Well-ordering on the integers: "every subset of $\mathbb{Z}$ has a least element", but what about $\mathbb{Z}$ itself?

Question

The integers are well-ordered.

If I take the entire set of integers though, there is no least element! Isn't the entire set of integers a valid subset of the integers?

Or (and I suspect this is the case), subset here is really in the very strictest of senses (i.e. $\mathbb{Z} \not\subset \mathbb{Z}$)?

Answer 1

The integers are not well-ordered. Take, for example, the subset of all even integers - there is no least element.

I think you're thinking of the naturals, $\mathbb{N}$, in which case this is true.

Answer 2

On $\mathbb{Z}$ (as on any set, if you accept the axiom of choice) there of course exists a well-order (let $0$ be the least element, with $0 < 1 < -1 < 2 < -2 < ...$). However, the typical order is not a well-order; it's merely a total order.

Answer 3

With the standard "<" order the integers are not well ordered. But if you use a different order the can be.

One such order I'll call "$\langle $which I'll define as. $a\langle b $ if $|a|<|b|$ and $-a \langle a $. That's a well-ordering.

Another is so that $0$ is the last number, every positive number is before any negative number, and otherwise they are sorted by absolute value.

Or you could put them in alphabetical order in english.