Re: Cantor's circular "proof" that evens = integers

On May 5, 5:11 am, Phil <toob-head...@xxxxxxxxxxxxx> wrote:
We will refer to the possibility that sets N and E
are equinumerous as property "P."

You CAN'T do that. "Property" is ALREADY in the dictionary.
Properties are properties OF things. If you are going to make
it a binary property, a property of TWO things, the equinumerosity
IS ALREADY the property in question.

You ALSO cannot do POSSIBILITIES around here.
EVERYthing just IS either true or false. You may not
know which, but that doesn't even matter.

The possibility that they are not equinumerous is "not-P."

The word you want INSTEAD of "property" is PROPOSITION.
I suppose you could think of a proposition as a property with
0 arguments, as a 0-ary or nullary property. The property of
equinumerosity is binary and is usually written infix. We would
be trying to prove N~E. You are defining P<=df=>N~E, which
entails -P<=df=>-(N~E).

Technically, not-P can have many values,

This IS JUST A DAMN LIE, Phool.
P is a proposition. It can have either of TWO values:
true or false. If P is true then -P is false, and conversely.

but for convenience, we will assume that not-P
means that set E has HALF as many elements as set N.

YOU CAN'T do that Phool: you have not DEFINED "half".
Even if you HAVE defined "half" FOR NATURAL NUMBERS,
which you in fact DID have to do, back when you were showing
that f was onto E, that STILL WON'T HELP *here*, because
THAT half was ONLY defined for ELEMENTS OF E.
You DO NOT KNOW whether the total NUMBER of elements
of N is or IS NOT an element of E! If that number is NOT
an element of E then that number DOES NOT *YET*, by YOUR
definitions, HAVE a "half"!

.