Lemma·L22
Equinumerosity is reflexive, symmetric, transitive
Same-size behaves like equality: every set matches itself, matches reverse, matches through.
For all sets
: (i)
; (ii)
; (iii)
, with
from D033.
In words
Every set is equinumerous with itself; equinumerosity read backwards still holds; and two sets equinumerous with a common third are equinumerous with each other.
Never needed: F08 · F10 · F11 · F12 · F13 · A03 · A04 · A05 · A06 · A07 · A08 (computed from the citation graph, not asserted).
Proof
- 1(i) The identity function is a bijection : it is injective, since means outright; and surjective, since each equals .
- 2
- 3(iii) If and are bijections, then is a bijection by L02 (iv).
∎
Remarks
The three laws are used constantly and silently: "let
; then also
" is (ii), and every chain of matchings is (iii). Note the subtlety of quantifying over all sets:
is a class relation, not a relation in the technical sense of a set of pairs, because its field would have to be the universal set forbidden by T02. Restricted to the subsets of any fixed set, it becomes an honest equivalence relation.
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