TheoremBase

An Injective Self-Map of a Finite Set is a Bijection

Statement

Let N\mathbb{N} be the set of natural numbers with successor map SS as in that definition, let n∈Nn\in\mathbb{N}, and let [n][n] be the initial segment determined by nn. Call a map u:X→Yu:X\to Y between sets injective if u(x)=u(x′)u(x)=u(x') implies x=x′x=x' for all x,x′∈Xx,x'\in X. The notions finite and has nn elements are those of the indicated definitions.

Then the following hold.

1. (Initial segments) Every injective map u:[n]→[n]u:[n]\to[n] is a bijection, and hence a permutation of [n][n].

2. (Finite sets) If XX is a nonempty finite set and u:X→Xu:X\to X is injective, then uu is a bijection from XX onto XX.

Proofs

Log in to submit a proof.

Loading...

Citations

Loading…

Dependencies

Loading…

Related

0 relations

Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.

No relations recorded yet.

Comments

Log in to comment.

Loading…