If all sets can be well-ordered, does this also mean that all sets can be partially ordered? Can someone give me an example of a set that is not partially ordered?
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A set isn't partially ordered until you put the structure of a partial order on it, in much the same way as a set isn't a group until you put a group structure on it. After all, posets (like groups) are sets-with-structure. If you give me a set and tell me nothing more then it is not partially ordered. Why? Because we haven't defined a partial order!
But every set can be equipped with a partial order; for example, if $A$ is a set, define a partial order $\le$ on $A$ by $a \le b$ if and only if $a=b$. (This is called the discrete order on $A$.)
In fact, most sets have many possible partial orders defined on them, and any one is as good as any other.
The point to take home is that "$X$ is partially ordered" and "$X$ can be partially ordered" are distinct statements. Let me know if you need any more clarification.
Aside: Whether or not every set can be well-ordered depends on (and is equivalent to) the axiom of choice. Since a well-order defines a partial order, if any set can be well-ordered then it can be partially ordered by the partial order induced by the well-order. (Specifically, if $<$ is a well-order on $A$, then for $a,b \in A$ put $a \le b$ if and only if either $a<b$ or $a=b$.) However, the assertion that every set can be partially ordered does not depend on the axiom of choice.
Yes. All sets can be well-ordered (every nonvoid subset has a first element in a linear order) if you are prepared to believe the Axiom of Choice.
Every set is partially ordered. This does not require the axiom of choice at all.
First we need to remember that in set theory everything is a set, so all the elements of a set are also sets.
We can now use the very natural $\subseteq$ ordering on the elements of its set.