Let $\mathcal{L}$ be a language. In David Marker's Model Theory: An Introduction, the set of $\mathcal{L}$-terms is defined to be the smallest set $\mathcal{T}$ containing all of the constant symbols of $\mathcal{L}$, the variable symbols $v_1$, $v_2$,… and such that for each function symbol $f$ (in $\mathcal{L}$) and $t_1,…,t_{n_f}∈\mathcal{T}$ (where $n_f$ is the positive integer associated to $f$), we must have that $f(t_1,…,t_{n_f})∈\mathcal{T}$.
Usually when one enlarges a set by including more and more points until it is in some way 'complete', it is contained in some larger set (e.g. smallest topology containing a given collection of subsets), which ensures that there are no weird 'Russell's Paradox' type issues, such as the final construction being a proper class, rather than a set.
But here there is no suggestion of any larger set. I can see that it follows by induction that any string that can be obtained by a finite number of applications of function symbols to the constant and variable symbols must be a set. Must these strings, by definition, only involve a finite nesting of function symbols i.e. $(v_1\cdot v_2)\cdot v_3$ or can it contain infinite ones such as $(((v_1\cdot v_2)\cdot v_3)\cdot v_4\cdot….))))$?
Many thanks
Best Answer
Upgrading my comments to an answer. I agree that Marker's definition appears slightly vague at first glance; what does the "smallest set" with that property mean? Let me flesh out this kind of construction a bit more generally. Given some property $P$, we often want to construct a set $\mathcal{T}$ that is the "smallest" set satisfying $P$. We mean by this that we want the following two things to hold:
Note that such a set is necessarily unique; indeed, suppose $\mathcal{T}'$ is another set satisfying both of these conditions. Since $\mathcal{T}$ satisfies condition $2$ and $\mathcal{T}'$ satisfies condition $1$, we have $\mathcal{T}\subseteq\mathcal{T}'$, and since $\mathcal{T}'$ satisfies condition $2$ and $\mathcal{T}$ satisfies condition $1$, we have $\mathcal{T}'\subseteq\mathcal{T}$. In other words, $\mathcal{T}=\mathcal{T}'$ as claimed.
Now, how do we actually "construct" this set $\mathcal{T}$? The idea is to take $\mathcal{T}$ to be the intersection of all sets satisfying property $P$. Now, this is not necessarily well-defined; for instance, if there does not exist a set satisfying $P$, then we cannot do this, since the intersection of the empty set is not defined. (Why not?) So we need for there to exist at least one set satisfying $P$; in practice, there is often a somewhat "natural" choice for this. Suppose for the purpose of the discussion that we have a set $X$ in mind that satisfies property $P$. By condition $2$ above, our desired set $\mathcal{T}$ will necessarily be a subset of $X$. So, in order to construct $\mathcal{T}$, it suffices to think exclusively about subsets of $X$. In particular, we can define $\mathcal{T}$ as the intersection of all subsets of $X$ that satisfy property $P$.
Now, does $\mathcal{T}$ satisfy condition $1$? Unfortunately, not necessarily; for instance, if $P$ is the property of being non-empty, then the intersection of all subsets of (say) $\mathbb{N}$ satisfying $P$ will itself be empty and hence not satisfy $P$. There is an implicit assumption lurking in the background; namely, that $P$ is stable under intersections. In other words, if $(A_i)_{i\in I}$ is a family of sets satisfying $P$, then $\bigcap_{i\in I}A_i$ also satisfies $P$. So, suppose $P$ is stable under intersections. Then we immediately have that $\mathcal{T}$ satisfies property $P$, since by definition it is an intersection of sets satisfying $P$! So we're done with condition $1$.
Does $\mathcal{T}$ also satisfy condition $2$? Indeed; let $Y$ be any set satisfying $P$. Then $Y\cap X$ satisfies $P$, since $P$ is stable under intersections, and $Y\cap X$ is a subset of $X$. Thus, by definition – since $\mathcal{T}$ is the intersection of all subsets of $X$ satisfying $P$ – we have $\mathcal{T}\subseteq Y\cap X$ and hence $\mathcal{T}\subseteq Y$, as desired. So we are done!
Now let's specialize to the example of Marker. In this case, $P$ is defined by taking $Y$ to satisfy $P$ if and only if $(1)$ $v_i\in Y$ for each $i$, $(2)$ $c\in Y$ for each constant symbol $c\in\mathcal{L}$, and $(2)$ $f(t_1,\dots,t_{n_f})\in Y$ for each function symbol $f\in\mathcal{L}$ and all elements $t_i\in Y$.
Exercise: Show that $P$ is stable under intersections, in the sense defined above. $\blacksquare$
Now, to define $\mathcal{T}$ by the process above, we first need to find some set $X$ satisfying $P$. First define $\mathcal{L}'$ to be the union of $\mathcal{L}$ with the set of variables symbols and the symbols "$($", "$)$", and "$,$". Now let $X$ be the set of all finite strings (ie finite sequences) of symbols of $\mathcal{L}'$.
Exercise: Show that $X$ satisfies property $P$. $\blacksquare$
Once you have shown these two exercises, you are done! Since we can now define $\mathcal{T}$ to be the intersection of all subsets of $X$ that satisfy property $P$. In particular however, note that $\mathcal{T}\subseteq X$. Thus every element of $\mathcal{T}$ is a finite string of symbols of $\mathcal{L}'$, and so, to answer your question – no, there are no elements of $\mathcal{T}$ with infinite nesting as you describe.