[Math] Set that is not well-ordered

Question

I want to find a set that is transitive, every non-empty subset of it has $\in$-maximum and is not $\in$-well ordered.

Any hint?

Answer 1

Let $S$ be a non-empty set such that every non-empty subset of it has an $\in$-maximum.

Then for $x,y \in S$ with $x \ne y$, we have either $x \in y$ or $y \in x$ because by hypothesis, $\{x,y\}$ has an $\in$-maximum. Therefore, $(S, \in)$ is a linear order. In the degenerate case that $S$ has only one element, clearly $(S, \in)$ is also a linear order.

Now suppose that a non-empty subset $T \subseteq S$ had no $\in$-smallest element. By the Axiom of Foundation, $T$ does have an $\in$-minimal element, i.e. an $x \in T$ with $x \cap T = \varnothing$.

But this then means that for all $y \in T$, we have $y \notin x$. Since $\in$ is a linear ordering on $S$, we conclude that $x \in y$ (if $x \ne y$). But then $x$ is an $\in$-smallest element; hence $S$ is well-ordered by $\in$.

Therefore, no set $S$ matching your conditions can exist. Note that the assumption that $S$ is transitive was not necessary to arrive at this conclusion.