Mapping whose Image of Natural Number n is Subset of Image of Successor/Corollary

Theorem
Let $f: \N \to A$ be a mapping from the set of natural numbers $\N$ to a class $A$. Let $f$ have the property that:


 * $\forall n \in \N: \map f n \subsetneq \map f {n^+}$

where $n^+$ is the successor of $n$.

Then:
 * $\forall n, m \in N: n < m \implies \map f n \subsetneq \map f m$

Proof
The proof will proceed by the Principle of Finite Induction on $\N$.

Let $S$ be the set defined as:
 * $S := \set {m \in \N: \forall n < m: \map f n \subsetneq \map f m}$

That is, $S$ is to be the set of all $n$ such that:
 * $\forall n < m: \map f n \subsetneq \map f m$

First we note that:
 * $\not \exists n \in \N: n < 0$

Thus $0 \in S$ vacuously.

Basis for the Induction
Consider the natural number $1$.

There is only one $n \in \N$ such that $n < 1$, and that is:
 * $n = 0$

By definition of $f$:


 * $\map f 0 \subseteq \map f {0^+}$

But by definition of the set of natural numbers:
 * $0^+ = 1$

So $1 \in S$.

This is the basis for the induction.

Induction Hypothesis
It is to be shown that if $k \in S$ where $k \ge 1$, then it follows that $k + 1 \in S$.

This is the induction hypothesis:
 * $\forall n < k: \map f n \subsetneq \map f k$

It is to be demonstrated that it follows that:
 * $\forall n < k^+: \map f n \subsetneq \map f {k^+}$

Induction Step
This is the induction step:

Let $k \in S$.

From the induction hypothesis:
 * $\forall n < k: \map f n \subsetneq \map f k$

Then from the definition of $f$:
 * $\map f k \subsetneq \map f {k^+}$

That is:
 * $\forall n \le k: \map f n \subsetneq \map f {k^+}$

That is:
 * $\forall n < k^+: \map f n \subsetneq \map f {k^+}$

So $k \in S \implies k^+ \in S$.

The result follows by the Principle of Finite Induction:


 * $\forall n, m \in N: n \le m \implies \map f n \subsetneq \map f m$