Proving a total-order is a well-order if and only if every initial segment is determined by an element I managed to prove the $\longrightarrow$ part, but I'm not entirely sure how to prove the second part. I can assume by contradiction that the total order $(X, \leq)$ is not a well-order, which means there is an infinite monotonically-decreasing sequence, but how do I arrive to a contradiction? Can I show there exists an initial segment which is not determined by any element in $X$?
Would appreciate any help or hints.
 A: By "determined by  an element" I presume you mean "is the set of elements $<$ some fixed element." If this is the case, then: suppose $a_1>a_2> . . .$ is a descending sequence. Is there an initial segment of $X$ corresponding, in some way, to this sequence? Why is this initial segment not determined by an element?

EDIT: As a further hint (and based on the comments below), think about, say, $X=$negative integers. Then:


*

*Is $X$ well-ordered?

*What is an initial segment of $X$ not "determined by" any element?

*How does that initial segment relate to the descending sequence $-1>-2>-3>...$?
Then, same question with $Y=$negative integers and $-\infty$:


*

*What is an initial segment of $Y$ not "determined by" any element?

*How does that initial segment relate to the descending sequence $-1>-2>-3> . . .$?
Hopefully this helps.
A: There is a straightforward proof, that is without contradiction, using the same arguments and ideas. This has the advantage that you do not need to use the axiom of choice (or the weaker axiom of dependent choice) to find this strictly decreasing sequence that is used there.
To be more specific, take a non-empty subset $A\subseteq X$ and consider the set
$$I:=\bigcap_{x\in A}I(x),$$
where $I(x):=\{y\in X:y<x\}$ is the initial segment determined by an element
$x\in X$. This is clearly a (proper) initial segment of $X$ and so it is determined by an element, say $z\in X$. Thus, we have that $I=I(z)$. This element $z\in X$ now is the least element of $A$. Indeed, if this is not the case then, there exists $x\in A$ such that $x<z$ and so $x\in I(z)=I$ and by the definition of $I$ it follows that $x\in I(x)$, which is a contradiction.
