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To prove: if $A$ is any well-ordered set of real numbers and $B$ is a nonempty subset of $A$, then $B$ is well-ordered.

My attempt: Suppose towards a contradiction that given any well-ordered set of real numbers $A$ and a nonempty set $B$ which is a subset of $A$, that $B$ is not well-ordered. Then by the well-ordering principle, $A$ has a least element, say $\mathit n$, while $B$ does not have a least element. So for all $\mathit b\in B$, there exists an element $\mathit m \in B$ such that $\mathit m < \mathit b$. Since $B$ is a subset of $A$, $\mathit m$, $\mathit b \in A$. Suppose $\mathit n = \mathit b \in A$. Then by our initial assumption on $B$, there exists an element $\mathit m \in B$ such that $\mathit m < \mathit n$. But $B \subseteq A$ and by definition of the well-ordering of $A$, $\mathit m \geq n$. Contradiction.

Model answer: Let $S$ be a nonempty subset of $B$. We show that $S$ has a least element. Since $S$ is a subset of $B$ and $B$ is a subset of $A$, it follows that $S$ is a subset of $A$. Since $A$ is well-ordered, $S$ has a least element. Therefore, $B$ is well-ordered.

I would like to ask if my proof by contradiction for the above question is valid and rigorous enough, since I have a few doubts such as letting $n=b \in A$. Is doing this step valid?

Greg Martin
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lohg
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  • If you take $I_1=[0,1]$ and $I_2=]0,1[$, then the minimum of $I_1$ is $0$ but $I_2$ has not a minimum. Something wrong in the question, I guess, or I'm missing something –  Oct 08 '23 at 15:48
  • You should avoid proofs by contradiction unless they are necessary. Your proof is hard to read and would require more work than this problem deserves to verify it. The subset relation is transitive so you shouldn't make simple proofs complicated. – John Douma Oct 08 '23 at 15:48
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    @Hola The usual ordering on $[0,1]$ is not a well-ordering. – Greg Martin Oct 08 '23 at 15:49
  • It is possible that $B$ is not well-ordered, but has a minimal element! In fact, even if the claimed statement were somehow not true, it would at least be immediate from the very definition of "well-ordered" that $B$ (being a nonempty subset of $A$) has a minimal element! – Hagen von Eitzen Oct 08 '23 at 15:51
  • @Ryan: You've made an error in your proof when you say "Suppose $n=b$". Both $n$ and $b$ have been given identities already, and so it's not valid to suppose that they are equal. I would say that you have a good idea about why the statement should be true; do you see how the model answer actually captures that idea and makes it rigorous? – Greg Martin Oct 08 '23 at 15:52
  • "Suppose towards a contradiction that given any well-ordered set of real numbers $A$ and a nonempty set $B$ which is a subset of $A$, that $B$ is not well-ordered." -- that's not even the right setup for a proof by contradiction, which would be "Suppose towards a contradiction that there exists a well-ordered set $A$ of real numbers and a nonempty subset $B$ of $A$ such that $B$ is not well-ordered." – blargoner Oct 08 '23 at 17:05
  • @Greg Martin Point taken, I had a feeling that was kind of fishy anyway. I understand how the model answer is rigorous by using the concepts of transitivity and creating an arbitrary set which relates to $A$ and $B$ which implies the well-ordering on $B$. – lohg Oct 08 '23 at 18:02

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