A Conversation for Bigger and Bigger Infinities
Dum and Dumber
Pirate Alexander LeGray Started conversation Feb 18, 2008
I'm sorry but on reflection I was naive in my previous response {}.
Adden-DUM
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.
Key: Complain about this post
Dum and Dumber
More Conversations for Bigger and Bigger Infinities
Write an Entry
"The Hitchhiker's Guide to the Galaxy is a wholly remarkable book. It has been compiled and recompiled many times and under many different editorships. It contains contributions from countless numbers of travellers and researchers."