Definition (Well Ordered Set)

A set $B$ with an order relation $<$ is well-ordered if every nonempty subset of $B$ has a smallest element

We know that $\mathbb{Z_+}$ is a well-ordered set.

But it doesn't seem clear to me why $\mathbb{Z}$ is not a well-ordered set as to me it seems that for every nonempty subset we can always find a smallest element.

I'll just illustrate with a few examples, (here $a, b, c \in \mathbb{Z_+}$)

  • $\min [a, b] = a$
  • $\min (a, c) = b$ where $ a < b < c$, for if there was no $b \in\mathbb{Z}$ between $a$ and $c$, then $(a, c) = \emptyset$
  • $\min \{-a, a\} = -a$
  • $\min [-a, b] = -a$
  • $\min [-a, -b] = -a$

I just can't seem to think of an example where a nonempty subset of $\mathbb{Z_+}$ doesn't have a smallest element.

This is obviously not true for $\mathbb{R}$ though as $(a, b) \subset \mathbb{R}$ doesn't have any smallest element, given the usual order relation on $\mathbb{R}$

So why is $\mathbb{Z}$ not a well-ordered set?

  • 4
    $\begingroup$ $\Bbb Z$ itself has no smallest element. $\endgroup$ – Brian M. Scott Dec 21 '16 at 20:38
  • 1
    $\begingroup$ @BrianM.Scott Ahhh, okay I see now. This is what lack of sleep does... $\endgroup$ – Perturbative Dec 21 '16 at 20:40
  • $\begingroup$ By the way, not all subsets are contiguous. $\endgroup$ – Rahul Dec 21 '16 at 20:40
  • $\begingroup$ @Rahul, Yep I realize that, I was just using intervals to illustrate some examples $\endgroup$ – Perturbative Dec 21 '16 at 20:42

What about $\mathbb{Z}$? That's a nonempty subset of $\mathbb{Z}$; what's its least element?

It is true that every bounded nonempty subset of $\mathbb{Z}$ has a least element (unlike $\mathbb{R}$), but this is a much weaker condition than compactness. (In fact, this is basically trivial in the case of $\mathbb{Z}$ - it's just the statement that bounded sets of integers are finite!)


Take the set of even integers. What is the smallest element with respect to $<$?


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