I am studying Halmos' Naïve Set Theory, and in Section 17, Well Ordering, he mentions the following exercise:
Prove that if $R$ is a partial order in a set $X$, then there exists a total order $S$ in $X$ such that $R\subset S$; in other words, every partial order can be extended to a total order without enlarging the domain.
Question: For any set $X$, isn't $X\times X$ a total order on it? And $X\times X$ definitely contains $R$.
(I know I must be doing a blunder here since this is definitely not so trivial, as I had a quick look on Wiki for this.)
No, $X\times X$ is generally not a total order, or even a partial order on $X$. If “order” means “strict order”, then it fails irreflexivity and asymmetry (provided $X$ is nonempty), and if it means “non-strict order” then it fails antisymmetry (provided $X$ has more than one element).