Proof by induction: For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition: :The set $\N_n = \left\{{0, 1, \ldots, n-1}\right\}$ of natural numbers less than $n$ is equivalent to none of its proper subsets. Here we use the definition of set equivalence to mean that $S$ is equivalent to $T$ if there exists a bijection between them. === Basis for the Induction === $P \left({1}\right)$ is the case: :The set $\N_1 = \left\{{0}\right\}$ is equivalent to none of its proper subsets. The only proper subset of $\left\{{0}\right\}$ is the empty set $\varnothing$, to which $\left\{{0}\right\}$ is trivially not equivalent. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :The set $\N_k = \left\{{0, 1, \ldots, k-1}\right\}$ is equivalent to none of its proper subsets. Then we need to show: :The set $\N_{k+1} = \left\{{0, 1, \ldots, k}\right\}$ is equivalent to none of its proper subsets. === Induction Step === Suppose that $P \left({k+1}\right)$ is not true. Let $\phi: N_{k+1} \to S$ be a bijection where $S \subsetneq N_{k+1}$. We aim to demonstrate that the existence of such a $\phi$ contradicts the induction hypothesis. We consider two cases: ;Case 1. Let $S \subseteq N_k$. Then we define the composite mapping $\phi \circ \phi$ defined as: :$\forall m \in \N_{k+1}: \phi \circ \phi \left({m}\right) = \phi \left({\phi \left({m}\right)}\right)$ From Composite of Bijections is Bijection, $\phi \circ \phi$ is a bijection. Now consider $T = \phi \circ \phi \left({\N_{k+1}}\right)$, the image set of $\N_{k+1}$ under $\phi \circ \phi$. Let $m, t \in \N_{k+1}$ such that: :$\phi \circ \phi \left({m}\right) = \phi \left({t}\right)$ Since $\phi$ is an injection it follows that: :$\phi \left({m}\right) = t$ As $S$ is a proper subset of $\N_{k+1}$ it follows from Set Difference with Proper Subset that $\N_{k+1} \setminus S \ne \varnothing$. Let $x \in \N_{k+1} \setminus S$. Suppose $\phi \left({x}\right) \in T$. Then that would mean that $\exists m \in \N_{k+1}$ such that: : $\phi \circ \phi \left({m}\right) = \phi \left({x}\right)$ which from above it would follow that: :$\phi \left({m}\right) = x$ and so $x \in S$. But this can not happen as $S \cap \left({\N_{k+1} \setminus S}\right) = \varnothing$ from Set Difference Intersection with Second Set is Empty Set. So it must be the case that $\phi \left({x}\right) \notin T$. Thus it can be concluded that $\phi \circ \phi$ is a mapping from $\N_{k+1}$ onto a proper subset $T$ of $S$. Now, consider the mapping $\psi: \N_k \to U$ where $U$ is a proper subset of $\N_k$. We define $\psi$ as follows: :$\forall a \in \N_k: \psi \left({a}\right) = \begin{cases} a & : a \notin S = \phi \left({N_{k+1}}\right) \\ \phi \left({a}\right) & : a \in S \end{cases}$ The next step is to show that $\psi$ is an injection. Consider $a, b \in \N_k$ such that $\psi \left({a}\right) = \psi \left({b}\right)$. Either $\psi \left({a}\right) = \psi \left({b}\right) \in \N_k \setminus S$, in which case: :$\psi \left({a}\right) = a = \psi \left({b}\right) = b$ or $\psi \left({a}\right) = \psi \left({b}\right) \in S$, in which case: :$\psi \left({a}\right) = \phi \left({a}\right) = \psi \left({b}\right) = \phi \left({b}\right)$ As $\phi$ is a injection it follows that $a = b$. So $\psi \left({a}\right) = \psi \left({b}\right) \implies a = b$ and so by definition $\psi$ is an injection. Now consider $\psi \left({\N_k}\right)$. We have that $\psi \left({\N_k}\right) = \N_k \setminus \left({S \setminus T}\right)$ so that $\psi$ is an injection from $\N_k$ to a proper subset of itself. This contradicts the induction hypothesis. ;Case 2. Let $S \not \subseteq N_k$. This means that $k \in S$. There exists $x \in \N_{k+1}$ such that $\phi \left({x}\right) = k$, as $\phi: \N_{k+1} \to S$ is a surjection. There also exists $y \in \N_{k+1} \setminus S$ as $\N_{k+1} \setminus S \ne \varnothing$ from Set Difference with Proper Subset. Let $\phi': \N_{k+1} \to S$ be defined as: :$\phi' \left({a}\right) = \begin{cases} \phi \left({a}\right) : & a \ne x \\ y : & a = x \end{cases}$ As $\phi$ is a bijection it follows that $\phi': \N_{k+1} \to S\,'$ is also a bijection, where: :$S\,' = \left({S \setminus \left\{{k}\right\}}\right) \cup \left\{{y}\right\}$ Proceeding as Case 1, we prove the same contradiction in the induction hypothesis. Thus by Proof by Contradiction, the assumption that $\phi: N_{k+1} \to S$ can be a bijection must be false. So the truth of $P \left({m}\right)$ for all $m \le k$ implies the proof of $P \left({k+1}\right)$. The result follows by the Second Principle of Mathematical Induction. Therefore: :For all $n \in \N$, the set $\N_n = \left\{{0, 1, \ldots, n-1}\right\}$ of natural numbers less than $n$ is equivalent to none of its proper subsets. {{qed}}