Curious about definition of well-ordered set We know that the definition of a well-ordered set guarantees the existence of least element in every non-empty subset of a set. 
This isn't a creative or good question, but I am just wondering why it using the existence of least element but not the greatest element or even both, just partially the least element. Anyone shed a bit light would be really appreciated. 
 A: It's just more natural that way, because it means that the smallest infinite well-ordered set is isomorphic to the natural numbers. If you changed the definition so that every subset had a greatest element, then you would get the negative integers instead. This is workable, I suppose, but why not do it the obvious way?
And if every subset of an ordered set $X$ had a least element and a greatest element, then $X$ would have to be finite, because a well-ordering (in the conventional sense) of an infinite set $X$ induces an obvious order-preserving bijection between the natural numbers and an initial segment of $X$ (and the natural numbers have no greatest element).
A: About greatest element, well ordering comes to generalize in a sense the natural number, which has no maximum, so it makes sense to take least element.
In addition, it makes sense when looking on well founded sets, a well founded set is a set where when order by $∈$ it has a minimal element(not minimum, it need not be unique), that means, if $A$ is non empty well founded set, then there exists an element $a∈A$ such that $a\cap A=\emptyset$. Well founded sets by $∈$ have the property that $A∉A$.
Well ordered set by $∈$ are just like well founded but with the property that the minimal element is unique.
If you change it to maximal element, we get that well founded(by $∈$) sets will have the property that there exists $a∈A$ such that $a$ is not in any other $b∈A$.
A set, $A$, being well founded(by $∈$) implies there is no sequence $x_i$ such that $x_0∋x_1∋\cdots∋x_n\cdots$ where all $x_i$ are in $A$, if you replace it to maximal element it will say that you have no sequence $x_i$ such that $x_0∈x_1∈\cdots∈x_n∈\cdots$ with $x_i∈A$ for all $i$, but by the axiom of infinity, $ω$(the set of natural numbers) exists and it is such that $0∈1\in\cdots$ and all of those are in $ω$.
So it doesn't make sense to change it to greatest element.  
