/sci/ - Science & Math

>>12431234
ok here's what i don't understand:
here's a paper of mochizuki and 4 of his friends, and supposedly there are more people working on iutt, let's say at least 20 of them
did those people also read the entirety of iutt and they believe it all works out?
or do they try to prove results using iutt as a black box?
or did they just jump on the bandwagon of a new memetheory and believe that mochizuki has his shit worked out?

>>11291201
i assume you meant A, B disjoint in (iii)
put A=B into (ii), you get that the function [math]\chi(A)[/math] can only have values 0 or 1
So you can write [math]f(A) := \{x \in X : \chi(A)(x) = 1\}[/math], then simply [math]\chi(A) = 1_{f(A)}[/math] (*). The four conditions can be now equivalently stated for f:
(i) [math]f(\emptyset) = \emptyset[/math],
(ii) [math]f(A \cap B) = f(A) \cap f(B)[/math],
(iii) [math]f(A), f(B)[/math] are disjoint and [math]f(A \cup B) = f(A) \cup f(B)[/math]
(iv) f injective
let [math]g: X \rightarrow P(X)[/math] be defined by [math]g(x) = f(\{x\})[/math] (**). So f and g are related by
[math]f(A) = \sum_{x \in A}g(a)[/math]. From (i-iv) it quickly follows that
(a) if [math]x \neq y[/math] then g(x) and g(y) are disjoint
(b) g(x) is nonempty
and vice versa, from any g satysfying (a-b) we obtain f using (*) and then chi using (**) which satisfies the original requirements

>>10990858
>>10990860
Let [math]G[/math] be a transitive subgroup of [math]S_n[/math] acts on [math]\{1 \dots n\}[/math], take any [math]x \in \{1 \dots n\}[/math].
Define [math]G_i := \{ g \in G : g(x) = i\}[/math] for i = 1...n.
For any i, j there is element h of G such that h(i) = j. Then the map [math]g \mapsto hg[/math] is an injection from [math]G_i[/math] to [math]G_j[/math].
Hence all [math]G_i[/math] have same size, so [math]|G| = n |G_1|[/math].