb R$ is not open. If it is not open, it means its complement is open i.e.

$$A:=\{p\not\in\mathbb Q : 0<p<\sqrt{2}\}\subset \mathbb R$$

where each rational number is surrounded by two real numbers and where $p\not\in\mathbb Q$ basically means $p\in\mathbb R/\ \mathbb Q$ (i.e. the rational section out). Now the culminating points are all in $A$ i.e. making it closed and it is also open because you can cover it with open balls (n.b. rational-real-rational -cover). So $A$ is clopen. Now what is $A^{C}$ then?

Is there some term for not-clopen?

-
If a set $S$ is not open, that does not mean that the complement of $S$ is open. For a simple example, let $S=(0,1]$. –  André Nicolas Mar 10 '12 at 7:20
@AndréNicolas: I must be messing this up with compactness?! Sorry have to go back to definitions... –  hhh Mar 10 '12 at 7:21
In the example given above, neither $S$ nor its complement is compact. –  André Nicolas Mar 10 '12 at 7:24
Basically, unlike doors (which are either open or closed, and never both) subsets of $\mathbb{R}$ can be neither open nor closed, and they can also be both open and closed. (Well, there are only two examples of the latter kind.) –  Arthur Fischer Mar 10 '12 at 7:29
@ArthurFischer: Sets which are both open and closed depends on the underlying topology. So it is incorrect to say that there are only two examples of the latter kind. –  user17762 Mar 10 '12 at 7:41

Let $S=\{p\in\Bbb Q:0<p<\sqrt2\}$. Not only is $S$ not open, it has empty interior: it does not contain any non-empty open set. Suppose that $U$ is a non-empty open subset of $S$; the open intervals form a base for the topology of $\Bbb R$, so $U$ must contain a non-empty open interval $(a,b)$. But then $a$ and $b$ are real numbers such that $(a,b)\subseteq S\subseteq\Bbb Q$, which is impossible: there is certainly an irrational number between $a$ and $b$.

For essentially the same reason your set $A$ is not open. If $(a,b)$ were a non-empty open interval contained in $A$, every real number between $a$ and $b$ would be irrational, which is of course false.

And since neither of these complementary sets is open, neither can be closed: if $S$ were closed, $A$ would be open, and if $A$ were closed, $S$ would be open. Thus, both $S$ and $A$ fail as badly as possible to be clopen.

I know of no single term that means not clopen; one simply says that $S$ is not clopen, unless one can specify more precisely that $S$ is not open, that $S$ is not closed, or (as in this case) that $S$ is neither open nor closed.

Finally, note that as André already mentioned, a non-open set does not necessarily have an open complement. A set with an open complement is closed, and there are many sets that are neither open nor closed. A few people have studied the class of door spaces, spaces in which every set is either open or closed (or both), but this is a very restricted class (and not really a very interesting one). Most spaces of interest are not door spaces.

-

A set $S$ of real numbers is open in $\mathbb{R}$ if for every $s$ in $S$, there exists a positive $\epsilon$ such that the interval $(s-\epsilon,s+\epsilon)$ is a subset of $S$.

A set $S$ of real numbers is then said to be closed in $\mathbb{R}$ if its complement is open in $\mathbb{R}$.

But it is possible that you are using a different (but equivalent) set of definitions. For example, one can define a subset $S$ of $\mathbb{R}$ to be closed in $\mathbb{R}$ if all the limit points of $S$ are in $S$. Then one says that a subset of $\mathbb{R}$ is open in $\mathbb{R}$ if its complement is closed in $\mathbb{R}$.

The two approaches described above are the two most common ones. Do check what is your official definition. With either one it should not be hard to verify that your set is not an open subset of $\mathbb{R}$, and neither is its complement.

-