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

G. Frege wrote:

On Thu, 12 Apr 2007 14:57:59 -0500, Phil <toob-headman@xxxxxxxxxxxxx> wrote:


N = {1, 2, 3, 4, ...}

E = {2, 4, 6, 8, ...}

have the same cardinality.



Here's a simple "proof" for that fact.

Let's start with a motivating "picture":

{1, 2, 3, 4, ...}
| | | |
{2*1, 2*2, 3*2, 4*2, ...}

With other words, the function

f: N --> E
n |-> 2*n

is a bijection between N and E (easy!). Hence E and N are equipollent (by
definition), or with other words, have the same cardinality. qed.


F.

Yes, but I need a bit more detail, if you can, especially if this is somehow different from my proof. You have a nice "bijection picture;" do you simply ASSUME that this bijection can be formed with sets that are already defined as specific infinite sets? (This issue is why I decided to first "partially define" E as a set of even positive integers with an unknown number of elements, THEN define it as having a bijection with N, thereby automatically "giving it" the correct number of elements.) Do you ASSUME that the set of integers is the set of ALL positive integers? Do you ASSUME that the set of even integers belongs entirely to N? What you have given is more of a shorthand reference TO the proof, as far as I can see.

Let me ask one question that can trump all of the above questions: do you disagree with the proof I gave, or think that we need to add to it or change it in some fundamental way, or should I proceed?

Thanks,
Phil
.

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