I want to prove this problem:
Prove the set of all circles in cartesian plane(Call that P) that they have not any intersection or none of them is in another one is denumerable.
And I must state that our coordinates is in R not Q!
For rational numbers we can denote a function such that all circles map to their center coordinates $(x,y)$ and radius$(z)$ so finally every circle will be mapped into $(x,y,z)$:$$f:P→ Q\times Q\times Q$$ $$f(c)=(x,y,z) ~~s.t~~c∈P$$
Its obvious that f is bijection.
So the function range is in $Q\times Q\times Q$.
And we know that $Q\times Q\times Q\sim N$.
So set of all circles with rational coordinates is denumerable and because we want no intersection , that is a subset of all circles and we know infinite subset of denumerable set is denumerable.
As you see I know how to prove it using rational numbers but for real numbers I have no any ideas.
Best Answer
It can be shown that for any circle $C$ there exists a point $A$ with rational coordinates (i.e. $A\in\mathbb{Q}^2$) lying inside $C$ (i.e. $A\in C$). Since the circles are disjunct such $A$ uniquely define $C$ inside which it lies. So, you can build an injection $P\rightarrow\mathbb{Q}^2$. Since $|\mathbb{Q^2}|=|\mathbb{N}|$ we are done.