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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?

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

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  • $\begingroup$ Thanks for redirecting me here. However, I don't see why the assumption that $ A $ is a well order is relevant with this argument. $\endgroup$ – Jytug Mar 7 '16 at 22:41
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    $\begingroup$ It's not relevant. It is a consequence, and a trivial one. $\endgroup$ – Asaf Karagila Mar 7 '16 at 22:42

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