Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Determine if it is countable or uncountable The set $E$ of all circle in $R$$^2$ with centers at rational coordinate points and positive rational radius.

I have no idea about this type of question.

share|cite|improve this question


1) If sets $A$ and $B$ are countable then their cartesian product $A\times B$ is countable.

2) Note that every element of $E$ is represented by a triple $(x,y,r)$ with $x,y\in \mathbb Q$ and $r\in \mathbb Q^{+}$

share|cite|improve this answer

Hint: if you can find a surjective mapping from a countable set to the set $E$ you are looking at, you will prove the set $E$ is countable.

Try finding a surjection from $\mathbb Q\times\mathbb Q\times\mathbb Q_+$ to $E$.

share|cite|improve this answer
Don't you need $\mathbb Q\times\mathbb Q\times\mathbb Q$ to $E$? As you have a rational radius as well. – Cruncher Feb 24 '14 at 15:53
Correct. Even more correct, it's $\mathbb Q\times \mathbb Q\times \mathbb Q_+$ – 5xum Feb 24 '14 at 19:15
Still not 100% understand... How about the $R$$^2$?? Is it mean real number? So it is uncountable?? – user109403 Feb 25 '14 at 2:04
No, because you can't have a circle with the center at $(\sqrt 2, 5)$. – 5xum Feb 25 '14 at 6:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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