cardinality of the set of points with one irrational coordinate, and one rational.

Question

Find the cardinality of the set of all points in the plane which have one rational and one irrational coordinate. Justify you answer.

My thoughts so far. We know that $\mathbb Q$, the set of all rational numbers, has a cardinality of $\aleph_0$. Also, since the set of irrational numbers is just the reals with the rationals removed, thus the set of irrational numbers has a cardinality of $\mathfrak c$ (continuum). Therefore, it seems logical to conclude that the ordered pair also has a cardinality of $\mathfrak c$.

I'm fairly sure this would justify the answer. However, is there a way to prove this using either a bijection or the Cantor-Bernstein theorem? Thanks in advance.

Answer 1

I would prove it as follows:

Let $A$ denote your set of points, and let $B=\{(0,x)\ |\ x\text{ irrational }\}$.

Then $B\subset A$, and $|B|=\mathfrak c$ (there is a bijection from the irrationals to $B$ given by $x\mapsto (0,x)$). Thus $\mathfrak c\leq |A|$.

Since $\mathbb{R}^2$ has the cardinality of the continuum, and since $A\subseteq\mathbb{R}^2$, we have also $|A|\leq \mathfrak c$. In total $|A|=\mathfrak c$.