User:Tungster24/Sector

In mathematics, sectors are a special number-like construction that are said to be equal to multiple values. They are shown as $$\ast A$$ where $$A$$ is a set.

Axioms of Sectors
$$\ast A := \forall x \in A( \ast A \rightrightarrows x)$$ (Axiom 0)
 * A sector is always built from a set. A sector is defined to be sector-number equal to all of it's elements. A sectors elements must be less than the successor fixed point.

$$\forall x( \ast\{x\} = x)$$ (Axiom 1)
 * A sector consisting of a single element is number-number equal to it's singular element.

$$f(\ast A) \leftrightarrows \ast \{f(x) | x \in A\}$$ (Axiom 2)
 * An unary function will output all the possible outputs if a sector is given as output.

$$f(\ast A_1,\ast A_2,\ast A_3...) \leftrightarrows \{f(x_1,x_2,x_3...) | x_i \in A_i \}$$ (Axiom 3)
 * An n-ary function will output all of the possible outputs for all combinations of the sectors. (Tungster's Theorem)

Notions of equality
Equality is defined in three "flavours". Sector to Sector ($$\leftrightarrows$$) Sector to Number ($$\leftleftarrows$$ and $$\rightrightarrows$$) and Number to Number (our regular equality $$=$$).

We can define these two new equality flavors more rigidly:


 * $$ \ast A \leftrightarrows \ast B \iff A = B$$ (or rarely $$ \ast A \equiv \ast B$$) (Axiom 4)
 * $$ \ast A \rightrightarrows x \iff x \in A$$ (Axiom 5)

This is to prevent a paradox that appears when we accept all of these flavours to be the same:

Assume a sector $$\ast\{1,2\}$$. Due to the definition of equality, $$\ast\{1,2\} = 2$$ and $$\ast\{1,2\} = 1$$ due to associativity, $$1=2$$. This is a contradiction.

Our flavours of equality prevent this, since Sector to Number equality and Number to Number equality are not one and the same. The proper version of this contradiction is as such:

Assume a sector $$\ast \{1,2\}$$. Number to Sector equality follows: $$\ast \{1,2\} \rightrightarrows 2$$ and $$\ast \{1,2\} \rightrightarrows 1$$. This however cannot imply $$1 = 2$$ since flavors of equality don't imply eachother. Contradiction is avoided.

This does not mean flavors can't interact:

$$ \xi \leftrightarrows \ast \{1,-1\} $$

$$ \xi^2+1 \leftrightarrows \ast \{1,-1\}^2+1 $$

$$ \xi^2+1 \leftrightarrows \ast \{1^2+1,(-1)^2+1\} $$ (Tungster's Theorem)

$$ \xi^2+1 \leftrightarrows \ast \{2,2\} \leftrightarrows \ast \{2\} $$ (last step is simple set simplification)

$$ \xi^2+1 = 2 $$ (Axiom 1)