MHB Two different circles in the plane with nonempty intersection

AI Thread Summary
The discussion revolves around proving that within two sets of circles in the plane, one that intersects the x-axis and another that is tangent to it, there exist at least two circles with a non-empty intersection. It is established that the set of circles that meet these criteria is at most countably infinite, while the x-axis contains uncountably many points. This leads to a contradiction, as if the circles did not intersect, the number of intersection points would also be countable. The conclusion drawn is that at least two circles must intersect due to the uncountable nature of the x-axis points. The argument effectively demonstrates the necessity of intersection among the circles in question.
Arnold1
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Hi.

Here is a problem I've been trying to solve for some time now. Maybe you could help me.
We have two sets
\mathcal {Q} is a set of those circles in the plane such that for any x \in \mathbb{R} there exists a circle O \in \mathcal {Q} which intersects x axis in (x,0).\mathcal {T} is a set of those circles in the plane such that for any x \in \mathbb{R} there exists a circle O \in \mathcal {T} which is tangent to x axis in (x,0).

We need to show that in each of these sets there exist at least two different circles whose intersection isn't empty.
It seems obvious that card (Q) \ge card (\mathbb{R}). Maybe we could somehow identify each circle with a different rational number?
 
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Arnold said:
Maybe we could somehow identify each circle with a different rational number?
Yes, every circle (including interior) contains a point with rational coordinates, just like every segment of the x-axis contains a point with a rational x-coordinate. Therefore, there is at most countably many disjoint circles on a plane.
 
So this is it? There are only countably many disjoint circles meeting the above specified conditions nut uncountably many points on x axis. Can we already deduce that at least two circles intersect?
 
Arnold said:
So this is it? There are only countably many disjoint circles meeting the above specified conditions nut uncountably many points on x axis. Can we already deduce that at least two circles intersect?
Yes, we can. If the circles don't intersect, then there is at most countably many of them. But each circle has at most two intersection points with the x-axis, so the number of intersection points is also countable, a contradiction.
 
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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