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EDIT: Completely overhauled the question, remove category theory from it (stated in plain set theory):

The original question Adding relations to a partial order

Let $X$ be a set and let $P: Pow(X) \rightarrow Partialorders(X)$ be a multivalued function ($\leq\in P(A)$ is read "$(A,\leq)$ satisfies property $P$").

The following are equivelent:

(1). The following statements are true:

(1a).For all $\leq\in P(X)$ and for all $A\subseteq X$ completely isolated (by $\leq$) subset there is a $\leq ' \in P(X)$ s.t. $\leq\subseteq\leq '$ (subsumption) and in $(X,\leq ')$: $A$ is continuous.

(1b). If For all $\leq\in P(X)$ and for all $A\subseteq X$ completely isolated (by $\leq$) subsets and for all $\leq ' \in P(A)$ s.t. $\leq|\substack{A}\subseteq\leq '$ (subsumption) and in $(A,\leq ')$ $A$ is continuous --------> then there is a partial order $\leq '' \in P(X)$ that subsumes both $\leq$ and $\leq '$.

(2). The following statements are true:

(2a). For all $A\subseteq X$ and for all $\leq \in P(A)$ if $B\subseteq A$ then $\leq\in P(B)$

(2b). For all $A\subseteq X$ every total order $\leq $ on $A$ satisfies: $\leq\in P(A)$.

Explanation in human language:

(1) basically means that every completely disconnected set can be extended to a continous set that preserves $P$ (this is 1a), and every extension of completely disconnected sets to continuous one that is locally in $P$ is globally in $P$ (this is 1b).

(2) gives a characteristic of extensions that satisfy this condition (the fact that (2)=>(1) is a generalization of the answer to the previous question which uses Zorn's lemma). It means that the inclusion map preserves the property $P$ and total orders satisfy it.

The other way around is much trickier.

Note that the exact question I'm asking is: (1)<=>(2) $\equiv$ axiom of choice

Showing that the axiom of choice implies (1)=>(2) is praticulary important and I belive it's more than enough (prooving the AC from (1)<=>(2) looks like a nightmare to even approach)

I'm going to use this extension with wildly bad behaved properties $P$ so I must know exactly how bad they can be, and I conjectured (2) is an exact requirement.

Thanks in advance for all attempting to help.

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  • $\begingroup$ So... you are asking if the fact that every set can be linearly ordered implies the axiom of choice? This was discussed on several threads before, and the answer is no. $\endgroup$ – Asaf Karagila May 19 '16 at 5:32
  • $\begingroup$ @AsafKaragila See the edit, I hope it's clearer now that I'm asking sonething much stronger (on the face of it). $\endgroup$ – Omer Rosler May 19 '16 at 8:11
  • $\begingroup$ @AsafKaragila How can the answer to that variant be no?? If I have an infinite set of pairs of elements and can linearly order the union of the set of elements, then from every pair I can pick the smaller element. That's the choice function in the axiom of choice! $\endgroup$ – btilly May 19 '16 at 17:44
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    $\begingroup$ @btilly: Yes, the axiom of choice for families of finite sets is implied by "Every set can be linearly ordered". Unfortunately, not much more than that. Not even countable choice. Not even countable choice from sets of reals. $\endgroup$ – Asaf Karagila May 19 '16 at 17:47
  • $\begingroup$ I'm afraid both of you don't understand what I'm asking: If a set can be lineary ordered iff some glueing axiom then do we have the choice axiom? $\endgroup$ – Omer Rosler May 20 '16 at 14:49

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