# Are all sets partially ordered?

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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The trivial order $x\leq y$ if and only if $x=y$, makes any set partially ordered. Don't need axiom of choice – Thomas Andrews Nov 8 '12 at 2:28

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.

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You mean $a\leq b$ iff $a=b$, I assume. – Thomas Andrews Nov 8 '12 at 2:37
@ThomasAndrews: Yikes! Yes I did, thanks. – Clive Newstead Nov 8 '12 at 9:33

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.

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A well ordering is a partial ordering. A set with no relation on it has no partial order (imposed) on it. – ncmathsadist Nov 8 '12 at 1:42

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.

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