Which of the following subsets of $\mathbb R^2$ is/are NOT countable?
(a){$(a,b)\in \mathbb R^2:a\le b$}
(b){$(a,b)\in \mathbb R^2:a+ b\in \mathbb Q$}
(c){$(a,b)\in \mathbb R^2:ab\in \mathbb Z$}
(d){$(a,b)\in \mathbb R^2:a, b\in \mathbb Q$}
Solution:
Argument for (a):Let us fix the first coordinate of $(a,b)$ i.e., $a$,since there are uncountable number of elements($b$) which are greater than $a$.Hence there are uncountable choices for $b$.But since $a$itself is arbitrary real number,therefore $a$ can also assume uncountable number of values.Hence,Set mentioned in (a) is an uncountable set.
Argument for (b): Let $a\in \mathbb Q^c$,now take $b=-a$.Hence $b$ is also irrational number.Since,$a$ is an arbitarary irrational number(which are uncountable).Hence,set mentioned in (b) contains all the irrational coordinates(both coordinates are irrational numbers).Hence,the set is an uncountable set.
Argument for (c): Let $a\in \mathbb Q^c$,now take $b=\frac{1}{a}$.Hence $b$ is also irrational number,making $ab=1 \in \mathbb Z$.Since,$a$ is an arbitarary irrational number(which are uncountable).Hence,set mentioned in (c) contains all the irrational coordinates(both coordinates are irrational numbers).Hence,the set is an uncountable set.
Argument for (d): Let $a,b\in \mathbb Q$.Both,$a$ and $b$ have countabe choices.Since the cartesian product of two countable sets is again a countable set.Hence the set mentioned in (d) is a countable set.
Are my arguments correct??