Set theory/Introduction

Notice: incomplete

What is a Set?
A set is a collection of other sets or nothing at all (under certain rules). All of mathematics comes from here. We're going to have to claim that some things are true purely from assertion, called axioms.

Demonstrate a set by a closed loop of string containing what the set contains. Use random objects to stand for arbitrary sets. Identical objects stand in for identical sets.

Axioms of Set Theory

 * 1) (Axiom of extensionality) Two sets are equal if they have the same elements. Demonstrate this by taking two loops of string and two copies of each object. Place one copy of each object in each loop of string. Point out that they're considered the same. Another thing to show if you have three copies of an object is to put one in one loop of string, and the other two in another loop of string.
 * 2) (Axiom of regularity) Every non-empty set contains something it has nothing in common with. Demonstrate this by trying to construct a counterexample and finding that you need infinitely nesting sets.
 * 3) (Axiom of union) Every non-empty set has elements. It's possible to take all elements of those elements and put them together to form a set. Place smaller loops inside a larger loop. You may put objects in some of the loops. Remove the smaller loops.
 * 4) (Axiom schema of replacement) Every non-empty set has elements. Replacing the elements with other sets will always still yield a set. Demonstrate this by placing objects in a loop and replacing the objects.
 * 5) (Axiom of infinity) The empty set exists (as demonstrated by an empty loop), as does a set with these properties: It a) contains the empty set and b) if it contains a set, it contains another set that contains the original set and all the elements in it.
 * 6) (Axiom of power set) A subset of a given set is a set that contains only the elements of the initial set. For any set, there exists a set containing every subset of the initial set, called the power set.
 * 7) (Axiom of choice) For any nonempty set with nonempty members, it's possible to pick out one member of each member set and put them together to form a non-empty set. Demonstrate this by placing loops of string inside a larger loop of string. Place objects inside the smaller loop. Then choose one object from each smaller loop and put it inside a separate loop of string. Note that the separate loop of string is non-empty.

Theorems
There were two assumptions in the axioms.

5. How do we know that there's a set containing the original and all its elements?

To prove this, I'll need to prove the axiom of pairing, which states: for any two sets, there's another set containing both. Here's how. Take the empty set {}. Take its power set: –  (this is the power set as the only subset is the empty set, so it's just the set containing the empty set). Then take its power set: {{}, – }. Replace the empty set with one of your sets and the set containing the empty set with another. Demonstrate this by demonstrating the empty set's power set's powerset, then replacing each element with a random object.

Now, let our two sets be the original set, and the set containing the original set. Then take the union over the set containing both. Demonstrate this by putting objects in a loop. This represents the original set. Then make a second copy of the original set and put a loop around it. Then put a loop around both sets. Removing the loops 'one layer down' will give a set containing the original set and all its elements.

6. How do we know that subsets exist?

Subsets exist for any set.

a) the empty set is a subset.

b) The set itself is a subset.

However, I'll prove a stronger claim: that you can make a subset from any property.

If you're afraid of circular reasoning, I used different axioms to the ones I was trying to justify.

Mathematics
This seems pointless. But it isn't. Here's why.

Russel's paradox
"What's the point of all this [expletive]?! Just make every collection of sets another set!" one may cry. However, this can't be done. Assume that it can. Then consider the set of all and only non-self-containing sets.

If this set contains itself, then, it's self-containing. But it only contains non-self-containing sets, meaning that it doesn't contain itself. So, the assumption that it contains itself is flawed.

If this set doesn't contain itself, then it's non-self-containing. But it contains all non-self-containing sets, including itself. So, the assumption that it doesn't contain itself is flawed.

It's clear that the only way that a set can neither a) contain itself, nor b) not contain itself is if it doesn't exist. But then this means that not all collections of sets are another set.

We call collections of sets classes. All sets are obviously classes, but, as has been demonstrated, not vice versa. Classes that aren't sets are called proper classes.

Ordered tuples
Two sets are the same if they have the same elements. But what if you care about the order? It's still possible to represent ordered tuples. Here's how:

Any ordered pair (a,b) is {{a},{a,b}}. Demonstrate this by taking two copies of the 'first' object and one copy of the 'second' object. Place a loop around one copy of the first object. Place another loop around the second copy of the first object and the copy of the second. Place a third loop around both loops.

Any ordered tuple that isn't an ordered pair (e.g., (a,b,c,d)) is just the ordered pair (a,(b,c,d)), or the set {{a},{a,(b,c,d)}}, where (b,c,d) is just (b,(c,d)), or {{b},{b,(c,d)}} and (c,d) is just {{c},{c,d}}. This means that (b,c,d) is {{b},{b,{{c},{c,d}}}}, and (a,b,c,d) is {{a},{a,{{b},{b,{{c},{c,d}}}}}}. Demonstrate this by doing the following:


 * 1) Take one copy of the last object and two copies of the other three.
 * 2) Place a loop around one copy of every object.
 * 3) Within this loop, place another loop around the contained copy of the first object.
 * 4) Place a third loop around each object that isn't the first.
 * 5) Introduce the second copy of the first object and place a fourth loop around the third loop and the second copy of the first object.
 * 6) Within the third loop, place a fifth loop around the contained copy of the second object.
 * 7) Place a sixth loop around each object that isn't the second.
 * 8) Introduce the second copy of the second object and place a seventh loop around the sixth loop and the second copy of the second object.
 * 9) Within the sixth loop, place an eighth loop around the contained copy of the third object.
 * 10) Introduce the second copy of the third object and place a ninth loop around the copy of the fourth object and the second copy of the third object.