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**SlipEternal** $\displaystyle A$ is not a valid element of $\displaystyle I$. Counting the number of possible sets for $\displaystyle A$ has absolutely no relationship whatsoever to the number of sets in $\displaystyle I$.

Here is why:

Assume $\displaystyle n_p=1$ for all $\displaystyle p$. Thus, $\displaystyle 2,3,5,7,11,... \in A$. However, because $\displaystyle 10=2+3+5 \notin A, f^{-1}(A)=\emptyset, card(\emptyset)=0<1$. Therefore, $\displaystyle A \notin I$. So, you just counted an uncountable number of sets, none of which are elements of $\displaystyle I$. Examples of infinite sets in $\displaystyle I$ are $\displaystyle \mathbb{N}$ where $\displaystyle f(\mathbb{N})=\mathbb{N}, f(\{2^n|n=0,1,2,...\})=\mathbb{N}$. Therefore, $\displaystyle I\text{ is nonempty}$. And, by the reason given in my last post, it is actually countable, as I indicated. The reason is subtle. You must first assume that you have an actual element of $\displaystyle I$ before trying to show that the number of possible elements of $\displaystyle I$ are uncountable. I assume that I have one such infinite set.

So, here's an examples. Let's assume that I have one of these infinite sets, $\displaystyle X \in I$. Assume that for two of its preimages, one contains the point $\displaystyle x$ and the other does not. (If each had equivalent containments, then they would be the same set, and therefore, contradict the notion that $\displaystyle X$ had multiple preimages, therefore some $\displaystyle x$ may be found. Let the minimal two elements of $\displaystyle X$ be $\displaystyle x_1,x_2$. The function $\displaystyle f$ guarantees that $\displaystyle x_1+x, x_2+x, x_1+x_2+x \in X$. However, one of the preimages does not include $\displaystyle x$, which implies that there must exist some other subsets whose sums are equal to those three elements. Keep in mind that we are dealing with some subset of natural numbers, so our sums only get larger. Not only that, but we are dealing with infinite sets, so we have an infinite number of finite subsets to sum up, and all of these sums are in $\displaystyle X$. When calculating these sums, it is worthwhile to calculate them in order of how large they are. $\displaystyle x_1+x_2$ are by definition the smallest elements of $\displaystyle X$. Therefore, they cannot be the sum of any other elements. If they are in $\displaystyle X$, it is because they are in EVERY preimage of $\displaystyle X$. If your numbers are too far apart, then $\displaystyle X$ will have a single distinct preimage.

For example: $\displaystyle A=\{3^n|n=0,1,2,3,...\}$. Its elements look like this: $\displaystyle \{1,3,9,27,...\}$.

$\displaystyle f(A)=\{1,3,4,9,10,12,13,27,28,30,31,36,37,39,40,.. .\}$. Now, for any preimage of $\displaystyle f(A)$, it must contain $\displaystyle 1,3$. The only way for them to be in $\displaystyle f(A)$ is if their preimage contains them. If $\displaystyle 4 \in f^{-1}(f(A))$, then $\displaystyle 1+4=5 \in f(f^{-1}(f(A)))$, which is not in $\displaystyle f(A)$, so 4 cannot be in its preimage. Continuing on, you can discover that the ONLY set that can be its preimage is $\displaystyle A$ itself. Therefore, the cardinality of its preimage is 1. This is NOT a valid set in $\displaystyle I$.

So, my construction of beginning with a set in $\displaystyle I$ is NOT assuming a countable set in the first place. It assumes that I have an arbitrary element of $\displaystyle I$ and makes use of the properties of what it means to be in $\displaystyle I$ to arrive at the conclusion that $\displaystyle I$ must be countable.