A Conversation for Bigger and Bigger Infinities

Dum and Dumber

Post 1

Pirate Alexander LeGray

I'm sorry but on reflection I was naive in my previous response {}.
I don't like this font,
Let the empty set be 'o',
It could be said that 'o' is a member of every set.(Can't be bothered to look up the axioms of set theory.)

If we let your X be a set of sets; {o} belongs to X:
Then S exists because f({o})=a, and if a={o} it is a set.
(The only problem I had was with the possible S={o} because this looks empty.)

Now, N can be constructed from {o} using the successor function. (I don't want to look this up either.)

So to another problem:

The above set N has a bijection between it and the natural numbers we use when we go shopping, but it aint the same.

Given Q is countable, it is reasonable to say Q<P(X); but from that you can't deduce R=P(X); because R is a complete field and P(X) is a set without operations ect, and you can't even say that it has the same number of elements.

I'm sure your right and it is, its just that a little more work is required to prove it.

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