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The same excerpt, starting from "One can also speak" describes how that "almost all" terminology is applied to integers (or, by extension, to any countable set,
by mynegation 12y ago
The same excerpt, starting from "One can also speak" describes how that "almost all" terminology is applied to integers (or, by extension, to any countable set, that is already measure zero). The example statement is "almost all prime numbers are odd", because there is a finite number of primes that are even (only one number 2).
Put another way, if we stick to the definition of "almost all" as elements not having the property forming measure zero set, then you do not need the video to prove the title: all integers are already measure zero. So, for subsets of integers, "almost all" means something else in jargon.
I could not watch the video, but I found (I believe) equivalent text. And "almost all" means that as we extend the reach the share of numbers not containing 3 goes to zero. It is mildly entertaining (albeit a bit too obvious) result, but my point was that usage of "almost all" is confusing because _I_ am used to a very specific meaning of it.
- drostie 12y agoIt's also a little confusing because there are sets which do not have a well-defined natural density among the integers, like: leastpow2 x = head $ dropWhile (\k -> 2*k <= x) $ map (2^) [0..] p k = k < 3 * (leastpow2 k `div` 2) limitless = [k | k <- [0..], k == 0 || p k] which oscillates between attaining 1/2 and then 2/3 for its partial densities.