Real Analysis, using Zorns Lemma

Question

Use Zorns lemma to prove that every finite totally ordered set has a unique minimal element.

I know how to use induction to prove it but I am asked to use Zorn's lemma.

what I have thus far:
Let $X$ be a finite totally ordered set with elements $x_1,...,x_n$. Let's arbitrarily pick an element $x_{1}$ to be the minimal element of $X$ itself, then an arbitrary element $x_{2}$ to be the minimal element of $X \setminus \{x_1\}$, and so on. I am just stuck on how I would incorporate Zorns lemma to prove that there exists a unique minimal element.

Please edit your question to include a description of what you've tried and where you're stuck. MSE is not intended as a place to simply post homework exercises. — Barry Cipra, Feb 13 '15 at 17:15
I think it would be easier to prove it by induction. Like here http://math.stackexchange.com/questions/24996/question-related-with-partial-order-finite-set-minimal-element or here http://math.stackexchange.com/questions/548806/a-finite-set-always-has-a-maximum-and-a-minimum — Martin Sleziak, Feb 13 '15 at 18:31
Wolfgang: What is a source of this problem. Was it phrased this way in some book? Is it from lecture you are attending? — Martin Sleziak, Feb 13 '15 at 20:01
Yes, I am aware of proving this by induction but I have to use Zorns lemma to prove it. — justanewb, Feb 14 '15 at 20:57
My professor, I know how to use induction to prove it but not with Zorns lemma — justanewb, Feb 15 '15 at 12:54

score 0 · Accepted Answer · edited Feb 15 '15 at 19:41

0

I am not sure if this is correct but I just attempted to prove this by contradiction:

Let $(T,\le)$ be a finite totally ordered set. Then $T$ has at least one unique minimal element. Suppose that $T$ has no minimal element. Then for each $t\in T$ there is a nonempty subset of $T$, $$S_{t} = \{u\in T:t \le u\}.$$ The axiom of choice says that consequently there exists a function $f:T\rightarrow T$ such that $f(t)\in S_t$ for all $t\in T$. That is, $$f(t) \ge t\text{ for all }t\in T.$$ This contradicts the fundamental lemma

As I've said not sure if this is right I didn't use Zorn's lemma but the axiom of choice which is equivalent so maybe that is ok?

edited Feb 15 '15 at 19:41

Martin Sleziak

53,687

answered Feb 15 '15 at 13:11

justanewb

1,179

What do you mean by "the fundamental lemma". – Martin Sleziak Feb 15 '15 at 19:45
BTW if $t$ is not minimal, it means that ${u\in T; u<t}$ is non-empty (not $u\ge t$). – Martin Sleziak Feb 15 '15 at 19:55
Fundamental Lemma as in Zorns lemma – justanewb Feb 15 '15 at 20:53

Real Analysis, using Zorns Lemma

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