# Question about whether the structures of the language are isomorphisms

from the Textbook $$P27$$ of Mathematics Logic by Christopher C. Leary Lars Kristiansen

$$\mathbf{Definition~of~isomorphic~and~isomorphism:}$$ Suppose that $$\mathfrak{A}$$ and $$\mathfrak{B}$$ are two $$\mathcal{L}$$-structures. We will say that $$\mathfrak{A}$$ and $$\mathfrak{B}$$ are isomorphic. Write $$\mathfrak{A} \cong \mathfrak{B}$$ if there is a bijection $$i : A \rightarrow B$$ such that for each constant symbol $$c$$ of $$\mathcal{L}$$, $$i(c^\mathfrak{A}) = c^\mathfrak{B}$$, for each $$n$$-ary function symbol $$f$$ and for each $$a_1, ..., a_n \in A$$, $$i(f^\mathfrak{A}(a_1, ..., a_n)) = f^\mathfrak{B}(i(a_1), ..., i(a_n))$$, and for each $$n$$-ary relation symbol $$R$$ in $$\mathcal{L}$$, $$(a_1, ..., a_n) \in R^\mathfrak{A}$$ if and only if $$(i(a_1), ..., i(a_n)) \in R^\mathfrak{B}$$. The function $$i$$ is called an isomorphism.

The Qustion is that:

Let $$(\mathbb{Z},0,1,<)$$, $$(\mathbb{Q},0,1,<)$$ and $$(\mathbb{R},0,1,<)$$ be the integers, rational numbers and real numbers, considered as structures in the language $$\mathcal{L} = \{ 0,1,< \}$$.

(a) Does there exist an isomorphism from $$(\mathbb{Z},0,1,<)$$ to $$(\mathbb{Q},0,1,<)$$? If so, explicitly define one. If not, why not?

Well my thought is there is $$\mathbf{NO}$$ isomorphism from structure $$(\mathbb{Z},0,1,<)$$ to structure $$(\mathbb{Q},0,1,<)$$. Since, the language $$\mathcal{L}$$ does not has the function symbol $$\div$$, so does the structure $$(\mathbb{Z},0,1,<)$$ and structure $$(\mathbb{Q},0,1,<)$$. So, there are some $$x,y$$ in $$\mathbb{Q}$$ that is not matched from $$\mathbb{Z}$$. which is not surjective, so it is impossible for $$i$$ to be bijective.

(b) Does there exist an isomorphism from $$(\mathbb{Q},0,1,<)$$ to $$(\mathbb{R},0,1,<)$$? If so, explicitly define one. If not, why not?

Well my thought is there is $$\mathbf{NO}$$ isomorphism from structure $$(\mathbb{Q},0,1,<)$$ to structure $$(\mathbb{R},0,1,<)$$. Since, it is quite similar with (a). Since, the language $$\mathcal{L}$$ does not have the function symbols like $$\sqrt{}$$ or $$^{\frac{1}{2}}$$, so does the structure $$(\mathbb{Q},0,1,<)$$ and structure $$(\mathbb{R},0,1,<)$$. So, there are some $$x,y$$ in $$\mathbb{R}$$ that is not matched from $$\mathbb{Q}$$. which is not surjective, so it is impossible for $$i$$ to be bijective.

Are my thoughts from $$(a)$$ and $$(b)$$ correct, or is there anything to add?

• These structures have only two constant symbols and one relation symbol. They don't have things like division (not even addition and multiplication). – ancientmathematician Nov 3 '19 at 9:08

There is no isomorphism between $$(\Bbb Z,0,1,<)$$ and $$(\Bbb Q,0,1,<)$$, but it has nothing to do with division being present or not. Furthermore, there are bijective functions from $$\Bbb Z$$ to $$\Bbb Q$$, so I'm not sure how you conclude that $$i$$ would not be surjective.

The trick is to take two elements $$n in $$\Bbb Z$$, then $$i(n), since $$i$$ is an isomorphism and thus preserves the relation $$<$$. However, there is some element $$i(n), as $$\Bbb Q$$ is dense, but then we would have $$n, and there exists no such integer.

For $$(\Bbb Q,0,1,<)$$ and $$(\Bbb R,0,1,<)$$ it is easier to show there exists no isomorphism, since there does not even exist a bijection between $$\Bbb Q$$ and $$\Bbb R$$. This is because the cardinality of $$\Bbb R$$ is uncountable, while $$\Bbb Q$$ has countable cardinality.

However, interestingly, the structures $$(\Bbb Q,0,1,<)$$ and $$(\Bbb R,0,1,<)$$ are non-isomorphic just because of their difference in cardinality, and not because they "behave" differently. For any (first-order) sentence that one could build using variables for the elements of $$\Bbb Q$$ or $$\Bbb R$$, the constants $$0$$ and $$1$$ and the relation $$<$$, it will be true in the structure $$(\Bbb Q,0,1,<)$$ if and only if it will be true in $$(\Bbb R,0,1,<)$$.

This is quite striking: if we can only use $$<$$, $$0$$ and $$1$$ in our language, it is impossible to "see the difference" between $$\Bbb Q$$ and $$\Bbb R$$. This notion is called elementary equivalence. The fact that $$(\Bbb Q,0,1,<)$$ and $$(\Bbb R,0,1,<)$$ are elementary equivalent is usually one of the first things that is proved in a text on model theory.

In fact, that $$(\Bbb Q,0,1,<)$$ and $$(\Bbb R,0,1,<)$$ are elementary equivalent shows that the square root operator $$\sqrt{\cdot}$$ can not be described using just the order relation.