Consecutive Subsets of N

Theorem

 * $$\mathbb{N}_0 = \varnothing = \mathbb{N}^*_0$$
 * $$\mathbb{N}_k = \mathbb{N}_{k+1} - \left\{{k}\right\}$$

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

In particular, $$\mathbb{N}_{k-1} = \mathbb{N}_k - \left\{{k-1}\right\}$$

Proof

 * $$\mathbb{N}_0 = \varnothing = \mathbb{N}^*_0$$:

First we look at $$\mathbb{N}_0$$:

$$\mathbb{N}_0 = \left\{{n \in \mathbb{N}: n < 0}\right\}$$ from the [[Definition:N_k|definition of $$\mathbb{N}_k$$.

From the definition of Zero, $$0$$ is the minimal element of $$\mathbb{N}$$]].

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

Thus $$\mathbb{N}_0 = \varnothing$$.

Next we look at $$\mathbb{N}^*_0$$:

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

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

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

Thus $$\mathbb{N}^*_0 = \varnothing$$.


 * $$\mathbb{N}_k = \mathbb{N}_{k+1} - \left\{{k}\right\}$$:

This follows as a direct application of Succeeding Set.