# The CH problem and the circle's area

## Main Question or Discussion Point

Hi,

L1 is a circle with a perimeter's length 1.

/ is "divided by"

n>2

RFC is the roughness of some polygon which is closed by L1 circle (all its vertices are ON the circle) and calculated as L1/n.

A circle is not a polygon, therefore we are talking about its roughness-magnitude, which is notated as RFA and calculated by L1/2^aleph0 (=0)

A 1 dimensional closed geometrical object is not a circle and it is not a polygon. Let us call it SC for semi-circle.

RFB is the roughness-magnitude of a SC which is closed by L1 circle (all its points are ON the circle) and calculated as L1/aleph0 (>0).

I think that SC must exist if the circle exists as a geometrical object.

More to the point:

Let us say that [oo] = 2^aleph0 = c (has the power of the continuum).

Therefore:

1/[oo] = 0

1/0 = [oo]

The first known transfinite cardinals are aleph0 and 2^aleph0.

2^aleph0 = c (has the power of the continuum) therefore RFA = 1/2^aleph0 = 0 (there are no "holes").

aleph0 < c (does not have the power of the continuum) therefore RFB = 1/aleph0 > 0 (there are "holes").

y = the size of the radius of L1 circle

The magnitude of 2^aleph0 is greater than the the magnitude of aleph0.

In the case of RfB there are aleph0 radii with r=y, but there are 2^aleph0 radii with r<y.

aleph0 < 2^aleph0, therfore r<y.

S2 is the area of a RFA geometrical object (a circle).

S1 is the area of a RFB geometrical object (a SC).

In case of RfA we know that S2 = 3.14... r^2 (where r=y).

But in case of RfB r<y, therefore S1 = 3.14... r^2 (where r<y).

S2 - S1 = x where x > 0.

A question: can we use x to ask meaningful questions on the CH problem ?

Organic