Given $X \subseteq \mathbb{R}, y \in \mathbb{R}$, define $y+X \triangleq \left \{ y+x : x\in X \right \}$
Let $X \subseteq \mathbb{R}$ s.t $X \cap (y+X) \neq \emptyset \ \forall y \in \mathbb{R}$.
Prove $X$ is not countable.
My first idea was trying to use something similar to Cantor's diagonalization but I'm pretty sure it doesn't work here, so I'm kind of clueless.
A hint on how I should approach this, rather than a solution, would be best.
Hint:
If $X$ is countable, then $Y=\{a-b \mid a,b \in X\}$ is countable, so there exists $y\in \mathbb{R}\setminus Y$.
Thanks #coxehj4142 Here is my sketch for proof:
If $X$ is countable, then Y defined above, which is all possible "differences in X" is countable:
Easy to show that $|X| \leq |\mathbb{N}| \Rightarrow |X \times X| \leq |\mathbb{N} \times \mathbb{N}|$ and $|\mathbb{N} \times \mathbb{N}|$ is countable. also $|Y| \leq |X \times X|$, thus $Y$ is countable.
So we can choose $y \in \mathbb{R} \setminus Y$, and by definition of $Y$ we get $x_1+y \neq x_2 \ \forall x_1,x_2 \in X$ and thus $X \cap (y+X) \neq \emptyset$ in contradiciton.
