Is the ordering principle preserved in generic extensions?

The ordering principle says that every set can be linearly ordered. In a previous question http://mathoverflow.net/questions/76995 Asaf Karagila asked which axioms are preserved in forcing extensions.
The proof that AC is preserved in generic extenxions seems to fail in the case of the ordering principle. Is it known whether or not the ordering principle is preserved by forcing?

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 This question got a lucky number! :-) – Asaf Karagila Oct 2 2011 at 21:46 Just added the link, for lazy people :P – David Roberts Oct 3 2011 at 3:57

The answer is no. The proof is in the paper cited in my question, Theorem 4.7

G. P. Monro, On Generic Extensions Without the Axiom of Choice. The Journal of Symbolic Logic Vol. 48, No. 1 (Mar., 1983), pp. 39-52

(I did not yet read the paper but just checked it out, so I cannot really give much of the proof).

Edit: I reviewed the proof, I will try to give an outline explanation.

Let $M\models ZF+V=L$. We then add $\omega$ many Cohen reals, and take a symmetric extension based on finite supports. The result is the Levy-Halpern model in which there exists a Dedekind-finite set of reals, and the ordering principle holds (a similar proof appear on Jech T., The Axiom of Choice). The symmetric extension is isomorphic to $\big(L(A)\big)^{M[G]}$.

Denote $A$ the new set of Cohen reals, which is the aforementioned Dedekind-finite set in the symmetric extension. Now take all the equivalence relations on finite subsets of $A$, ordered by extension. That is:

$\{(B,f)\mid B\subseteq A, |B|<\omega, f\colon B^2\to 2:\{(x,y)\mid f(x,y)=1\}\text{ is an equivalence relation on }B\}$

And $(B,f)$ is stronger than $(B',f')$ if $B'\subseteq B$ and $f$ extends $f'$.

Let $F$ be a generic filter over this poset. $F$ allows us to define $E$, as the generic equivalence relation on $A$ in $\big(L(A)\big)^{M[G]}[F]$. Denote by $Y$ the set of equivalence classes of $A/E$.

We have that $Y$ is amorphous. Since amorphous sets cannot be linearly ordered we are done (the proof for this theorem is pretty much straightforward when considering a linear order and then the least element that has infinitely many above it, or infinitely many below it).

As a side note I should add that V=L is not needed, since it is only used to assert that the symmetric extension is $\big(L(A)\big)^{M[G]}$. However we can use the theorem by Grigorieff which says that this is essentially $\big(HOD(A)\big)^{M[G]}$.

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