Self-duality of the ordered set $(\mathbb R,<)$ together with additional structure

Question

For which structure $s$ (by which I mean relations or functions on $\mathbb R$) is the structure $(\mathbb R, <, s)$ isomorphic to $(\mathbb R, >, s)$ (for example by $x\mapsto - x$)?

More specifically, is the $3$-ary relation $|\cdot - \cdot| < \cdot$ such an $s$?

Answer 1

One simple way to establish negative results is to look at what's definable in the reduct $(\mathbb{R};s)$. For example, we have:

Suppose $X\subseteq\mathbb{R}$ is bounded below but not bounded above. Then $(\mathbb{R};<,X)\not\equiv (\mathbb{R}; >,X)$. Consequently, if such an $X$ is definable in $(\mathbb{R};s)$, then $(\mathbb{R};<,s)\not\equiv(\mathbb{R};>,s)$.

• The relation "$\equiv$" is elementary equivalence: $\mathcal{A}\equiv\mathcal{B}$ iff $\mathcal{A}$ and $\mathcal{B}$ satisfy the same first-order sentences, that is, iff $Th(\mathcal{A})=Th(\mathcal{B})$. Elementary equivalence is trivially implied by isomorphism, so establishing non-elementary-equivalence is a strong way to establish non-isomorphism. On that note, elementary equivalence is generally much coarser than isomorphism; for example, by compactness (in the form of the upwards Lowenheim-Skolem theorem) we have that every infinite structure is elementarily equivalent to some infinite structure of strictly greater cardinality.

Proof. For the initial claim, consider the sentence "$\forall x\exists y(x\triangleleft y\wedge y\in X)$." This is true in $(\mathbb{R};<,X)$ but not in $(\mathbb{R};>,X)$. (Here "$\triangleleft$" is the ordering in the respective structure - that is, the symbol whose interpretation is $<$ or $>$ respectively.) The "consequently" then follows immediately by replacing in the sentence above the symbol $X$ with the appropriate definition of $X$ in $(\mathbb{R};s)$.

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The claim above resolves your more particular question: taking $s$ to be the ternary relation $$s(a,b,c)\iff\vert a-b\vert<c,$$ note that the set of positive reals is definable in $(\mathbb{R};s$) since $c$ is positive iff we have $s(a,a,c)$ for all/some $a$.

It's worth noting that in the claim above and its application we only used the fact that first-order logic is isomorphism-invariant. In general, we can use arbitrary isomorphism-invariant logics and their equivalence notions to prove non-isomorphism results. First-order logic is merely the most important of these.