0

I know that there are infinities of different magnitudes for countable things (e.g. via Cantor's diagonal argument), and that the ratios of the sizes of two such infinities may itself be infinite. However, how do ratios of infinities work for uncountable things? And can the ratio of two infinities of uncountable things be finite?

Consider an infinitely large two-dimensional plane, consisting of a point $p$ and two unit vectors $v_1$ and $v_2$ rooted at $p$. Divide the infinite plane into two regions: $A$, the region of space spanned by the "wedge" centered at $p$, the "short way around" between $v_1$ and $v_2$, and $B$, the remainder of the plane (the "long way around"). The areas $|A|$ and $|B|$ are infinite, however it would seem that $|A| \over |B|$ is finite, since if a circle of any nonzero radius centered at $p$ is intersected with $A$ and $B$, the ratio of the areas of the two intersected regions would always be be the constant value $\theta \over {2\pi - \theta}$, where $\theta = cos^{-1}{{|v_1 . v_2|} \over {|v_1||v_2|}}$.

(Actually maybe I am mistaken, and points on a plane are countable, by extension of Cantor's argument to two dimensions...)

Luke Hutchison
  • 361
  • We have to be careful with ratios of infinite cardinalities. We do not have a usual ratio , we must interprete the ratio as a limit (already in the infinite-countable case). This is defined as the "density" of a set within another set. I am not sure how this is done in the unconutable case and when such ratios are finite. – Peter Jul 23 '20 at 07:31
  • Why do you take intersection of the areas with the same circle ? Considering the area between $v_1$ and $v_2$ bounded by a circle of radius $R_1$ and the outer area bounded by a circle of radius $R_2=mR_1, m\in \mathbb{N}$ then as $R_1 \to \infty$ , we get the unbounded areas but the ratio of the areas $\frac{|A|}{|B|}= \frac{\theta}{2\pi-\theta}\frac 1 {m^2}$ which is dependent on $m$ and so the ratio is not well-defined. – user-492177 Jul 23 '20 at 07:43
  • @user710290 I don't understand why you have introduced $m$. The point of using a single circle was to show that no matter what radius of circle is used, the ratio of the area of a "slice of the pie" to "the whole pie" is constant. If you then take the limit as the radius of that circle goes to infinity, you should arrive at the conclusion that the same ratio is also constant and finite with the circle radius at infinity. – Luke Hutchison Jul 24 '20 at 10:25
  • @LukeHutchison Actually my intention was to show that there are several absurdities while comparing two infinities. The way you have taken the ratio of the areas differs from that of mine , so we get different constants as the limits. But yes if your point was just to show that the ratio is a finite constant for two infinities then you can ignore my work. – user-492177 Jul 24 '20 at 10:48

0 Answers0