Unless the probability of the terrible scenario (taken for example as an annual probability) is constantly decreasing, it will indeed happen almost surely.
Not true. For example, if the probability of the terrible scenario at some point drops to zero, that is sufficient.
What is the current probability of Nikita Khruschev ordering a nuclear strike on the United States? What was this probability 55 years ago and what happened to it?
The current probability is pretty low. It was higher before and dropped pretty dramatically in 1971. It is also currently getting lower since it is becoming more difficult to resurrect him. But it is nowhere near zero yet.
On the other hand, this may satisfy the condition of constantly getting lower, so it may never actually happen.
"summation of {i = 0} to n of (n combination i) = 2^n"
This is not a proof that "2^{aleph_0}" is the cardinality of the set of the subsets of natural numbers. You assume it works in the infinite cardinal case (without proving it), and then say that you thus proved it. You got confused by notation.
"I shall proffer a mathematical proof to show that for any infinite set of cardinality aleph_0 (the cardinality of the set of natural numbers) there are aleph_1 (2aleph_0) distinct infinite subsets."
No. 2^{aleph_0} is /by definition/ the cardinality of the set of the subsets of the natural numbers. It's named that way to allow the intuition of "summation of {i = 0} to n of (n combination i) = 2n" to work with cardinalities.
"aleph_1 (2^{aleph_0})"
aleph_1 = 2^{aleph_0} has been shown to be independent from ZFC. ie, if we haven't worked within inconsistent math for that past 60 years, what you just said is unprovable. You might have confused aleph and beth numbers.
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).
There is another fallacy: that an infinite path includes the infinite number of elements. However, we could imagine circular path, which has the finite number of elements. Eternal return by Nietzsche is a (wrong) example of such path.
It may help reject some false ideas about immortality and create the more positive image of it, without entrapping into black holes or need to remember infinite number of things.
It may help reject some false ideas about immortality and create the more positive image of it
I don't think that most people would think circular immortality is especially positive.
If we have linear and circular immortality, we could create morу complex constructions of pathways in the space of possible minds with merges, attractors, dead ends, two-dimensional figures etc. It completely dissolves fear of bad infinity associated with the naive understanding of immortality.
I don't see a convincing argument here. In fact, I don't see any argument at all, convincing or otherwise.
It is not a heuristic; it is a mathematical proof, given the condition I mentioned (namely that the probability does not diminish indefinitely. A circular path which does not contain the element has a probability of zero for the element, but we have no reason to expect circular paths.)