A structure is composed of the domain, constants, relation symbols, and function symbols. I understand all of the ingredients of a structure immediately except for the constants. The definition in my book (A Shorter Model Theory) is

A set of elements of $A$ called constant elements, each of which is named by one or more constants. If $c$ is a constant, we write $c^A$ for the constant element named by $c$.

An example of a structure with constants is

$$\langle \mathbb{R}, +, -, \cdot, 0,1, \le \rangle$$

I understand that $0$ and $1$ are special because they are the identity elements for addition and multiplication, respectively, but why are they the constants, and in what sense do we need to specify them? I understand that I get a totally different structure if I where to use $<$ instead of $\le$, but I don't see what is so special about the constants. Why not have a structure of the following type:

$$\langle \mathbb{R}, +, -, \cdot, \frac{1}{2},\pi, \le \rangle$$

Is this different and if so, in what sense? Viewed as a group $\mathbb{R}$ has many choices of elements but only one identity element so I can make sense of specifying the identity element when talking about a group, but in what sense are the symbols related when we write out the structure (or signature)?

  • 3
    $\begingroup$ Unfortunately, when people are introduced to abstract algebra, they are introduced to the structure "a semigroup that happens to have an identity and inverses", but they are told that the structure is a "group". The theory of groups is weird in that you can actually do group theory while suffering from that misconception. If you want to get an idea about what's going on, you should instead look at monoids. In particular, you should pay attention to the difference between "monoid" and "semigroup with unit". $\endgroup$
    – user14972
    Dec 1, 2017 at 6:29
  • $\begingroup$ constatnts are symbols, i.e. part of the language. They are called so because they refer to "firm and fixed" elements of the domain. $\endgroup$ Feb 23, 2018 at 8:38

3 Answers 3


I've always thought the treatment of constants as a separate class of symbols is rather silly. A constant is just a special kind of function symbol: a $0$-ary function symbol. There's only one way to give no inputs to a function, so such a function must always give the same output.

So, if you understand the role of function symbols, you also understand the role of constant symbols! The roles are exactly the same; constants are just functions that require no inputs in order to give an output. Changing the interpretation of a constant symbol from $0$ to $\frac{1}{2}$ as you suggest very much gives you a different structure, just as much as changing the interpretation of the function symbol $+$ to some other binary operation would. Both $\langle \mathbb{R}, +, -, \cdot, 0,1, \le \rangle$ and $\langle \mathbb{R}, +, -, \cdot, \frac{1}{2},\pi, \le \rangle$ are perfectly good structures, but they are different structures, just as much as $\langle \mathbb{R},+\rangle$ and $\langle \mathbb{R},\cdot\rangle$ are different structures.

As for your final question:

in what sense are the symbols related when we write out the structure (or signature)?

The answer is "not at all"! The relations and functions that are part of a structure need not have any relationship at all, other than that they are relations and functions on the same underlying set.

  • $\begingroup$ Well.... When I say "in what way are they 'related' ", I don't mean they have some $ a priori$ relation, I mean once we write down the form of a structure, what do the symbols tell us. But it seems you've answered this: Namely, the constants are 'related' to the elements of the structure as they are a function of the elements (in the sense you outlined above). $\endgroup$
    – Squirtle
    Nov 30, 2017 at 22:28

I think the real meat of your question is when you write:

Why not have a structure of the following type: $$\langle \mathbb{R}, +, -, \cdot, \frac{1}{2},\pi, \le \rangle$$ Is this different and if so, in what sense?

The answer is: yes, they are very different - specifically, they are not isomorphic. Let me describe a bit how the choice of (interpretations of) constants (symbols) in particular can affect the isomorphism type of a structure; I'll use simpler examples, but it should be clear how to address your specific example the same way.

Constants - or rather, the interpretations of constant symbols - can be anything you want (that is, any elements of the structure in question), and precisely how a constant symbol is interpreted affects the isomorphism type of the structure.

Specifically, let's look at the language $L=\{\color{red}{ +}, \color{red}{ \times}, \color{red}{ c}\}$ for simplicity where $\color{red}{ +},\color{red}{ \times}$ are binary functions and $\color{red}{ c}$ is a constant symbol (and for clarity I'll use red text for symbols, and usual black text for actual functions/elements - that is, for the interpretations of the symbols). Let's consider two structures:

  • $\mathcal{A}=(\mathbb{R}; \color{red}{ +}^\mathcal{A}=+, \color{red}{ \times}^\mathcal{A}=\times, \color{red}{ c}^\mathcal{A}=0)$

  • $\mathcal{B}=(\mathbb{R}; \color{red}{ +}^\mathcal{A}=+, \color{red}{ \times}^\mathcal{A}=\times, \color{red}{ c}^\mathcal{A}=17)$.

(Note that $17$ is a really silly choice of interpretation for $\color{red}{ c}$ - but that's fine! There's no rule on what sort of element of the structure a constant symbol can be interpreted as - we can use a constant to denote any element whatsoever. We could equally well have chosen $\pi$ or $\sqrt{2}$.)

There is a natural bijection between these two structures, namely the identity $id: r\mapsto r$.

Is $id$ an isomorphism?

Well, to be an isomorphism is to preserve all "basic" formulas (and be bijective); in this case, the "basic" formulas are:

  • $x\color{red}{+}y=z$.

  • $x\color{red}{\times}y=z$.

  • $x=\color{red}{ c}$.

It's easy to see that the last of these is not preserved by $id$: we have $$\mbox{$\mathcal{A}\models 0=\color{red}{ c}^\mathcal{A}\quad$ but $\quad\mathcal{B}\models\neg id(0)=\color{red}{ c}^\mathcal{B}$.}$$

This means that $id$ is not an isomorphism.

Exercise: there is no isomorphism between $\mathcal{A}$ and $\mathcal{B}$. HINT: show that such an isomorphism must send $0$ to $0$ ...

Now, you might object that $\mathcal{A}$ and $\mathcal{B}$ are "really" the same, the only difference is in the silly choice of evaluation of $\color{red}{ c}$. But isomorphism is a precise notion; it doesn't take into account whether or not we think part of the structure is "silly." That said, here are a couple relevant facts:

  • $\mathcal{A}$ and $\mathcal{B}$ do have isomorphic reducts (when we forget the constant symbol $\color{red}{ c}$): letting $L_0=\{\color{red}{ +}, \color{red}{ \times}\}$, $\mathcal{A}$ and $\mathcal{B}$ have reducts $\mathcal{A}_0=(\mathbb{R}; \color{red}{ +}^\mathcal{A}=+, \color{red}{ \times}^\mathcal{A}=\times)$ and $\mathcal{B}_0=(\mathbb{R}; \color{red}{ +}^\mathcal{A}=+, \color{red}{ \times}^\mathcal{A}=\times)$. These are obviously isomorphic - in fact, they are literally the same structure.

  • Moreover, in a precise sense $\mathcal{A}$ can be "recovered from" $\mathcal{A}_0$ and $\mathcal{B}$ can be "recovered from" $\mathcal{B}_0$ - that is, $0=\color{red}{ c}^\mathcal{A}$ is definable in $\mathcal{A}_0$ and $17=\color{red}{ c}^\mathcal{B}$ is definable in $\mathcal{B}_0$ (exercise - note that this is not true for all possible choices of constants, e.g. interpreting $\color{red}{ c}$ as $\pi$).

  • So $\mathcal{A}$ and $\mathcal{B}$ are "equivalent" in a certain sense (namely they are bi-interpretable). But this isn't isomorphism.

Here's a further observation which I think will help clarify just how "rigid" the notion of isomorphism is:

Look at the language $S=\{\color{red}{ c_0}, \color{red}{ c_1}\}$ consisting only of two constant symbols, and think about the structures $$\mathcal{C}=(\mathbb{R}; \color{red}{ c_0}^\mathcal{C}=0, \color{red}{ c_1}^\mathcal{C}=1)\quad\mbox{ and }\quad \mathcal{D}=(\mathbb{R}; \color{red}{ c_0}^\mathcal{C}=1, \color{red}{ c_1}^\mathcal{C}=0).$$ Then - perhaps surprisingly - the identity map $id$ is not an isomorphism between them since

  • It doesn't "send $\color{red}{c_0}$ to $\color{red}{c_0}$" - we have $id(\color{red}{c_0}^\mathcal{C})=id(0)=0\color{green}{\not=}1=\color{red}{c_0}^\mathcal{D}$.

  • It doesn't "send $\color{red}{c_1}$ to $\color{red}{c_1}$" - we have $id(\color{red}{c_1}^\mathcal{C})=id(1)=1\color{green}{\not=}0=\color{red}{c_1}^\mathcal{D}$.

(Crucial inequalities highlighted in green.)

Being an isomorphism is a really really strict property: you have to really preserve all the "basic facts" about the structure exactly as written.

(Incidentally, $\mathcal{C}$ and $\mathcal{D}$ above are isomorphic e.g. via the map swapping $0$ and $1$ and leaving everything else unmoved - as an exercise, check that this is the case and then cook up an example of two structures, which differ only in "swapping" a pair of constant symbols, which are not isomorphic at all.)

  • $\begingroup$ Thank you for this very detailed answer. Can you explain what $\models$ means in your answer above (I have a hunch because the LaTeX code is "\models", LOL). I am $very$ new to model theory and have just completed an intro course to logic, so for me $A \models B$ just means that from $A$, I can (semantically) prove $B$. $\endgroup$
    – Squirtle
    Nov 30, 2017 at 22:24
  • $\begingroup$ @Squirtle "$\mathcal{M}\models\varphi$" means "$\varphi$ is true in the structure $\mathcal{M}$." This is used in the definition of semantic entailment of one set of sentences from another: "$A\models B$" iff every model of $A$ is also a model of $B$ iff for every $\mathcal{M}$, if $\mathcal{M}\models A$ then $\mathcal{M}\models B$. $\endgroup$ Nov 30, 2017 at 22:36

One good reason to specify constants in a signature for models is that you want to make sure that these are preserved by homomorphisms. Consider, as in your example, the signature of ordered fields $\{ +, -, \cdot, 0,1, \leq \}$. A homomorphism $f:A \to B$ of structures of this signature will always satisfy $f(0_A) = 0_B$, and similar for $1$, whereas this need not be the case for the signature $\{ +, -, \cdot, \leq \}$.

Moreover, you can refer to constants in formulas that are then quantifier-free. For example, consider a field $K$. Within the first signature, the formula $\varphi(x) = (x = 0)$ is quantifier-free, whereas it is not possible to express the same statement in the second signature without the use of quantifiers. This can be quite important when proving properties of certain classes of models.

  • $\begingroup$ This is a surprisingly good answer. So if we are talking about a group, I want to specify the identify because I want a group homomorphism to preserve the identity.... but what you are saying is more general it seems. It seems to me that you are saying that when we do model theory we are interested in specifying special elements that are preserved by $any$ homomorphism in $any$ sense of the word "homomorphism" (be it algebraic, vector space, etc.). $\endgroup$
    – Squirtle
    Nov 30, 2017 at 22:31

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