To preface this question, it it important to say that the exercise included below is taken from lecture notes of an introductory course in the foundations of (undergraduate) mathematics, and that before this exercise (which is added as an aside, really) there has been no rigorous definition of boundedness.

Leading up to the exercise, the text reads:

The proof [not relevant to this question] I give below uses the Well-Ordering Principle, in the form

Every subset of $\mathbb Z$ which is bounded above has a greatest element.

For reference purposes, the original 'form' in which the Well-Ordering Principle was given is as follows:

Any subset $T$ of the integers $\mathbb Z$ which is bounded below has a least element.

Exercise: Is this form of the Well-ordering principle something new, or does it follow from the old version?

I thought that we could say that

For, say $S \subset \mathbb Z$:

If $S$ is bounded above then there exists a greatest element in $S$ as being equivalent to the statement that if there is not a greatest element in $S$ then $S$ is not bounded above. In this reasoning I have attempted to form the contrapositive version of the original statement.

I am not sure if the statement that, for $S \subset \mathbb Z$,

$S$ is bounded above $\implies$ there exists in $S$ a greatest element

should actually be biconditional, i.e:

$S$ is bounded above $\iff$ there exists in $S$ a greatest element (perhaps this may be the case depending on the actual definition of boundedness).

At any rate, I struggle to see how this is equivalent to the original framing of the Well-Ordering Principle that: if $S \subset \mathbb Z$ is bounded below then there exists in $S$ a least element. That's to say, that my gut would have me respond to this exercise and argue that this form is new, but the fact that the author refers to both as being forms of the Well-Ordering Principle leads me to believe I am wrong. In any event I remain unsatisfied either way. Please also edit/add tags to help me place this question correctly.

  • $\begingroup$ I am not sure I see any useful way to consider whether these are equivalent. They are simply true statements about the integers. On the other hand, given an ordered set $X$, the statements would not be equivalent for $X$. $\endgroup$ – Tobias Kildetoft Aug 13 '18 at 19:05
  • $\begingroup$ @TobiasKildetoft are they both essentially axioms? Would you say the question posed in the exercise is that it doesn’t ‘follow’ from, the old version? Why does the author refer to both as the ‘Well-Ordering Principle’? $\endgroup$ – Benjamin Aug 13 '18 at 19:08
  • 2
    $\begingroup$ On the other hand, if $X$ has an order-reversing involution (such as unary negation on $\mathbb{Z}$) then in fact the two are equivalent for $X$. $\endgroup$ – Daniel Schepler Aug 13 '18 at 19:08

Given a subset $T \subseteq \mathbb{Z}$, define $-T := \{ -n \mid n \in T \}$. Then what you need to check to establish that the original version of the Well Ordering Principle implies the alternate version is:

  • If $T$ is bounded above, then $-T$ is bounded below.
  • If $-T$ has least element $m$, then $T$ has greatest element $-m$.

Proving the other implication is very similar.


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