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Prove that class of all structures that are isomorphic structure of the form $\mathbb{A}=\langle\mathcal{P}(A),\cup^{\mathbb{A}},\cap^{\mathbb{A}},\subseteq^{\mathbb{A}}\rangle, $ is not axiomatizable.$$\mathbb{A}=\langle\mathcal{P}(A),\cup^{\mathbb{A}},\cap^{\mathbb{A}},\subseteq^{\mathbb{A}}\rangle$$ should be understand as "normal" operations on set.

It can be proved by Lowenheim-Skolem's theorem. If there existed a such set $S$ thet $|\mathcal{P}(S)| = \aleph_0$ then Lowenheim-Skolem's theorem wouldn't apply here, right?

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    You should be more careful with how you write and define things (is $A$ fixed ? And the way you wrote your proof isn't very precise) but this is the idea, indeed. – 2017-01-27
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    Please present your proof by the L-S theorem to give us some idea of what you are trying to prove. – 2017-01-27

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Yes, that's right.

In a bit more detail: by Lowenheim-Skolem, we would have to have a countably infinite model of that theory. But then the underlying set of that model, $M$, would have the same cardinality as some powerset - that is, for some $B$ we'd have $\vert M\vert=\vert\mathcal{P}(B)\vert$.

But then $B$ must be of cardinality strictly smaller than that of $M$. What sets have cardinality less than $\aleph_0$? The finite ones! So $B$ is finite. But then $M$ is finite too, contradiction.

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    Noah, how do you know that you can use Lowenheim-Skolem theorem? – 2017-02-10
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    @Logic What do you mean? It's a theorem; it applies. LS says (actually this is a special case of LS), "If $T$ is a theory in a countable language with an infinite model, then $T$ has a countably infinite model." The language here - $\{\cup, \cap, \subseteq\}$ - is definitely countable, and there is definitely an infinite structure of the kind considered (e.g. $\mathcal{P}(\mathbb{N})$). So - if $T$ axiomatizes this class of structures - $T$ has a countably infinite model. – 2017-02-10
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    ok, the problem was that you used special case of LS. Now, everything is clear. – 2017-02-10