Finding cardinality of sets composed of a difference set $\bar A= \{x-y \ | \ \forall x,y\in A \}$ 
For a non empty set $A$ of real numbers, let $\bar A= \{x-y \ | \ \forall x,y\in A  \}$ be the difference set of $A$.
Find the cardinality of the following sets:

*

*$B= \{A\in \mathcal P(\mathbb Z) | \bar A\subseteq(-1,1)  \}$


*$C= \{A\in \mathcal P(\mathbb R) | \bar A=\{-1,0,1\}   \}$


*$D= \{A\in \mathcal P(\mathbb R) | \bar A\subseteq(-1,1)\}$

Okay, so first let's see if I understand the notations, $B$ has a cardinality of $\aleph_0$ because all the elements must be equal to themselves $(x=y)$ and we know that $|\mathbb Z|=|\mathbb N|=\aleph_0$.
For $C$ for all $x,y$ we must have $x-y\lt 2$, I see that the cardinality must be $\frak c$ but I don't know how to show it.
The last one, $D$, I'm not sure how to read it but it must be at least $\frak c$. I don't know how to show it.
 A: Ok, I see that $B$ is just the set of singletons from $\mathbb{Z}$. It seems that $C$ is the collection $\left\{\{x,x+1\}|x\in\mathbb{R}\right\}$, so you're right about those two cardinalities.
The third one, $D$, should include any subset of $\mathbb{R}$ that has diameter less than $1$. (It also includes sets such as $(0,1)$ with diameter $1$, but those won't affect the cardinality.) Therefore, every subset of the interval $(0,1-\epsilon)$ is included in $D$, so the cardinality of $D$ is at least $2^{\frak{c}}$. It can't be greater than that, because that's the cardinality of the entire power set of $\mathbb{R}$, so we've got that the cardinality of $D$ is $2^{\frak{c}}$.
A: Any element of $C$ have at most 2 elements:
suppose it has 3 distinct elements $x > y > z$, $x - y$ and $y - z$ must be $1$ but then $x - z = 2$ which is absurd. So it's easier to count them now.
For $D$, $(0, 1) \in D$ so $|D| \ge \mathfrak c$. In fact, every subset of $(0, 1)$ is in $D$, so $|D| \ge 2^{|(0,1)|} = 2^{\mathfrak c}$ so you can deduce the result.
