What is the point of constant symbols in a language?

For example we take the language of rings $(0,1,+,-,\cdot)$. What is so special about $0,1$ now? What is the difference between 0 and 1 besides some other element of the ring?

I am aware, that you want to have some elements, that you call 0 and 1 which have the desired properties, like $x+0=0+x=x$ or $1\cdot x = x\cdot 1=x$.

Is there something else, which makes constants 'special'?

Other example: Suppose we have the language $L=\{c\}$ where $c$ is a constant symbol. Now we observe the L-structure $\mathfrak{S}_n$ over the set $\mathbb{Z}$, where $c$ gets interpreted by $n$.

Is there any difference, between $c$ and $n$? Or are they just the same and you can view it as some sort of substitution?

For $\mathfrak{S}_0$ we would understand $c$ as $0$. Since there are no relation- or functionsymbols, we just have the set $\mathbb{Z}$ and could note them as

$\{\dotso, -1, c, 1, \dotso\}$

If we take the usual function $+$ and add it $L=\{c,+\}$ now $\mathfrak{S}_0$ has the property, that $c+c=c$ for example.

I hope you understand what I am asking for.

I think it boils down to:

Is there a difference between the structure $\mathfrak{S}_n$ as L-structure and $\mathfrak{S}_n$ as $L_\emptyset$-structure, where $L_\emptyset=\emptyset$ (so does not contain a constant symbol).

But I want to get as much insight here as possible. So if you do not understand what I am asking for, it might be best, if you just take a guess. :)

Thanks in advance.

  • $\begingroup$ In the empty language, given any structure with more than one element, then there are no definable members in your structure. But once you add a constant, you add a definable element. $\endgroup$ – Asaf Karagila Apr 15 '19 at 23:53

An $L$-structure is not just a set, it is a set together with interpretations of the constant symbols, function symbols and relation symbols in $L$. You need to keep track of the interpretations as additional data so that you can do things like define homomorphisms of $L$-structures: namely, they're those functions that respect the interpretations of the symbols.

For example, "$\mathbb{Z}$ as a group" and "$\mathbb{Z}$ as a set" have the same underlying set, but the former additionally has (at least) a binary operation $+ : \mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$, which must be preserved by group homomorphisms.

In your example, a homomorphism of $L$-structures $f : \mathfrak{S}_n \to \mathfrak{S}_m$ would be required to satisfy $f(n) = m$, since $n$ and $m$ are the respective interpretations of the constant $c$, but a homomorphism of $L_{\varnothing}$-structures would not.

So while "$\mathfrak{S}_n$ as an $L$-structure" and "$\mathfrak{S}_n$ as an $L_{\varnothing}$-structure" have the same underlying set, they are not the same object.

Fun fact: the assignment from "$\mathfrak{S}_n$ as an $L$-structure" to "$\mathfrak{S}_n$ as an $L_{\varnothing}$-structure" is an example of a forgetful functor.

  • $\begingroup$ Now I wonder, if structures $\mathfrak{S}_n, \mathfrak{S}_m$ can be isomorphic (for $n\neq m$) when you take $L=\{c,<\}$. Then it has to be $n<m\Leftrightarrow f(n)<f(m)$. But such an isomorphism must preserve the constant symbols. So $f(n)=m$ and $f(m)=n$ and for every other $z\neq m,n$ it is $f(z)=z$. But then you get $n<m\Leftrightarrow f(n)<f(m)$ so $n<m\Leftrightarrow m<n$? $\endgroup$ – Cornman Apr 15 '19 at 23:44
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    $\begingroup$ @Cornman: Right, so what that tells you is that if $m \ne n$ then there is no isomorphism of $\{ c, < \}$-structures between $\mathfrak{S}_n$ and $\mathfrak{S}_m$ (assuming that $<$ is interpreted as the usual order relation on $\mathbb{Z}$ in both $\mathfrak{S}_n$ and $\mathfrak{S}_m$). $\endgroup$ – Clive Newstead Apr 16 '19 at 0:15

Clive's answer is already a good one, I just wanted to add another important point about constants. They give us the power to say infinitely many things about one element.

For example, if we consider Peano Arithmetic then obviously $\mathbb N$ is a model. Now, add a constant $c$ to our language and add sentences $c > \bar n$ for all $n \in \mathbb N$ (where $\bar n$ stands for 1 added $n$ times: $1 + 1 + \ldots + 1$). This new theory is consistent by compactness, so it has a model $M$. In $M$ we have an interpretation for $c$, which is bigger than all natural numbers. So we obtain a nonstandard model of arithmetic. Something similar can be done to create a model that looks like the reals, but has infinitesimals.

  • $\begingroup$ Thank you. That is insightful. I was wondering if you take $\mathbb{R}$ with the usual language of a field and add the constant symbols of $\pi$ and $e$, where $\pi=3,1415\dotso$ and $e=2.71828\dotso$ as usual. Then there would be no point in adding those constants? Because these 'symbols' already exist in the structure with the same behavior. $\endgroup$ – Cornman Apr 16 '19 at 9:10
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    $\begingroup$ Actually, there would be a point. These are not algebraic numbers, so they cannot be defined in $\mathbb R$ in the language of a field. In the language of a field, there is a model of $Th(\mathbb R)$ (all the sentences that are true in the reals) that consists just of the algebraic numbers. If we add $\pi$ and $e$ as constants, then the algebraic numbers would no longer be a model because we cannot interpret $\pi$ and $e$ in them. (btw, this theory has a name: "real closed fields" or "RCF") $\endgroup$ – Mark Kamsma Apr 16 '19 at 9:28
  • $\begingroup$ Thank you, that makes it much more clear. I did not think about them beeing trancendentel. So constants make a lot of sense, even to me now! $\endgroup$ – Cornman Apr 16 '19 at 9:31

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