I know that a set to be totally ordered and for example $A,B \in P(X)$ must either be $A \le B$ or $B \le A$. And also $\le$ is equivalent to $\subset$ for sets.
But I am not sure how I would prove that this is untrue for both.
2026-03-29 20:36:50.1774816610
How to prove that a set is not totally ordered?
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You need to find two elements of the order which are incomparable. That what it means that the order is not total.
In the case of a power set ordered by inclusion, this means $A$ and $B$ such that $A\nsubseteq B$ and $B\nsubseteq A$. Do note that this requires that $X$ has at least two elements.