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Find the cardinality of $A =\{(x,y) \in \mathbb R \times \mathbb Q : > x^2 - y = 0 \}\}$

This question may seem trivially easy but I have just started learning about cardinality and still find some things confusing.
My attempt: First of all, I rewrote the definition of the set in a simpler way, adding $y$ to both sides and making necessary assumptions
$$A =\{(x,x^2) : x \in \mathbb R \land x^2 \in \mathbb Q \} = \{(x,x^2) : x\in \mathbb Q \} \cup \{(\pm\sqrt q , q):q \in \mathbb Q \}$$ Now, the cardinality of both of these sets is $\aleph_0$, thus the cardinality of the set in question is $\aleph_o$
Is my solution correct? I not, where have I made a mistake?

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  • $\begingroup$ You made a mistake rewriting $A$. For example $(2, \sqrt{2}) \in A$ but it's not in your set. $\endgroup$
    – Stefan Mesken
    Jan 11, 2018 at 17:24
  • $\begingroup$ $A = \{(\pm \sqrt{y},y):y \in Q \land y \ge 0\}$. $\endgroup$
    – Math Lover
    Jan 11, 2018 at 17:25
  • $\begingroup$ @StefanMesken Why $(2, \sqrt 2) \in A$? $2^2 \ne \sqrt 2$ $\endgroup$
    – Aemilius
    Jan 11, 2018 at 17:26
  • $\begingroup$ I messed up the order. I meant $(\sqrt{2},2) \in A$. $\endgroup$
    – Stefan Mesken
    Jan 11, 2018 at 17:27
  • $\begingroup$ Sorry, but I still can't see what's the problem with my definition of the set. $x^2 - y = 0 \iff y = x^2$ $\endgroup$
    – Aemilius
    Jan 11, 2018 at 17:29

2 Answers 2

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Yes $x^2 = y$, but since $\mathbb{R}$ contains all the square roots of elements of $\mathbb{Q}$, perhaps $\{(\pm\sqrt{y}, y) \in \mathbb{R} \times \mathbb{Q} \}$ would be a bit simpler. Then this set has one element per element of $\mathbb{Q}$...

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  • $\begingroup$ You should put a $\pm$ before $\sqrt y$ as the negative ones are in the set as well. $\endgroup$
    – Ross Millikan
    Jan 11, 2018 at 17:35
  • $\begingroup$ Thanks, this solution is in fact simpler. But what is the problem in mine? $\endgroup$
    – Aemilius
    Jan 11, 2018 at 17:35
  • $\begingroup$ @RossMillikan : What? Not everyone uses the multivalued function $\sqrt{\cdot}$? (Thanks. Fixed.) $\endgroup$
    – Eric Towers
    Jan 11, 2018 at 17:38
  • $\begingroup$ @Aemilius : While your equality of $A$ with a union of two sets is valid, it twice includes every element (the two sets on the right are identical). This isn't fatal, but it is a longer/harder story to tell and to explain clearly. $\endgroup$
    – Eric Towers
    Jan 11, 2018 at 17:43
  • $\begingroup$ They are not identiclal. For example, the first set does not include $(\sqrt 2, 2)$ $\endgroup$
    – Aemilius
    Jan 11, 2018 at 17:45
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The cardinality of the set $$A =\{(x,y) \in \mathbb R \times \mathbb Q : > x^2 - y = 0 \}\}$$ is $\aleph_0$.

Note that for any positive rational number $y$, you get two elements of $A$ , namely,( $\sqrt y ,y)$ and $ (-\sqrt y , y)$.

Therefore your set is infinitely countable.

You also have $(0,0)$ in your set which does not change the cardinality.

Thus the cardinality of $A$ is $\aleph_0$.

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