how can we define real number to form of a (first-order) structure of type $\mathcal L$?

Question

We know that :

Let $\mathbb{R}$ denote the set of all real numbers. Then:

1-) The set $\mathbb{R}$ is a field, meaning that addition and multiplication are defined and have the usual properties.

2-)The field $\mathbb{R}$ is ordered, meaning that there is a total order ≥ such that, for all real numbers $x, y$ and $z$:

if $x ≥ y$ then $x + z ≥ y + z$;

if $x ≥ 0$ and $y ≥ 0$ then $xy ≥ 0$.

The order is Dedekind-complete; that is: every non-empty subset $S$ of $\mathbb{R}$ with an upper bound in $\mathbb{R}$ has a least upper bound (also called supremum) in $\mathbb{R}$ .

And also we study in universal algebra that :

If $\mathcal{L}$ is a first-order language, then a (first-order) structure of type $\mathcal L$(or $\mathcal{L}$-structure) is an ordered pair $\mathbf{A}=\langle A , L \rangle $ , with $A = ∅$, where $L$ consists of a family $R$ of fundamental relations $r^{\mathbf{A} }$ on $A$ indexed by ${\mathcal R }$ (with the arity of $r^{\mathbf{A} }$ equal to the arity of $r$ , for $r ∈ \mathbf{R} $) and a family $F$ of fundamental operations $f^{\mathbf{A}}$ on $A$ indexed by $\mathcal{F}$ (with the arity of $f^{\mathbf{A}}$ equal to the arity of $f$, for $f \in \mathcal{F}$). A is called the universe of $A$, and in practice we usually write just $r$ for $r^{\mathbf{A}}$ and $f$ for $f^{\mathbf{A}}$ A. If $\mathcal R = ∅$ then $\mathbf A$ is an algebra; if $\mathcal F = ∅ $then $\mathbf A$ is a relational structure. If $\mathcal{L}$ is finite, say $\mathcal F = \lbrace {f_1, . . . , f_m}\rbrace $, $\mathcal R = \lbrace r_1, . . . , r_n \rbrace$, then we often write $\langle A, f_1, . . . , f_m, r_1, . . . , r_n \rangle$ instead of $\langle A, L \rangle $.

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Now how can we define real number to form of a (first-order) structure of type $\mathcal L$ ?

Answer 1

$(\mathbb{R};+,\times,<)$ (or similar) is already a perfectly satisfying first-order structure. There are also various other natural expansions, like $(\mathbb{R}; +,\times,\sin)$ - e.g. we can even consider $\mathbb{R}$ together with all real functions (this very-huge-language picture is relevant e.g. in nonstandard analysis) - and which specific first-order structure is appropriate to put on $\mathbb{R}$ depends on context.

EDIT: Let me make the point of the above explicit: there is no "unique right" way to view $\mathbb{R}$ as a first-order structure. Sometimes we care about $\mathbb{R}$ as an ordered field; then $(\mathbb{R};+,\times)$ - or any of the definitionally-equivalent variants, like $(\mathbb{R};+,-,\times,<)$, or even $(\mathbb{R};+,-,\times,^{-1},<)$ if we either view $^{-1}$ as a relation or allow partial functions in our semantics - is the right approach. In other situations we want more (as in nonstandard analysis mentioned above), or less (if we just want to consider $\mathbb{R}$ as a group), or just ... different (maybe we want to think about addition and $e^x$, but not multiplication!).

• Let me mention one particular type of reason we might actively want to work with a more limited language. When we limit the language, we often see powerful "tameness" phenomena. Most importantly, the structure $(\mathbb{R};+,\times)$ (and many of its variants) is o-minimal, but $\mathbb{R}$ with all functions isn't o-minimal. Precisely what o-minimality is isn't important right now; the point is that knowing that a structure is o-minimal lets us prove interesting things about it, and so we can prove things about $\mathbb{R}$ as a field (that is, we can prove results in real algebraic geometry) by thinking about $(\mathbb{R};+,\times)$ (and its relatives) rather than the more "expanded" structures. So more expressive power isn't always a good thing!

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Now you might object that Dedekind-completeness doesn't show up anywhere in the above. This is because Dedekind-completeness isn't a first-order property, so it won't be folded into the definition in any explicit way; rather, it will be a property this particular structure happens to have, but others like it may not - e.g. there will be ordered fields elementarily equivalent to $\mathbb{R}$ which aren't Dedekind-complete. Put another way, the "least upper bound" function is infinitary, and so doesn't fit into the framework of first-order structures.