Language structure of $\mathbb{R}$ and $L_{\mathbb{R}}$

Question

Ok, so, I'm reading up on some first order logic and am now studying languages and structures. If we define the language $L:= \{0,1,+,\cdot\}$, with $0,1$ the constants and $+, \cdot$ the $2$- place function symbols, then we should have that $\mathbb{R}$ is an $L$-structure. I'm having some problems with that.

Obviously, from $L$, we can achieve the natural numbers by setting $1+1 = 2$, $2+1=3$ and so on. The integers can now be achieved by working with variables (which is permitted in any language): $x + 1 = 0$, $x + 2 = 0$ and so on. In a similar way, one can obtain the rationals, again by using variables: $2 \cdot x = 1$, $\ldots$ you get the idea. I can even see how you get some irrational numbers (like $x \cdot x = 2$ gives you $\sqrt{2}$), but I fail to see how one can obtain numbers like $\pi$ or $e$. So that's one thing I do not understand.

Another problem to me is: Given that $L$-structure, one can consider the language $L_{\mathbb{R}}$, i.e. the language of the structure $\mathbb{R}$: $L_{\mathbb{R}}$ is $L$ together with, for each element $r$ of $\mathbb{R}$, an extra constant (also denoted by $r$). Here we assume that constants$(L) \cap \mathbb{R} = \emptyset$. So that there are no repetitions. So far so good. However, someone said to me that $L_{\mathbb{R}} = \{0,1,+,\cdot, \pi, e, \sqrt{2}, \ldots\}$. I understand why $\pi, e$ are in there, but not $\sqrt{2}$, as $\sqrt{2}$ can be achieved by $x \cdot x = 2$. I'm probably mixing some stuff up, but can someone help me shine some light on this?

Thanks in advance!

Answer 1

A structure is not a syntactic object: we don't expect to be able to express every element of a structure for a language $L$ as a term in $L$ or even define every element by a formula of $L$.

To say that $\mathbb{R}$ is a $L$-structure for the language $L$ defined by the signature $(0, 1, +, \cdot)$ means that we can interpret $0$ and $1$ as elements of $\mathbb{R}$ and $+$ and $\cdot$ as binary operations on $\mathbb{R}$, i.e., functions from $\mathbb{R}\times\mathbb{R}$ to $\mathbb{R}$. If we do this in the obvious way: identifying $0, 1, +$ and $\cdot$ with the zero element, the unit element, addition and multiplication respectively, the resulting structure has many interesting properties. One of these interesting properties is that $\mathbb{R}$ contains many elements that cannot be defined in $L$: this doesn't stop $\mathbb{R}$ being an $L$-structure.

There is no constant for $\sqrt{2}$ in the signature of $L$, however we can extend that signature to an uncountably infinite signature which contains constants for every real number, not just $0$ and $1$. Your $L_{\mathbb{R}}$ is the language defined by this signature. Some of these constants like $\sqrt{2}$ are definable in $L$ and some like $e$ are not.

Answer 2

Take a simpler example. Just consider the language of equality (without any symbols), then every non-empty set is a structure, but you can't define any element in any structure with more than two elements.

This is not a problem, and shouldn't be a problem. Remember that in order to define $e$ we use things far stronger than the first-order language of fields. And the definition of $\pi$ is also far outside the reach of first-order language of fields. It involves circles, which themselves involve a notion of distance, and so on.

We can define $\pi$ and $e$ or some other constants in the mathematical universe. Using way more than just the first-order logic over $\Bbb R$. Then, if we want to return into the realm of first-order logic somehow, we turn this into a first-order definition in set theory. But that's a whole other thing about the foundations of mathematics, and I'd rather not get into it here.

(Let me also add that without $\leq$ you cannot really define the element $\sqrt2$ using $x\cdot x=1+1$, since $-\sqrt2$ will also satisfy that formula. This can be fixed, but it is something to notice.)