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

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.

• "an ordered pair" is not the correct expression in your context. You are talking about a set of ordered pairs, not just one ordered pair. Each ordered pair $\langle x,y\rangle$ usually represented (following Kuratowski) as $\{\{x\},\{x,y\}\}$ has cardinality $2$ (or $1$, if $x=y$). The current title "Cardinality of an ordered pair" is misleading, a more suitable one might be "Cardinality of the set of points with one irrational coordinate" – Mirko Dec 11 '15 at 16:57
• Valid point, I took that into account now. – Alex Dec 11 '15 at 17:43

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 $|A|=\mathfrak c$.