0
$\begingroup$

[ edited 9th april 2019]

I'd like to solve this question using only basic concepts of geometry ( without analytic geometry) and of elementary set theory.

If I am correct, the following process allows me to " build" the desired set. But, how could I express, in a single formula,the result of this process, using the proper logical and set theoretical symbolism?

(1) I choose an arbitrary point O (as first "center").

(2) I define an equivalence relation : the set (of points) A is equivalent to the set (of points) B iff all the elements of A are at the same distance from O as are all the elements of B ; I obtain an infinity of equivalence classes ( one for each possible " orbit")

(3) inside each equivalence class ( that is , for each " orbit") I choose the greatest set using the inclusion relation ( for the circle is the greatest set amongst sets whose points are all at a given distance from a given "center"); I obtain an infinity of circles for the first center O.

(4) I repeat (1)-(3) for each point in the plane, that is, for each possible "center".

(5) Using, maybe, the union operation, I gather all my ( infinite) collections of circles for a given center in a new set , which would be the set of circles in a plane P.

Remark - This is not homework; it is a question I ask myself, as a gratuitous exercise in logic/ elementary set theory

$\endgroup$
6
  • 1
    $\begingroup$ Pay attention to the difference between a set and its points. The set CIRCLE you defined is a set of subsets. The union of all subset is all the plane (every point is contained in some circle) but the set CIRCLE is not the plane $\endgroup$
    – karmalu
    Apr 8, 2019 at 13:56
  • 1
    $\begingroup$ Why not $C=\{(x,y,r): r\gt 0\}$? $\endgroup$
    – John Douma
    Apr 8, 2019 at 13:59
  • $\begingroup$ @ John Douma - Is a triple sufficient? Would we need a quadruple in order to take the center into account ? Or not? $\endgroup$
    – user654868
    Apr 8, 2019 at 14:05
  • 1
    $\begingroup$ @Eleonore Saint James - I think (x,y) is the centre. $\endgroup$
    – Paul
    Apr 8, 2019 at 14:08
  • $\begingroup$ I think I understand in which way a given circle can be identified by a triple < x, y, r> x and y being the coordinates of the center and r the radius. But I am wandering whether this triple is " able" to define the circle as a set of points. Is it the case? $\endgroup$
    – user654868
    Apr 8, 2019 at 14:16

1 Answer 1

4
$\begingroup$

Mathematicians already have a symbol for a circle of radius $r$ centred at $(x_0 ,y_0 )$, it is $$B((x_0 ,y_0 ),r)=\{ (x,y) | (x-x_0 )^2 + (y - y_0 )^2 = r \}$$ Then, the set of all circles is $$C=\{B((x_0 ,y_0 ),r) | r>0, (x_0 ,y_0 ) \in Plane \}$$

$\endgroup$
2
  • $\begingroup$ @ NazimJ - Thanks for your answer. ( I upvoted). I think this answer is correct. Do you think it would be possible to define the desired collection without using analytic geometry? $\endgroup$
    – user654868
    Apr 8, 2019 at 14:58
  • 1
    $\begingroup$ You could let $\vec{x}$ denote a point on the plane, then as mentioned in previous comments, every open ball $B(\vec{x}, r)$ can be "labelled" by an ordered pair $(\vec{x},r)$. And the set $\{(\vec{x},r) |r>0 \}$ is isomorphic to the set of all circles, since every circle has a label, and every label corresponds to a circle (it is a bijection). Is that what you are looking for? $\endgroup$
    – NazimJ
    Apr 8, 2019 at 15:34

You must log in to answer this question.