My proof is not going to be very formal (and I do not need the countinuum hypothesis

) I am going to be working with a set of sets having single elements. This is what CaptainBlack proposed because I think it would be easier.

Let $\displaystyle \mathbb{Z^+}=\{1,2,3,...\}$ the natural numbers. Thus, $\displaystyle P(\mathbb{Z^+})$ is larger (the property of a power set). Thus, $\displaystyle \bigcup\{x\} ,x\in P(\mathbb{Z^+})$ having the cardinality greater than $\displaystyle \aleph_0$. But the previous demonstration is the the uninon of all sets having single elements. Further, by the properties of the power set we have that $\displaystyle P(P(\mathbb{Z^+}))$ is even larger. Thus, $\displaystyle \bigcup\{x\},x\in P(P(\mathbb{Z^+}))$ is the union of sets having a single element. This is even larger then the previous one. Thus, using this construction we can construct the cardinality of "the sets of all finite sets" to be as large as we like. Thus there is no such cardinal number. Now the conclusion I draw from this that some sets can be

**so** large that you cannot use cardinals numbers anymore, and thus there is no problem with this set.

I asked some people about this problem, and they have no idea. I hope someone finds a solution to my problem. Perhaps, the problem is that no one ever considered this set.

CaptainBlack, you mention a "well-defined set" I heard this term many times before and wanted to ask what it mean? Further I happen to know that the "Super-set" the set of all sets is not well defined and thus we cannot use it. Further, my set is the set of all FINITE sets and it itself is infinite, thus it is not an element of itself. I see no problem with my set.