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This is question 6.17 from Mathematical Proofs by Chartrand/Polimeni/Zhang.

Prove: For each integer $m$, the set $S = \{i ∈ Z : i ≥ m\}$ is well-ordered.

The given proof in the back of the book states:

Proof: We need only show that every nonempty subset of $S$ has a least element. So let $T$ be a nonempty subset of $S$. If $T$ is a subset of $N$, then, by the Well-Ordering Principle, $T$ has a least element. Hence we may assume that $T$ is not a subset of $N$. Thus $T − N$ is a finite nonempty set and so contains a least element $t$. Since $t ≤ 0$, it follows that $t ≤ x, \forall x ∈ T$ ; so $t$ is a least element of $T$.

Here's where I get lost: I get that $T - N$ is a finite, nonempty set. I do not get why this implies it must contain a least element $t$. Is it because of the fact it is finite? Are we using the knowledge that, while $Z$ does not contain a least element, any finite subset of $Z$ does contain a least element?

Also: I get why this $t \in T - N \leq 0$, but I don't get why we then know $t \leq x, \forall x \in T$. Is it because we've have concluded that this element $t$ is the least element of $T - N$? I feel like I'm on the verge of getting it, but it isn't quite all adding up.

chrisT
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  • Yes, indeed a finite set of integers contains a least element. That's clear, isn't it? – Dietrich Burde May 23 '20 at 19:00
  • This is not a duplicate "per se", but the actual question is the same. One more reason why the question should be the question, rather than the context. (I'm not complaining about the set up, well done on explaining the issue.) – Asaf Karagila May 23 '20 at 19:31
  • @AsafKaragila Yes, thank you; I should have made the question the actual question, not the context. I will keep that in mind for future questions. Thanks again! – chrisT May 23 '20 at 19:44

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