No Bijection between Finite Set and Proper Subset/Proof 2

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
 * The set $\N_n = \set {0, 1, \ldots, n - 1}$ 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
$\map P 1$ is the case:
 * The set $\N_1 = \set 0$ is equivalent to none of its proper subsets.

The only proper subset of $\set 0$ is the empty set $\O$, to which $\set 0$ is trivially not equivalent.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * The set $\N_k = \set {0, 1, \ldots, k - 1}$ is equivalent to none of its proper subsets.

Then we need to show:
 * The set $\N_{k + 1} = \set {0, 1, \ldots, k}$ is equivalent to none of its proper subsets.

Induction Step
Suppose that $\map P {k + 1}$ 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:
 * $\forall m \in \N_{k + 1}: \map {\paren {\phi \circ \phi} } m = \map \phi {\map \phi m}$

From Composite of Bijections is Bijection, $\phi \circ \phi$ is a bijection.

Now consider $T = \map {\paren {\phi \circ \phi} } {\N_{k + 1} }$, the image set of $\N_{k + 1}$ under $\phi \circ \phi$.

Let $m, t \in \N_{k + 1}$ such that:
 * $\map {\paren {\phi \circ \phi} } m = \map \phi t$

Since $\phi$ is an injection it follows that:
 * $\map \phi m = t$

As $S$ is a proper subset of $\N_{k + 1}$ it follows from Set Difference with Proper Subset that $\N_{k + 1} \setminus S \ne \O$.

Let $x \in \N_{k + 1} \setminus S$.

Suppose $\map \phi x \in T$.

Then that would mean that $\exists m \in \N_{k + 1}$ such that:
 * $\map {\paren {\phi \circ \phi} } m = \map \phi x$

which from above it would follow that:
 * $\map \phi m = x$

and so $x \in S$.

But this can not happen as $S \cap \paren {\N_{k + 1} \setminus S} = \O$ from Set Difference Intersection with Second Set is Empty Set.

So it must be the case that $\map \phi x \notin T$.

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: \map \psi a = \begin{cases}

a & : a \notin S = \map \phi {N_{k + 1} } \\ \map \phi a & : a \in S \end{cases}$

The next step is to show that $\psi$ is an injection.

Consider $a, b \in \N_k$ such that $\map \psi a = \map \psi b$.

Either $\map \psi a = \map \psi b \in \N_k \setminus S$, in which case:
 * $\map \psi a = a = \map \psi b = b$

or $\map \psi a = \map \psi b \in S$, in which case:
 * $\map \psi a = \map \phi a = \map \psi b = \map \phi b$

As $\phi$ is a injection it follows that $a = b$.

So $\map \psi a = \map \psi b \implies a = b$ and so by definition $\psi$ is an injection.

Now consider $\map \psi {\N_k}$.

We have that $\map \psi {\N_k} = \N_k \setminus \paren {S \setminus T}$ 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 $\map \phi x = k$, as $\phi: \N_{k + 1} \to S$ is a surjection.

There also exists $y \in \N_{k + 1} \setminus S$ as $\N_{k + 1} \setminus S \ne \O$ from Set Difference with Proper Subset.

Let $\phi': \N_{k + 1} \to S$ be defined as:
 * $\map {\phi'} a = \begin{cases}

\map \phi a : & 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:
 * $S\,' = \paren {S \setminus \set k} \cup \set y$

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 $\map P m$ for all $m \le k$ implies the proof of $\map P {k + 1}$.

The result follows by the Second Principle of Mathematical Induction.

Therefore:
 * For all $n \in \N$, the set $\N_n = \set {0, 1, \ldots, n - 1}$ of natural numbers less than $n$ is equivalent to none of its proper subsets.