When K is a cardinal, 2^K is defined as the cardinality of the set of subsets of K (or alternatively, the cardinality of the set of functions from K into {0,1}, but it is easy to show that these are equivalent). Your “proof” of this is completely wrong. For any finite k, (aleph_0 choose k) = aleph_0, so the summation (k=0 to infinity) of (aleph_0 choose k) is (aleph_0)^2, which is also aleph_0. That’s because you only added together the subsets with finite complement, of which there are only countably many. Also, aleph_1 is not necessarily the same as 2^(aleph_0). I’m also not sure why you even brought this up.
I don’t think anyone has actually committed the fallacy that you describe, and when people say things like that, they mean something more like your alternative formulation (described more precisely by entirelyuseless).
When K is a cardinal, 2^K is defined as the cardinality of the set of subsets of K (or alternatively, the cardinality of the set of functions from K into {0,1}, but it is easy to show that these are equivalent). Your “proof” of this is completely wrong. For any finite k, (aleph_0 choose k) = aleph_0, so the summation (k=0 to infinity) of (aleph_0 choose k) is (aleph_0)^2, which is also aleph_0. That’s because you only added together the subsets with finite complement, of which there are only countably many. Also, aleph_1 is not necessarily the same as 2^(aleph_0). I’m also not sure why you even brought this up.
I don’t think anyone has actually committed the fallacy that you describe, and when people say things like that, they mean something more like your alternative formulation (described more precisely by entirelyuseless).