Originally Posted by

**ynj** well....you can prove it like this..

we can define the exponation of cardinals like that:

if $\displaystyle k_1,k_2$are cardinals of two sets$\displaystyle A,B$, then $\displaystyle k_1^{k_2}$is the cardinal of the set of all functions mappings $\displaystyle B$to $\displaystyle A$.

Let the cardinal of the natural numbers to be $\displaystyle \aleph_0$, and the cardinal of real numbers to be $\displaystyle \aleph$.

Then let $\displaystyle S$ be set of the all finite subsets of a continuum.

For every$\displaystyle x\in S$,since $\displaystyle x$is finite, there is a bijection $\displaystyle f_x$ from$\displaystyle n$(meaning the set $\displaystyle \{0,...,n-1\}$)to $\displaystyle x$, thus every $\displaystyle y\in x$can be written as $\displaystyle f_x(m)$. But we can also see $\displaystyle f_x$ as mapping from $\displaystyle \mathbb{N}$ to $\displaystyle x$, letting all other natural number map to a strange element $\displaystyle a$different from all real number. Thus now we have mapped all $\displaystyle x\in S$ to some function from $\displaystyle \mathbb{N}$ to$\displaystyle \mathbb{R}\cup \{a\}$(which have the same cardinality as $\displaystyle \mathbb{R}$).And clearly this mapping is one-to-one.

Clearly we have $\displaystyle S\succeq \aleph$. From above we also have $\displaystyle S\preceq\aleph^{\aleph_0}$.

But $\displaystyle S\preceq\aleph^{\aleph_0}=(2^{\aleph_0})^{\aleph_0 }=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=\aleph$

So $\displaystyle S\sim\aleph$, it have the cardinality of a continuum.