The smallest set of well formed formulas in Tourlakis' Mathematical Logic In page 13 of Mathematical Logic, Tourlakis defines the set of all well-formed-formulae as:

... the smallest set of strings, WFF, that satisfies: 1) All Boolean variables are in WFF, and so are the symbols ⊤ and ⊥. 2) If A and B are strings in WFF, then so are the strings (¬A), (A∧B), (A∨B), (A→B), (A≡B).

Capture from the book
But as I understand it, the set of Boolean variables is infinite.
So, how can a set be a smallest one when the cardinality is infinite?
In this question somebody asked about it and the answer was

The "smallest set" condition there is crucial: if we omit it, there are lots of sets satisfying the definition

and I would go even further as to say, indeed there are infinite sets that satisfy the condition of being sets of Well Formed formulas, it's not unique because I can always find an element (Boolean variable) that belongs to another set of WFF.
 A: The natural numbers $\mathbb{N} = \{0,1,2,3,\ldots\}$ is the smallest set $I$ with the property that $0 \in I$ and whenever $n \in I$ then $n+1 \in I$ ($I$ is an inductive set). What do we mean when we say 'smallest' though? In this context we aren't talking about cardinality, but about set containment. So saying that $\mathbb{N}$ is the smallest set with the aforementioned property means that if $I$ were another such set then $\mathbb{N} \subseteq I$.
This is the same sort of thing with $\mathsf{WFF}$ - we specify some initial set of things that must belong to it (the Boolean variables and $\bot,\top$ in this case) and then require that it be closed under a certain set of operations (in this case, the Boolean connectives followed by surrounding with parentheses). $\mathsf{WFF}$ is the smallest (with respect to containment, not cardinality) set with the aforementioned properties.
How do we know there is such a set? One can show that the intersection of all such sets (either all inductive sets in the case of $\mathbb{N}$, or all sets containing the Boolean variables and so on in the case of $\mathsf{WFF}$) also has the corresponding properties. Since we took the intersection of every set with those properties, it must be the smallest in terms of containment.
