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So, I need to find the cardinality of set of all subsets of $\mathbb R$ that are neither open nor closed and are disjoint with integers (i.e. if $A$ is such a set, then $A \cap \mathbb Z=\emptyset$).

Intuitively, there are $c$ such sets.

My reasoning is as follows (not mathematically strict):

I can prove there are $c$ sets of the form $(a,b]$ that are disjoint with $\mathbb Z$. Obviously, those sets are neither open nor closed. There's also $c$ sets of form $[a,b)$ that are disjoint with $\mathbb Z$. So, all together, $c$ of those sets.

Now (and that's where I'm being cautious) every set that is neither closed nor open (and disjoint with $\mathbb Z$) can be represented as countable union of sets of form $(a,b]$ or $[a,b)$ and hence the sought cardinality is $c\aleph_0=c$.

I'm unsure about the last argument, it definitely looks a bit shaky.

Any help is appreciated.

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    $\begingroup$ It is certainly not true that every set that is neither closed nor open (and is disjoint with $\mathbb{Z}$) can be represented in the way you say. One counterexample is $\{0.5\} \cup (1.25,1.75)$, but there are probably much weirder examples out there. $\endgroup$ – kccu Nov 16 '17 at 19:33
  • $\begingroup$ @kccu thanks for the comment, can't believe I missed that $\endgroup$ – windircurse Nov 16 '17 at 19:41
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every set that is neither closed nor open (and disjoint with $\mathbb{Z}$) can be represented as countable union of sets of form $(a,b]$ $[a,b)$.

As kccu commented, this is false. Note that in fact there are $2^c$-many subsets of $\mathbb{R}$ in general, and only $c$-many sets which can be written as a countable union of half-open intervals (exercise), so most sets cannot be represented in such a way (and disjointness from $\mathbb{Z}$ is irrelevant here, so I'll ignore it going forward). In fact, this same reasoning basically shows that there is no good classification of the sets of reals, at all.

For example, consider all the sets we can build from open intervals using complements, countable unions, and countable intersections; this is a huge family of sets, called the Borel sets, but there are still only $c$-many of them, so most sets aren't Borel. Throwing more closure operators into the mix - e.g. closing under continuous images too - doesn't affect anything (unless you throw $2^c$-many closure operators into the mix!); you're never going to get all the sets of reals in such a nice construction.


Here's an explicit proof that there are $2^c$-many such sets:

Let $A$ be the set of dyadic rationals and $B$ be the set of non-dyadic rationals $(B=\mathbb{Q}\setminus A$). If $X$ is any set containing $A$ and disjoint from $B$, then $X$ is neither open nor closed (exercise).

Fix a bijection $b$ between $\mathbb{R}$ and the irrationals, $\mathbb{I}$. For $S\subseteq\mathbb{R}$ let $Y_S=b[S]$, and let $X_S=b[S]\cup A$. What does the map $S\mapsto X_S$ tell you about the number of sets of reals which are neither closed nor open? (And, do you see how you can tweak this to get the same thing for sets of reals which are neither closed nor open and are disjoint from $\mathbb{Z}$?)

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Let $N=\{$all the subsets of $\mathbb{R}$ which are neither open nor closed$\}$, obviously since $N \subseteq \mathcal{P}(\mathbb{R})$ we have $k(N) \leq 2^c$.

To show that also $2^c \leq k(N)$ we define $f : \{0, 1\}^{[0, 1]} \to N$ by $X \mapsto X \cup (2, 3]$.

$k(\{0, 1\}^{[0, 1]}) = 2^c$ and $X_1 \neq X_2 \implies X_1 \cup (2, 3] \neq X_2 \cup (2, 3]$, so $f$ is an injection.

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