Consecutive Subsets of N

Theorem

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

where $$\N_k = \left\{{0, 1, 2, 3, \ldots, k-1}\right\}$$ as defined here.

In particular, $$\N_{k-1} = \N_k - \left\{{k-1}\right\}$$

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

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


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

This follows as a direct application of Succeeding Set.