Is a total order formed by set membership necessarily a well order?

Question

If $(X,\in)$ is a totally ordered set, then is it necessarily well ordered? ($\in$ denotes the usual set membership.)

Answer 1

Yes, assuming the Axiom of Regularity (or Foundation, as it is sometimes called). The axiom literally states that $\in$ is well-founded, so every linearly order it induces is a well-ordering by that virtue alone.

Of course, it is possible that the Axiom fails and we have $x\in y\in x$, in which case, $\{x,y\}$ is not even linearly ordered by $\in$ (it is not antisymmetric); or that we have for each $n\in\Bbb N$, $x_n=\{x_{n+1}\}$, which means that $\{x_n\mid n\in\Bbb N\}$ forms a decreasing $\in$-chain, so while linearly ordered, it is not well-ordered.