The set of subsets of $\mathbb R$ which satisfy almost everything of the axioms for $\mathbb R$ and question of cardinality Suppose that $A$ is the set of all subsets of the set $\mathbb R$ which are not closed under multiplication or addition, as ordinarily defined in the case of $\mathbb R$.
So, $S \in A$ if and only if there either exist $x,y \in S$ such that $x+y \notin S$ or if there exist $x,y \in S$ such that $xy \notin S$.
Is cardinality of $A$ greater than the cardinality of $\mathbb R$?
 A: I'd like to make my statements a bit more rigorous, but comments are too short so I let it to be an answer.
For start, let's consider addition only (multiplication can be treated pretty the same way).
Let $A =\{S \subset \mathbb{R} | S 
\text{_not closed under } +\}$ and $B =\{S \subset \mathbb{R} | S 
\text{_closed under } +\}$.


*

*$\text{card}(A) = \text{card}(2^\mathbb{R})$. 
$\blacktriangleleft$ Consider $S = \{\{2,3\} \cup S' | S' \subset (-\infty;0]\}$. Since $2+3 = 5 \notin S$, it is subset of $A$.
Now $2^\mathbb{R} \supset A \supset S$ and $\text{card}((-\infty;0]) = \text{card}(\mathbb{R})$, thus $\text{card}(S) = \text{card}(2^\mathbb{R})$.
$\blacktriangleright$

*$\text{card}(B) = \text{card}(2^\mathbb{R})$. 
$\blacktriangleleft$ Consider $S = \{[20;+\infty) \cup S' | S' \subset [10;11]\}$.
Since $\forall x,y \in [10;11]: x+y \in [20;+\infty)$, elements of $S$ are closed under $+$, thus it is subset of $B$.
Now $2^\mathbb{R} \supset B \supset S$ and $\text{card}([10;11]) = \text{card}(\mathbb{R})$, thus $\text{card}(S) = \text{card}(2^\mathbb{R})$.
$\blacktriangleright$

*For any binary operation $f(x;y)$ on $\mathbb{R}$ such that $f(z;z) \neq z$ it is true that cardinal number of sets not closed under $f$ is $\text{card}(2^\mathbb{R})$. 
$\blacktriangleleft$ 
Let me use designations $A$ and $B$ as previously.
Since $A \cup B = 2^\mathbb{R}$, it is enough to show that $\text{card}(A) \geq \text{card}(B)$, i.e. construct an injection from $B$ to $A$. 
Consider any set $\beta \in B$.
I take any element $b \in \beta$ and remove $f(b;b)$ from $\beta$. 
Obviously $(\beta \setminus \{f(b;b)\}) \in A$.
If there exists any different $\beta'$ with an element $b'$ such that $\beta \setminus \{f(b;b)\} = \beta' \setminus \{f(b';b')\}$, then $b \in \beta'$ and $b' \in \beta$, which means $f(b';b') \in \beta$ and $f(b;b) \in \beta'$.
Thus $\beta = (\beta \setminus \{f(b;b)\}) \cup \{f(b;b), f(b';b')\} = (\beta' \setminus \{f(b';b')\}) \cup \{f(b;b), f(b';b')\} = \beta'$.
$\blacktriangleright$
