Proper Subset of Finite Ordinal is Equivalent to Smaller Ordinal

Theorem

Let $n$ be a finite ordinal.

Let $x \subsetneq n$.

Then for some finite ordinal $m < n$:

$m \sim x$

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

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

Proof

Proof by induction:

For all finite ordinals $n$, let $\map P n$ be the proposition:

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

$\map P 0$ is vacuously true, as there are no proper subsets of $0 = \O$.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ 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 finite ordinal $j \in k$ such that $j \notin x$.

Then we define a mapping $f$ on $x$ as:

$\forall i \in x: \map f i = \begin {cases} i & : i \ne k \\ j & : i = k \end {cases}$

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 $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.

Therefore:

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

$\blacksquare$

Also see

• No Bijection between Finite Set and Proper Subset

Sources

• 1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 13$: Arithmetic