Consecutive Subsets of N

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Theorem

Let $\N_k$ denote the initial segment of the natural numbers determined by $k$:

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


Then:

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


In particular:

$\N_{k - 1} = \N_k \setminus \left\{{k - 1}\right\}$


Proof

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

$\blacksquare$