**KOCH SNOWFLAKE**

This object has the property of having a finite area enclosed by an infinite perimeter.

**Don't believe it? Then see the proof below.**

The recursive construction of this fractal is achieved by extending the middle third of each straight line by an additional equilateral triangle. The first three iterations (plus the starting – zeroth iteration) are shown below.

And each additional triangle

*is 1/9th of the area of the one it appears on*

**tn***, as shown in the diagram below for*

**tn-1***and*

**t0***.*

**t1**

The entire area

*A***of the entire Koch’s Snowflake can be calculated as follows, where**

*∞**is the area of the initial triangle:*

**A0**So the area is finite, as it is obvious by inspection. But how about the perimeter of this object?

Let

*L***be the perimeter at the n’th iteration where***n**is the perimeter of the initial triangle. Let***L0***be the length of the side of the initial triangle.***l****THE CANTOR SET**

This object has the property of countably infinite cardinality and measure zero.

Below are seven iterations of the Cantor Set, n+1, below n.

The Cantor Set shares similar odd properties, since its cardinality is countable infinity, but it’s measure is zero, as shown below. The construction of the set, proceeds recursively, starting with the unit interval [1,0]. At each iteration the middle 1/3

^{rd}open interval is subtracted from any previous intervals. So the first three iterations are the following sets: ℵ ∪∈

Cantor has shown, in his famous diagonal argument, that the cardinality of infinite binary sequences ∆ is uncountable. Below we present a bijection between ∆ and the Cantor set, defined as:

Thus demonstrating that the Cantor set cardinality is uncountable and hence by the continuum hypothesis of cardinality of the continuum i.e. ℵ1

Define the functions on the closed intervals:

*([a,b]) = [a, a + 1/3(b-a)],*

**L***([a,b]) = [a + 2/3(b-a), b]*

**R**And with each binary sequence

*d*∈∆ we associate a sequence of closed intervals:

**F***d0*

**,****F***d1*

**,****F***d2*

**,**

**…**Defined recursevly as follows

**F***d0*=

*Cn*= [0,1]

**F***dn+1*=

**(**

*L**), if*

**F**dn*dn*= 0

**F***dn+1*=

**(**

*R**), if*

**F**dn*dn*= 1

Now,

*f*is clearly*surjective*by construction since for each*d*we find a unique intersection of the sequence**F***d0*,**F***d1*,...To complete the proof we need to show that f is

*injective*. Consider have 2 binary strings*red*and*blue*. Suppose*n*is the least number such that*re**dn*≠*blue**n*. and suppose w.l.o.g.*re**dn*= 0. To complete the proof it needs to be shown that*f*(*re**d*) ≠*f*(*blue*).The illustration below we see how

*red*and*blue*are mapped onto*C*, and*f*(*red*) ∩*f*(*blue*) is denoted with purple (violet?) color.We have

**F***bluen*

*=*

**F***redn*from the choice of

*n*,

*f*(

*red*) ∈

**F***redn*+1 =

*(*

**L**

**F***redn*),

*f*(

*blue*) ∈

**F***bluen*+1 =

**(**

*R*

**F***redn*), but

*(*

**L**

**F***redn*) ∩

*(*

**R**

**F***redn*) = ∅

Hence

*f*(

*re*

*d*) ≠

*f*(

*blue*). Which proves that

*,*

*ℵ*

**C**=c**1**

*C***? The**

**measure of the Cantor Set**. One way to answer that is to subtract from the measure of the unit interval:

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