MHB Is the Axiom of Choice Necessary to Well-Order Finite Sets?

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The discussion focuses on demonstrating that every finite set can be well-ordered without the need for the Axiom of Choice. The approach suggested involves using induction on the cardinality of the set, starting with a well-ordered set of size k and extending it to k+1. An ordering is defined for k+1 that incorporates the ordering from k, ensuring that the additional element is greater than all elements in k. It is established that any nonempty subset of k+1 has a lowest member, confirming the well-ordering property. The conclusion emphasizes that every total order on a finite set qualifies as a well-ordering.
Csharp
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Hi,

I want to show that there exists a well ordering for every finite set.

(I know if you add axiom of choice you can prove this theorem for infinite sets too but I think the finite sets do not need axiom of choice to become well ordered)
 
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Have you tried using induction on the cardinality of the set?
 
Good idea.

Suppose that k is well ordered.

k+1= k U {k}

First of all I'll define an ordering on k+1.
If s and t are both in k then I use the ordering from k.
If one of s and t is k then k>s.

Suppose that S is a nonempty subset of k+1.
Then if it doesn't contain k it has a lowest member.
If it contains k then S-{k} has a lowest member which is also lower than k itself.
 
Csharp said:
I want to show that there exists a well ordering for every finite set.
Every total order on a finite set is a well-ordering.
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...

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