Question

Use a cardinality argument to prove that there is no universal set for all sets

Answer 1

this should be good

Answer 2

This relies on the idea that the cardinality of a set A is strictly less than the cardinality of the power set of A, P(A).

Answer 3

So is it smth like this ?
\( card(U) \leq card(P(U)) \) but \( P(U) \subset U\)

Answer 4

so... that means that for the set of supposedly every possible element A we can construct a power set P(A) that has a higher cardinality, and therefore contains elements not in A

???
(just taking a stab at it)

Answer 5

P(U)⊂U ??? That's clearly false!

Answer 6

I was gonna say...

Answer 7

well its the universal subset

Answer 8

let me try again

Answer 9

but James just said that P(A) has a higher cardinality than A
how could P(U)⊂U ??

Answer 10

However, it is true that if U exists, then P(U) = U. Now you can obtain a contradiction.

Answer 11

Turing how wld u show that its a proper subset in latex

Answer 12

what do you mean? I'm not sure what symbol you are referring too

Answer 13

with that equal sign underneath

Answer 14

\subseteq\[\subseteq\]

Answer 15

ohhhh ok thanks

Answer 16

http://omega.albany.edu:8008/Symbols.html

Answer 17

for future reference
and welcome for my minor contribution :P

Answer 18

Thanks James