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.

  • 1
    $\begingroup$ "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" $\endgroup$ – Mirko Dec 11 '15 at 16:57
  • $\begingroup$ Valid point, I took that into account now. $\endgroup$ – Alex Dec 11 '15 at 17:43

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 $|A|=\mathfrak c$.

  • $\begingroup$ Nice job! I would've never thought to construct a B that is contained within A. Thank you for your help. $\endgroup$ – Alex Dec 11 '15 at 17:51

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