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According to the well ordering theorem "Any set can be well ordered". Whenever we have a well ordering on a set, it is not difficult to construct a new well ordering with a largest element.
My question is:

For a given infinite set, can we find a well ordering on it such that there is no largest element?

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    $\begingroup$ Set of natural numbers. $\endgroup$ – Wuestenfux Jul 2 '17 at 18:54
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    $\begingroup$ Induce the well-order by a bijection with a cardinal number. $\endgroup$ – Daniel Fischer Jul 2 '17 at 18:55
  • $\begingroup$ Use Choice + transfinite induction to define it one at a time? $\endgroup$ – Chappers Jul 2 '17 at 19:02
  • $\begingroup$ math.stackexchange.com/questions/6501/… $\endgroup$ – Count Iblis Jul 2 '17 at 19:16
  • $\begingroup$ @CountIblis That question is not related to this one at all. $\endgroup$ – Noah Schweber Jul 2 '17 at 20:02
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Yes. In fact, there is a canonical way to transform a well-ordering into a well-ordering of the same set with no greatest element.

Suppose I have an infinite well-order $X$. Let $F_X$ be the set of elements of $X$ with finitely many things bigger than them - for instance, the greatest element of $X$ (if such an element exists) is in $F_X$, while (if $X$ is infinite) the least element of $X$ (which always exists, since $X$ is a well-ordering) is not in $F_X$.

It's not hard to show that $F_X$ is finite, since otherwise we could build an infinite descending chain in $X$ (exercise). Now let $I_X=X\setminus F_X$; this is a well-ordered set with no greatest element.

If $X$ is infinite, $I_X$ and $X$ have the same cardinality. If you want an explicit isomorphism, we can "move $F_X$ to the back" - define a new ordering $\prec$ on $X$, given by $a\prec b$ iff

  • $a, b\in I_X$ and $a<b$ in the original sense of $X$; or

  • $a, b\in F_X$ and $a<b$ in the original sense of $X$; or

  • $a\in F_X, b\in I_X$.

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If $A$ is infinite then $A$ can be put in bijection with $A \cup \mathbb{N}$. Now from the well ordering of $A$ you can ge a well ordering of $A \cup \mathbb{N}$ without a largest element ( in the obvious way ) and then back, a new well-ordering of $A$ without a largest element.

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    $\begingroup$ As an interesting aside, note that for the first sentence, we either need a weak form of the axiom of choice or the assumption (which is present here) that $A$ is well-orderable: it is consistent with ZF that there is an infinite set $A$ which is not in bijection with $A\cup\mathbb{N}$ (namely, any infinite Dedekind-finite set). $\endgroup$ – Noah Schweber Jul 2 '17 at 21:48
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    $\begingroup$ @Noah Schweber: Very interesting.! To do away with the bijection, I guess we could just move $\mathbb{N}$ from the beginning to the end, similarly to the move in your solution. $\endgroup$ – Orest Bucicovschi Jul 2 '17 at 23:44
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In the usual development of ZFC, a cardinal is defined to the the least ordinal with a given cardinality. Infinite cardinals are "limit ordinals" --- as ordered sets they have no largest element.

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