Well Ordering Principle Proof without mathematical induction with different approach

Question

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.

Related: https://matheducators.stackexchange.com/questions/10021/why-are-induction-proofs-so-challenging-for-students/10057#10057 — Ethan Bolker, Mar 29 '23 at 14:07
Among many related posts: https://math.stackexchange.com/questions/4517260 https://math.stackexchange.com/questions/1532738 https://math.stackexchange.com/questions/2932074 — Anne Bauval, Mar 29 '23 at 14:10
Does this answer your question? How do you prove Well-Ordering without Mathematical Induction? (and vice-versa) It gives a simpler proof (based on the same tool: induction) than the answer below. — Anne Bauval, Mar 29 '23 at 14:17
Unfortunately, they are not answer my question. So, why closed??? — math404, Mar 29 '23 at 14:58

score 1 · Answer 1 · answered Mar 29 '23 at 14:05

1

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.

answered Mar 29 '23 at 14:05

Lee Mosher

135,265

Here are direct proofs that (on natural numbers) induction $\implies$ well ordering. Without introducing $k$ and $T$ like the OP did. I think it might be considered as a duplicate. – Anne Bauval Mar 29 '23 at 14:15
Any method without using induction? Like contradiction? – math404 Mar 29 '23 at 14:57
I mean, suppose that $T$ do not have a least element. Then, for any $t \in T$, there is an $s \in T$ such that $t>s$. But, $s < t \le n$. I got stucked here. – math404 Mar 29 '23 at 15:33
Since you were stuck, I showed you how how to make your method work by using induction. There are other proofs that use induction by other methods, for example at the link provided by @AnneBauval. But no, you cannot avoid induction on this problem. – Lee Mosher Mar 29 '23 at 16:22

Well Ordering Principle Proof without mathematical induction with different approach

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