# A well order on a subset of a power set

I'm struggling with one problem which touches upon ordering of a power set.

Assume I have a well order $\langle X,\preceq\rangle$ where $|X|=|\mathbb{R}|$. Is it possible to find (i.e. give a construction) a well ordering on an uncountable subset of $\langle P(X),\subseteq\rangle$? There is a well order there, that I know, but is there a way to explicitly show where it is? We might as well use $\langle P(\mathbb{R}),\subseteq\rangle$, but I can't come up with any well order there.

Perhaps you've got some hints?

• Not in general. You can not find a well-ordering on $\mathcal{P}(\mathbb{R})$ from a well-ordering on $\mathbb{R}$. – Hanul Jeon Mar 6 '16 at 11:56
• I doesn't have to be on the whole $\mathcal{P}(\mathbb{R})$, I'm fine with some subordering. – Jules Mar 6 '16 at 12:00
• If you find a sub-well-ordered subset, you can find it by just taking some finite linearly ordered subset. – Hanul Jeon Mar 6 '16 at 12:02
• I think the set of initial segments is fit for your question. – Hanul Jeon Mar 6 '16 at 12:12
• @Jules right. Empty set is the least initial segment. – Hanul Jeon Mar 6 '16 at 12:22

Every partial order embeds into its power set. Moreover this embedding is only cardinality dependent. So if $|A|\leq|B|$, every partial order of $A$ embeds into $\langle\mathcal P(B),\subseteq\rangle$.
To see this, if $\langle A,\leq\rangle$ is a partial order, $a\mapsto\{a'\in A\mid a'\leq a\}$ is an order embedding.
So yes, if $X$ is an uncountable well-ordered set, it embeds into its power set. The assumption that $|X|=|\Bbb R|$ is entirely irrelevant here.
• Thanks for redirecting me here. However, I don't see why the assumption that $A$ is a well order is relevant with this argument. – Jytug Mar 7 '16 at 22:41