Prove that a set is infinite if and only if it is equipotent to a proper subset.

Question

Prove that a set $X$ is infinite if and only if it is equipotent to a proper subset $H$.

Proof.

Suppose $X$ is equipotent to a proper subset $H$. Then there exists a bijection $f: H \to X$. Toward a contradiction, suppose $X$ is finite.

Since $X$ is finite, $H$ is finite. Moreover, we can write

$$X = \{x_1, x_2, \dots, x_n\} \hspace{1cm} H = \{h_1, h_2, \dots, h_m\}$$

Since $f$ is injective, $h_j \neq h_k$ implies $f(h_j) \neq f(h_k)$. But recall that $H$ is proper, so $\exists \, x_i \in X \setminus H$.

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But at this point, I get stuck. Intuitively, I understand that this can not be a bijection, but I'm having trouble showing this.

I have already read Proof that a set is infinite if and only if it has an infinite proper subset and this post does not provide an actual proof answer or assumes that the set is countable.

Answer 1

We want to prove that:
(1) X is equipotent to a proper subset H if and only if (2) X is an infinite set.

Note: X is equipotent to H $\begin{smallmatrix} def \\ \iff \\ \space \end{smallmatrix}$ there is a bijection between X and H $\begin{smallmatrix} def \\ \iff \\ \space \end{smallmatrix}$ |X|=|H| (they have the same cardinality)

H is a proper subset of X $\begin{smallmatrix} def \\ \iff \\ \space \end{smallmatrix}$ H $\subsetneq$ X.

Proof

(1)$\Rightarrow$(2) Assume by contradiction that X is finite. Since there exist a bijection $f:X\rightarrow H$ then by definition of cardinality $|X|=|H|$. But H is a subset of X which is proper so $\forall h \in H : h \in X$ but $\exists x \in X : x \notin H$. Hence H has at least one element less than X, so $|H|<|X|$ - contradiction.

(2)$\Rightarrow$(1) To prove this direction in (ZF) we need the condition "X is a Dedekind-infinite set" instead of just an infinite set. However, we can prove it in (ZF + ACw) applying the axiom of countable choice:

In wikipedia you can find the proof that every infinite set is Dedekind-infinite in (ZF + ACw).

Answer 2

I am going to assume that the definition of a finite set is a set equipotent to $\{1 , \cdots ,n\}$ for some natural number $n$, because that seems to be the one that is being used in the question.

Notation: We denote $[|1,n|]$ the set of integers between $1$ and $n$.

Proposition: Let $n \in \mathbb{N}$, $A \subseteq [|1,n|]$ and $f: [|1,n|] \longrightarrow A$ and injective map. Then $A=[|1,n|]$ and $f$ is a bijection.

We will prove this statement by induction on $n$.

For $n=0$, if $A$ is subset of an empty set, then $A$ has to be equal to the empty set and there is only one map from the empty set to itself and it is a bijection.

Now assume the statement true for $n\in \mathbb{N}$. Let us prove it is true for $n+1$. Take an injective map $f$ from $[|1,n+1|]$ to $A\subseteq [|1,n+1|]$.

By contradiction, assume that $k \not \in A$ for some $k \in [|1,n+1|]$. Let $\sigma$ denote the transposition $(k \ n+1)$, that is to say the permutation on $[|1,n+1|]$ that swaps $k$ and $n+1$ and fixes everything else. Denote $A'= \sigma (A)$. The map $\sigma \circ f$ restricted to $[|1,n|]$ then is an injective map from $[|1,n|]$ to $A'$, so by assumption $A'=[|1,n|]$ and $\sigma \circ f$ is a bijection from $[|1,n|]$ to $A'$.

That means that $\sigma \circ f(n+1)=n+1$, since $\sigma \circ f$ is injective. That means that $f(n+1)=k$ , which is a contradiction.

Now let us prove that $f$ is surjective. By contradiction if there exists a $k \in [|1,n+1|]$, such that $\forall x \in [|1,n+1|], f(x) \neq k$, then $f$ is an injective map from $[|1,n+1|]$ to $[|1,n+1|] \setminus \{k\}$, which we have already shown cannot be the case.

Proposition: If a set $X$ is equipotent to a proper subset, then it is not finite.

By contradiction assume that $X$ if finite and that there exists $Y \subseteq X$ a proper subset and $g$ a bijection from $X$ to $Y$. Let us denote $\phi$ a bijection from $X$ to $[|1,n|]$ for some $n \in \mathbb{N}$. The map $\phi \circ g \circ \phi^-1$ is then an injective map from $[|1,n|]$ to $\phi (Y) \subseteq [|1,n|]$, therefore by the previous proposition: $\phi (Y)=[|1,n|]$. By applying $\phi^{-1}$ to both sides of this equation, we get that $Y=X$, which is a contradiction.