Re: Scott and George's Teaching Thread


On Sep 28, 3:21 pm, george <gree...@xxxxxxxxxx> wrote:

0s. EgAw[weg<->(we(p(n)Xn)Xp(n)) ^ (w_3=w_1) ] {Separation}

Yeah, I know, "w_3" is "cheating" but we CAN define ways
of getting back inside ordered n-tuples.

I am about to do this now.
This is not the answer to the question you ask below but it
does prove that there was a right way to do this.
This was an instance of Separation, so after the ^ ,
it needs a (notationally correct) predicate. That predicate,
instead of w_3=w_1, would be
Exep(n)EyenEzep(n)[w=(x,y,z) ^ x=z].
The fact that I knew I was going to need that was why I called it w.
I trust you can also see why I decided to write w_3=w_1 instead.

Anyway, getting to the question you actually asked,
we DEFINE functions as sets of ordered pairs.
It just is that way because we say so; we simply insist
that that's what "function" MEANS, over here in ZFC-land.
Of course, not EVERY set of ordered pairs is a function.
The general term is "relation". A set is<df> a relation iff
every element of it is an ordered pair. A relation is a function
iff its pairs match only 1 second element to any given first element.

This is a little bit ambiguous in that at least two functions
(u and p) are built-in to the language and defined over the WHOLE
domain. But in set theory, we can have more; we can also have
functions
that we ourselves define. Some logic texts go so far as to give
the built-in functions a different name (some call them "functors",
but that word has a more prominent meaning from a whole 'nother
discipline, category theory). In either case, regardless of whether
the function is a set or is built-in, it is applied by writing its
name
immediately followed by parens around the argument, and that
conglomeration evaluates to the function's value for/at that argument.

So what is the general method of defining a function from a set of
ordered pairs?

You don't *have* to define the function FROM the set of ordered
pairs; the set of ordered pairs IS the function, so once you have
defined the set, you HAVE defined the function.
The "g" whose existence is alleged by 0s. above IS a function.

Maybe your tension headache comes from g's being a set of
ordered triples, when I told you a function was set of ordered pairs.
But the cure is, set-theoretically, there is simply nothing wrong
with defining an ordered n+1 tuple as an ordered pair whose first
element is an ordered n-tuple and whose second element is the
n+1st element (of the "original" tuple). So g is, in addition to
being
a set of ordered pairs, ALSO a set of ordered triples. And g is, in
addition
to being a relation, ALSO a function.


.

Relevant Pages