Motivation for set theory axioms

This assumes you have some "general idea" of what a set is. For example you are happy that and that this set (I use singular because they are identical) contains the elements [ilmath]1[/ilmath], [ilmath]2[/ilmath] and [ilmath]3[/ilmath].

The goal is to be rigorous, need to introduce some axioms (things we can assume "just are") that are as simple as possible so we can say "from these axioms, we know this set exists" - for example we use a letter ([math]\emptyset[/math]) to denote the empty set, what if there are 2 empty sets? We would like to justify this.

From this we could say as an axiom: there is only 1 empty set - but this is overly-strong (it is indeed correct, but there are simpler axioms that can lead to it) - let us dive in.

Quick note

This will follow the same format as the start of the set theory axioms page

Membership

We assume "is a member of" is a property we can talk about

The axiom of existence

We have to assume at some point that sets exist. I hate to sound like a philosopher (truly I do) but if we have the set containing a stone and a twig how do we know what makes a twig and where it ends and all that silly stuff. So let us sidestep that by simply assuming:

there exists a set with no elements

Which is reasonable, in our system so far we have no assumptions about what sets can contain, just there exists an empty one, I am happy with this, it's not a big leap to assume this, to deny it would be to deny empty shopping lists (as they are a set of things you wish to buy, or tuples of things and quantities and crossed-off status)

The axiom of extensionality (equivalence)

If every element of [ilmath]X[/ilmath] is an element of [ilmath]Y[/ilmath] and also every element of [ilmath]Y[/ilmath] is also an element of [ilmath]X[/ilmath] then - that is [ilmath]X[/ilmath] is identical to [ilmath]Y[/ilmath].

This looks a lot like - infact this is the axiom we use to to get to this.

The empty set is unique

This is our first result. It's usually a lemma because it's not really important enough to warrant "theorem" status.

Proof

Let us assume we have two distinct sets [ilmath]A\text{ and }B[/ilmath] - this is okay since all we are saying is "they are not identical" - we've got an axiom which tells us what it is to be identical, we need not come up with stuff to put in sets; and of course that [ilmath]A\text{ and }B[/ilmath] are empty - contain nothing.

By the implies and subset relation and noting the truth table of implies, we see [math]x\in A\implies x\in B[/math] as false[math]\implies[/math]false, this is technically true. Thus we have [math]A\subset B[/math]

Similarly we have [math]B\subset A[/math] and by the axiom of extensionality we see [math]A=B[/math] - contradicting that they didn't satisfy this.

Thus it is false that there is more than one empty set. We now know (from just these two axioms) there is only one empty set and we assign it the symbol [math]\emptyset[/math]