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 is not open. If it is not open, it means its complement is open i.e. $$A:=\{p\not\in\mathbb Q : 0 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?
Elaborate on $A^{c}:=\{p\in\mathbb Q : 0 not open and not closed in $\mathbb R$
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general-topology
definition
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3If a set $S$ is not open, that does *not* mean that the complement of $S$ is open. For a simple example, let $S=(0,1]$. – 2012-03-10
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0@AndréNicolas: I must be messing this up with compactness?! Sorry have to go back to definitions... – 2012-03-10
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0In the example given above, neither $S$ nor its complement is compact. – 2012-03-10
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0Basically, 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.) – 2012-03-10
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0@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. – 2012-03-10
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0@Sivaram: It is not incorrect in context, since the discussion clearly concerns the usual topology on $\Bbb R$. – 2012-03-10
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0@Sivaram: I was talking about subsets of $\mathbb{R}$, and it is natural to assume that it is given the metric topology. Here, $\mathbb{R}$ clearly has only two clopen subsets. You are of course right: with the discrete topology all subsets are clopen; with the trivial (anti-discrete) topology $\emptyset,\mathbb{R}$ are the only subsets which are either open or closed (and each is both); under the Sorgenfrey (lower-limit) topology the answer is different yet again. But this seems to be overly complicated based on the question asked. – 2012-03-10