Proper Subset of Finite Ordinal is Equivalent to Smaller Ordinal

Theorem
Let $\N$ be the natural numbers, defined as the minimal infinite successor set $\omega$.

Let $n \in \N$.

Let $x \subseteq n$.

Then:
 * $\exists m \in \N: m < n: m \sim x$

where $m \sim x$ denotes that $m$ is (set) equivalent to $x$.

That is, every proper subset of a natural number $n$ is equivalent to some natural number smaller than $n$.

Proof
Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $x \subsetneq n \implies \exists m \in \N: m < n: m \sim x$

$P \left({0}\right)$ is vacuously true, as there are no proper subsets of $0 = \varnothing$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 0$, then it logically follows that $P \left({k^+}\right)$ is true.

So this is our induction hypothesis:
 * $x \subsetneq k \implies \exists m \in \N: m < k: m \sim x$

Then we need to show:
 * $x \subsetneq k^+ \implies \exists m \in \N: m < k^+: m \sim x$

Induction Step
This is our induction step:

Let $x \subsetneq k^+$.

Then either:
 * $(1) \quad x \subsetneq k$, in which case the induction hypothesis applies

or:
 * $(2) \quad x = k$, in which case the result is trivially true

or:
 * $(3) \quad k \in x$.

In case $(3)$, we find a number $j \in k$ such that $j \notin x$.

Then we define a mapping $f$ on $x$ as:
 * $f \left({i}\right) = \begin{cases}

i & : i \ne k \\ j & : i = k \end{cases}$

Clearly $f$ is injective and $f$ maps $x$ into $k$.

So the image of $x$ under $f$ is either equal to $k$ or by the induction hypothesis equivalent to some element of $k$.

Consequently, $x$ is always equivalent to some element of $k$.

So $P \left({k}\right) \implies P \left({k^+}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N: x \subsetneq n \implies \exists m \in \N: m < n: m \sim x$

Also see

 * Proper Subset of Finite Set No Bijection