Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to prove the following statement:

In well-ordered set $\langle\mathbb{N},<\rangle$, moving $0,1,2,3,...,n-1$ to the end, retaining that order, results in a well-ordered set $\langle \Bbb N,<^{(n)}\rangle$.

My work:

Saying that the relation obtained is $<^{(n)}$. I proved that the relation is a total order, but how one can prove that to every non-empty subset of $\langle\mathbb{N},<^{(n)}\rangle$ has least element (first element).

Thank you.

share|cite|improve this question
You can consider two cases. Let $A$ be a nonempty subset of $\mathbb N$. Then either $A\setminus\{0,1,\dots,n-1\}\ne\emptyset$ or $A\subseteq\{0,1,\dots,n-1\}$. Can you show in each of these cases that $A$ has smallest element w.r.t. the given order? – Martin Sleziak May 4 '12 at 18:28
I think elementary-set-theory tag is more suitable. I've also added ordinals, since the question deals with well-ordered sets. – Martin Sleziak May 4 '12 at 18:30
Thank you for editing! – Salech Alhasov May 4 '12 at 18:40
up vote 2 down vote accepted

It is easy to divide to cases here. Suppose $A\subseteq\mathbb N$ is non-empty.

  • If $0,\ldots,n-1\notin A$, show that $\min_< A$ (the minimal element of the usual order) is still minimal in the new order.

  • If $A\subseteq\{0,\ldots,n-1\}$ then the same as above, $\min_<A$ is the same.

  • Lastly, if $A$ contains both elements from $\{0,\ldots,n-1\}$ and elements from $\{n,n+1,\ldots\}$, show that $\min_< \Big(A\cap\{n,n+1,\ldots\}\Big)$ is the new minimum.

share|cite|improve this answer

HINT: Let $A$ be a non-empty subset of $\Bbb N$. It's convenient to let $M=\Bbb N\setminus\{0,1,\dots,n\}$.

There are two cases.

  1. $A\cap M\ne\varnothing$. In this case the $<^{(n)}$-least element of $A$ must belong to $M$ (why?), and $<$ and $<^{(n)}$ give the same ordering of $M$, so ... ?

  2. $A\subseteq\{0,1,\dots,n\}$. This case is easy: why?

share|cite|improve this answer
@Martin: Yes, it was a typo; I caught and fixed it right away, but apparently not before you caught it. – Brian M. Scott May 4 '12 at 18:46
@Martin: Aargh. You're right: it was a different typo that I fixed before. (Both were holdovers from an earlier version that was harder to read.) – Brian M. Scott May 4 '12 at 19:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.