Proposition: Let X be a totally ordered non-empty set such that whenever a subset A⊆X satisfies ∀x [(∀y<x ⟹ y∈A)⟹x∈A]; x,y∈X then A=X. Then X is well-ordered.
This is the proposition that I'm trying to prove, my reasoning is the following:
Let B⊆X, B being a not empty set. Suppose B hasn't a minimum and let A be its complement. We want to show A = X.
Case 1: If X has a minimum, then this minimum, call it x0, cannot be in B, for then B would have a minimum. Now let n∈X, assume ∀m∈X (m<n ⟹ m∈A). It follows x∉B, because if it belonged to B then it would be its minimum since all previous elements are in A. Thus n∈A and by hypothesis, we have A = X, which implies B is the empty set. Therefore there cannot be a subset of X with no minimum.
Case 2: If X hasn't a minimum
This is the case where I'm confused because if my supposed subset B, was equal to X, then its complement A would be the empty set, I don't see how I can use a similar argument as the above mentioned. I've seen a proof of this in https://math.blogoverflow.com/2015/03/10/when-can-we-do-induction/
In which they use the same argument as I, with the base case implicitly proved. But It seems to me that assuming in this case, even A has an element is a very strong assumption since we have A equals the empty set.
Could you help me understand why their proof is valid or how to prove this case. Thanks in advance for your answers.
Suppose $\emptyset\ne A \subset X$ and $A$ has no least member. Let $B=X\setminus A.$
Now if $x\in X$ and $\{y\in X: y<x\}\subset B$ then $x\in B$; otherwise $x$ would be the least member of $A.$ So strong induction does not work for $B$ because $$\forall x\in X \, (\,[\forall y<x (y\in B)]\implies x\in B)$$ but $B=X\setminus A\ne X$.