Denote $\Bbb Z_0$ be the set of all non-negative integers.
Well Ordering Principle for $\Bbb Z_0$. Every non-empty subset $S$ of $\Bbb Z_0$ has a least element; that is, there exists $m \in S$ such that $m \le s$ for all $s \in S$.
Note: $0$ is a least element of $\Bbb Z_0$.
Proof: Let $\emptyset \ne S \subseteq \Bbb Z_0$ be arbitrary given. Let $k$ be an element of $S$. Then $k$ is a non-negative number. Define a subset $T$ of $S$ by $T=\{x \in S \mid x \le k\}$. Then $T \ne \emptyset$ and $T \subseteq \{0,1,2,\ldots,k\}$. Clearly, $T$ is a finite subset of $\Bbb Z_0$ and therefore it has a least element, say $m$. Then $0 \le m \le k$. Now, we must to show that $m$ is the least element of $S$. Indeed, let $s$ be any element of $S$. If $s>k$, then the inequality $m \le k$ implies $m<s$. If $s \le k$, then $s \in T$. Since $m$ being the least element of $T$, then $m \le s$. Hence, $m$ is the least element of $S$. This completes the proof. $\qquad \square$
Does this correct? I'm had a little bit confusing on the sentence "therefore it has a least element, say $m$". It seems obviously true, but I couldn't to prove it yet. Any ideas? Thanks in advanced.
That sentence is true, but it does need proof, and I would say one proves it by induction on $k$. So you don't get a free ride here, induction has to come in somewhere.
We must prove that for all $k \in \mathbb Z_0$, every nonempty subset of $\{0,...,k\}$ has a least element.
Basis step: if $k=0$ and $T \subseteq \{0\}$ is a nonempty subset, then $T = \{0\}$ and therefore $0$ is the least element of $T$.
Inductive step: suppose that every nonempty subset of $\{0,...,k-1\}$ has a least element. Now let $T \subset \{0,...,k-1,k\}$ be a nonempty subset. To prove that $T$ has a least element there are two cases.
Case 1: If the intersection $T \cap \{0,...,k-1\}$ is nonempty then by induction it has a least element $j$. It follows that $j$ is also the least element of $T$, because the only other possible element of $T$ is $k$ but $j < k$.
Case 2: If $T \cap \{0,...,k-1\}$ is empty then $T=\{k\}$ and so $k$ is its least element.