Why is "totally ordered" necessary in this implication of the Axiom of Foundation

The Axiom of Foundation was stated as:

Every non-empty set contains an element disjoint from it.

Then the following:

Every non-empty set $S$ contains an element $x$ such that no element $y\in S$ satisfies $y\in x$. This means that $x$ is an $\epsilon$-minimal element of $S$.

Thus the Axiom of Foundation implies that ever set that is totally ordered under $\epsilon$ is well ordered under $\epsilon$.

My question, please, is why is "totally ordered under $\epsilon$" necessary?

After all, if, as stated earlier, every non-empty set $S$ contains an $\epsilon$-minimal element and every non-empty subset is also a set, is that not enough (i.e., without the stipulation that it is also totally ordered) to assert that $S$ is well- ordered?

Thanks

Solution 1:

Take set $\{0,\{1\}\}$. It is not totally ordered under $\in$ because neither $0\in\{1\}$ nor $\{1\}\in 0$, and thus not well-ordered under $\in$. Still, it satisfies Axiom of Foundation.

Solution 2:

This comes down to the way "well-ordering" is defined.

Any non-empty set $S$ is well-founded - a well-founded partial order is one in which every (nonempty) set has a (possibly not unique) minimal element. A well-order, by contrast, is a particular kind of linear order; and $S$ need not be linearly ordered by $\in$!

Here's a way to rephrase the bolded statement:

Foundation says that the $\in$-relation on any set $S$ is well-founded; and so, in case $S$ is linearly ordered by $\in$, we will have that $\in$ is a well-ordering on $S$ (since the well-founded linear orders are exactly the well-orderings).