# The Zermelo-Frankel Axioms

The ZF-Axioms are kind of a bed rock in Mathematics. A place where everything can begin. Now this is not the only way to formalise mathematics. There many theories around. I haven't got there yet. Category theory and Homeotopic type theory come to mind here.Now, let the caveat be stated here clearly.

**I am not a mathematician, I am not formally trained in the topic and I do understand there may be errors here. If you are so disposed and would like to free the Internet of error. Please let me know! I will be happy to be schooled in this manner. What appears below is how I have understood the subject and it may be helpful to you.**

## Axiom of Extensionality

Means that any two sets with the**same members**is the

**very same set.**

∀x∀y (∀z z ∈ x ≡ z ∈ y) ≡ (x = y)Lets hold this true that for

**every set x & y**, where for

**all z**such that

**z is member of x if & only if it is also a member of y**, this is the same as saying

**x = y**

## Axiom of the Empty Set

There is a set with no members.∃x∀y y ∉ xThere exists a set

**x**that for all

**y**,

**y**is

**not a member**of

**x**.

## Axiom of Unordered Pairs

From any**two sets x & y**, we can

**construct a set**that contains both

**x & y**.

The notation for that set is {x,y}.

∀x∀y ∃z ∀w w ∈ z ≡ (w = x ∨ w = y)This says for

**every pair of sets x & y**there exists a set

**z**(for all

**w**) that contains x and y and no other members. Here

**w**act like a dummy variable, it cannot equal anything but

**x or y**.

## Axiom of Union

∀x∃y ∀z z ∈ y ≡ (∃t z ∈ t ∧ t ∈ x)Given any set

**x**, there is a set

**y**such that, for any element

**z**,

**z**is a member of

**y**if and only if there is a set

**t**such that

**z**is a member of

**t**and

**t**is a member of

**x**.

## Axiom of Infinity

∃x∅ ∈ x ∧ [∀y (y ∈ x) → (y ∪ {y} ∈ x)]

## Axiom Scheme of Replacement

B(u, v) ≡ [∀r(r ∈ v ≡ ∃s[s ∈ u ∧ A n (s, r)])]

[∀x∃!yA n (x, y)] → ∀u∃v(B(u, v))

## Power Set Axiom

∀x∃y∀z[z ∈ y ≡ z ⊆ x]

## Axiom of Choice - An add on Axiom

For every,e ∈ C f (e) ∈ e.

∀C∃f∀e[(e ∈ C ∧ e = ∅) → f (e) ∈ e]

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29nov16 | admin |