What?
This argument seems way less convincing to me than the diagonalization argument, because as n is going to infinity that length does become 1.
Even using 1/2, the set that remains is nonempty due to the Cantor intersection theorem. The total length of the intervals is 1, which means that the remainder has no "interior" (i.e., contains no open interval), but the converse is not true: removing intervals whose lengths sum to less than one does not mean that the remainder will contain any interval. This is the consideration that allows you to create what are called "fat Cantor sets" -- the middle thirds Cantor set has Lebesgue measure zero, but by removing smaller intervals you can get other, homeomorphic sets that have positive measure.
I've never made peace with Cantor's diagonaliztion argument because listing real numbers on the right side (natural number lhs for the mapping) is giving a real number including transedentals that pre-bakes in a kind of undefined infinite.
Maybe it's the idea of a completed infinity that's my problem; maybe it's the fact I don't understand how to define (or forgot cauchy sequences in detail) an arbitrary real.
In short, if reals are a confusing you can only tie yourself up in knots using confusing.
Sigh - wish I could do better!
The diagonalization argument is an intuitive tool, IMHO. It is great if it convinces you, but it's difficult to make rigorous in a way that everyone accepts due to the use of a decimal expansion for every real number. One way to avoid that is to prove a little fact: the union of a finite number of intervals can be written as the finite union of disjoint intervals, and that the total length of those intervals is at most the total length of the original intervals. (Prove it by induction.)
THEOREM: [0, 1] is uncountable. Proof: By way of contradiction, let f be the surjection that shows [0, 1] is countable. Let U_i be the interval of length 1/2*i centered on f(i). The union V_n = U_1 + U_2 + ... + U_n has combined length 1 - 1/2*n < 1, so it can't contain [0, 1]. Another way to state that is that K_n = [0, 1] - V_n is non-empty. K_n also compact, as it's closed (complement of V_n) and bounded (subset of [0, 1]). By Cantor's intersection theorem, there is some x in all K_n, which means it's in [0,1] but none of the U_i; in particular, it can't be f(i) for any i. That contradicts our assumption that f is surjective.
Through the right lens, this is precisely the idea of the diagonalization argument, with our intervals of length 2*-n (centered at points in the sequence) replacing intervals replacing intervals of length 10*-n (not centered at points in the sequence) implicit in the "diagonal" construction.
> Because dwm is customized through editing its source code, it's pointless to make binary packages of it. This keeps its userbase small and elitist. No novices asking stupid questions.
...sucks less than what? :) Simple is good, but simpler does not necessarily mean better.
- Is this a "traditional" RPN calculator?
- Does it have bonus features, like symbolic processing?
- Is it programmable?
I believe outcomes would be better if kids used RPN calculators when learning, and programmable is definitely a plus.