Cardinal of a finite set is unique

Question

I have been thinking about how I can prove whether a cardinal of a finite set is unique, after some thoughts I figured out it would be better to prove it in such a way,
If $n \in \mathbb{N}$, then there is no one-to-one mapping of $\mathbb{N}$ onto a proper subset $X \subset \mathbb{n} $
I tried to prove it by induction on $\mathbb{n}$ that for $\mathbb{n}=0$ it is trivial.
if we assume that it is true for $\mathbb{n}$ then we should prove it for $\mathbb{n+1}$ but I don't know how to go further.

Answer 1

Intuitively, the cardinality of a set is the "number of its elements". But what does it mean exactly? Are we allowed to talk of the cardinality of a set, at least when it is finite? Said differently, "cardinality" can be defined functionally, as a function defined over all finite sets? (Actually, this is your question). And what does it mean that a set is finite? As many basic questions, it is important to set clear definitions beforehand.

For every $n \in \mathbb{N}$, we set $\underline{n} = \{0, \dots, n-1 \}$; in particular, $\underline{0} = \emptyset$. (Actually, in a set-theoretic foundation of mathematics, this could be the definition of the elements of $\mathbb{N}$, so that $n = \underline{n}$; this is irrelevant with respect to your question, anyway for more details see here).

By definition, a set $S$ is finite if it is equipotent to some $\underline{n}$, i.e. if there is a bijection $f \colon S \to \underline{n}$ for some $n \in \mathbb{N}$.

In order to well define the notion of cardinality of a finite set $S$, we have to show beforehand that if $S$ is equipotent to $\underline{n}$ and to $\underline{m}$ (for some $m,n \in \mathbb{N}$), then $n = m$. The proof is simple: by hypothesis, there are bijections $f \colon S \to \underline{n}$ and $g \colon S \to \underline{m}$, hence $f^{-1} \circ g \colon \underline{n} \to \underline{m}$ is a bijection (because the inverse of a bijection is a bijection and the complosition of two bijections is a bijection). As suggested by @AsafKaraglia, using the pigeonhole principle, it is immediate to prove that if $\underline{n}$ and $\underline{m}$ are equipotent then $n = m$ (pedantically, you can prove it by induction on $m \in \mathbb{N}$).

Answer 2

It's easy to show bijections are composible and cardinality is transitive. That is if $\phi: A \to B$ is a bijection and $\psi: B \to C$ is a bijection then $\psi \circ \phi: A \to C$ is a bijection.

So if bijections from $\mathbb N_m = \{1,2, ..m\} \to A$ and from $\mathbb N_n = \{1,2,....n\}\to A$ exist that then bijections from $\mathbb N_m\to \mathbb N_n$ exist.

If $\phi:\mathbb N_m = \{1,2,.....m\} \to \mathbb N_n\{1,2,....n\}$ is a bijection then define $\psi: \mathbb N_{m-1}=\{1,2,...., m-1\} \to \mathbb N_{n-1} =\{1,2,.....,n-1\}$ via $\psi(k) = \phi(k)$ if $k < \phi^{-1}(n)$ and $\psi(k) = \phi(k+1)$ if $k\ge \phi^{-1}(n)$. It's easy to verify $\psi$ is a bijection.

By induction: we can repeat this $\min(m,n)$ to find a bijection (or an inverse of a bijection) between $\emptyset=\mathbb N_0$ and $\mathbb N_{|n-m|}$. But the only bijections to or from the empty set is from or to the empty set. So $\mathbb N_{|n-m|} = \emptyset$ and $n=m$.

So if a bijection exists between $\mathbb N_m \to A$ then $\mathbb N_m$ and $m$ are a uniquely determined and thus cardinality is well-defined and uniquely determined..

Answer 3

I think that is the definition of what a natural number is: We define an equivalence relation $\cong$ between sets by saying, two sets $A \cong B \iff \exists f \colon A \rightarrow B$, which is a bijection. This partitions the collection of finite sets into mutually disjoint classes, and we then choose a representative of each class as the cardinality of the sets in that class.

As a side remark, this is in fact a fundamental mechanism in all abstraction (or at least in most): you identify some sort of equivalence, and then throw out the details within the equivalence classes, so you can concentrate on what is left over. So, the 'uniqueness' of the cardinality, for example, follows from your decicion: sets with the same size are by definition (or choice, if you will) the same. It is the same principle that is in use when we talk about isomorphic groups, rings, ...: they are essentially the same, so we only need to study a single representative, and can choose to study whichever one feels most comfortable for our purpose.