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If there are $\prod_{i=1}^{\infty} i$ elements in a set where $i$ is a natural number, does the cardinality of a set equal to $\aleph_0$?

Also, if there are $\Sigma_{i=1}^{\infty} 2^{i} $ elements in a set, what does the cardinality of a set equal to?

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The product has cardinality continuum. –  azarel Feb 15 '12 at 0:36
    
@azarel Thank you. I edited a question to add one more question. Can you also help me here? Thanks. –  user24996 Feb 15 '12 at 0:53
    
$\sum_{i<\aleph_0}2^i=\aleph_0$ if that is what you mean. –  azarel Feb 15 '12 at 0:56

2 Answers 2

First, it really ought to be written $\prod_{i=1}^\infty i$. As azarel said in the comments, this is $\mathfrak{c}$, the cardinality of $\mathbb{R}$ and of $\wp(\mathbb{N})$ (among many other things).

More generally, if $a_n$ is a positive integer for each $n\in\mathbb{N}$, there are two possibilities. Either only finitely many of the $a_n$ are greater than $1$, in which case $\prod_{n\in\mathbb{N}}a_n$ is really just the ordinary product of the finitely many $a_n>1$, or infinitely many of the $a_n$ are greater than $1$, in which case $\prod_{n\in\mathbb{N}}a_n=\mathfrak{c}$. No such product can be equal to $\aleph_0$.

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It is not totally obvious what you mean, but if you are asking whether the cardinality of the permutations of the natural numbers (in a sense $\aleph_0 !$) is $\aleph_0$, then no. It is $2^{\aleph_0}$.

It is at least that much since one example of a subset of the permutations of the natural numbers are those where they are divided into pairs where each pair can swapped or not.

It is no more since the permutations of the natural numbers is a subset of the set of countably infinite sequences of natural numbers, which has cardinality $\aleph_0 ^ {\aleph_0} $. But $\aleph_0^{\aleph_0} \le (2^{\aleph_0})^{\aleph_0} = 2^{{\aleph_0}^2} = 2^{\aleph_0}.$

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