2
$\begingroup$

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.

$\endgroup$
  • 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
3
$\begingroup$

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 $|A|=\mathfrak c$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.