Lemma·L27
Counting a partition of equal classes
k boxes with m things each: k times m things in total.
Let
be a partition of a set
, and
. If
and every
satisfies
, then
so
is finite with
, with
from D033.
In words
For natural numbers k and m: if a set is split into k classes and every class has exactly m elements, then the whole set is equinumerous with k times m, so it is finite with k times m elements.
Never needed: F10 · F13 · A03 · A04 · A05 · A07 (computed from the citation graph, not asserted).
Proof
- 1We induct on , quantifying over everything else: let be the set of such that for every set and partition of with and all classes , one has (Separation; is fixed throughout).
- 2
- 3
- 4: the restriction is injective and its image is : every equals for exactly one (bijectivity), and iff iff (D006: and leave ). So is a bijection , and its inverse is a bijection (L03): .
- 5is a partition of : its members are non-empty and pairwise disjoint because those of are (D023), and by definition of .
- 6, disjointly: every lies in some class (covering); if then , otherwise and ; conversely and . So by Extensionality. Disjointness: if , then for some , and with , contradicting pairwise disjointness of (D008, negation).
- 7
∎
Remarks
Multiplication as repeated addition, upgraded to sets:
boxes of
marbles hold
marbles. The statement quantifies over all pairs
inside the induction: peeling off one class changes both the set and the partition, so the induction hypothesis must be portable. This is the counting engine of Lagrange's theorem, where the boxes are the cosets of a subgroup, all of size
by L30.