Halmos writes in "Naive Set Theory":
Given the set ${X}$, consider the collection $W$ of all well-ordered subsets of $X$. Explicitly: an elememt of $W$ is a subset $A$ of $X$ together with a well ordering of $A$. We partially order $W$ by continuation.
Then he goes on and writes:
The collection $W$ is not empty, because, for instance $\emptyset\,
\epsilon \,W$. If $X \neq \emptyset$, less annoying elements of $W$ can be exhibited; one such is {($x, x$)}, for any particular element $x$ of $X$.
My problem is that Halmos defined earlier in section 6 about Ordered Pairs that an ordered pair is defined as follows:
$(a, b) = \{\, \{a\}, \{a, b\}\}$.
And if $a=b$ then we have
$(x,x)= \{\{x\}\}$.
My problem/question is that if $W$ contains subsets of $X$ and if $x$ is an element of $X$, then why is
{($x, x$)} in $W$? By the definition of an ordered pair that Halmos gave we can only conclude that $(x,x)\, \epsilon \,W$.
Best Answer
It seems that a "well-ordered set" means just a well-ordering relation.
That is, when we would normally write $(E,R)$ where $E$ is the set and $R$ is the order relation on $E$, in fact it is enough to write only $R$ (since we can recover $E$ from $R$). So, for example, $R=\{(x,x)\}$ is the unique well order relation on the set $\{x\}$, so (in this understanding) $R$ is a well-ordered set.