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Lemma·L23

Uniqueness of cardinality

A set matches at most one natural number.

For every set and all : with from D033.
In words
For a set A and natural numbers m and n: if the set is equinumerous with m and also equinumerous with n, then m and n are the same number.
Never needed: F10 · F13 · A03 · A04 · A05 · A07 (computed from the citation graph, not asserted).

Proof

  1. 1
    Let and be bijections. Then is a bijection by L03, and is a bijection by L02 (iv).
  2. 2
    Suppose . By L16, either or . If : is in particular an injective function , which T12 forbids. If : is a bijection (L03), in particular injective, again forbidden by T12.
  3. 3
    Both cases are absurd (falsum), so (F04, excluded middle).

Remarks

This is what makes counting meaningful: however a finite set is enumerated, the same number comes out. Nothing analogous is being claimed for infinite sets here; that theory needs ordinals and choice. The lemma licenses the notation of D035.

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