Consecutive Subsets of N

Theorem

 * $\N_0 = \varnothing = \N^*_0$
 * $\N_k = \N_{k+1} \setminus \left\{{k}\right\}$

where:
 * $\N_k = \left\{{0, 1, 2, 3, \ldots, k-1}\right\}$

defined as a initial segment of the natural numbers.

In particular:
 * $\N_{k-1} = \N_k \setminus \left\{{k-1}\right\}$

Proof

 * $\N_0 = \varnothing = \N^*_0$:

First we look at $\N_0$:

$\N_0 = \left\{{n \in \N: n < 0}\right\}$ from the definition of $\N_k$.

From the definition of zero, $0$ is the minimal element of $\N$.

So there is no element $n$ of $\N$ such that $n < 0$.

Thus $\N_0 = \varnothing$.

Next we look at $\N^*_0$:

$\N^*_0 = \left\{{n \in \N^*: n \le 0}\right\}$ from the definition of $\N^*_k$.

From the definition of One, the minimal element of $\N^*_0$ is $1$.

From Zero Strictly Precedes One we know that $0 < 1$, so there is no element of $n$ of $\N^*_0$ such that $n \le 0$.

Thus $\N^*_0 = \varnothing$.


 * $\N_k = \N_{k+1} \setminus \left\{{k}\right\}$:

This follows as a direct application of Strict Lower Closure of Sum with One.