Read-back: the free group on any set is residually finite

The statement

Let $\alpha$ be any set whatsoever (any type, in any universe; it may be empty, finite, or infinite of any cardinality). Let $F(\alpha)$ denote the free group on $\alpha$, with its usual group structure. Then

where the intersection ranges over all normal subgroups $N$ of $F(\alpha)$ of finite index.

Because the intersection of subgroups is again a subgroup and always contains $1$, this is equivalent to the elementwise form:

or, contrapositively: for every $g \in F(\alpha)$ with $g \ne 1$ there exists a normal subgroup $N$ of $F(\alpha)$ of finite index with $g \notin N$.

No hypothesis is placed on $\alpha$; the only binder is $\alpha$ itself, universally quantified.

What each ingredient means, spelled out

The free group $F(\alpha)$. Its elements are equivalence classes of finite words in the alphabet $\alpha \times \{\text{true}, \text{false}\}$, i.e. finite lists of pairs $(x, b)$ with $x \in \alpha$ and $b$ a boolean, where $(x,\text{true})$ is read as the letter $x$ and $(x,\text{false})$ as $x^{-1}$. Two words are identified when one can be obtained from the other by a finite chain (in either direction) of single reduction steps, a single step being the deletion of an adjacent pair $(x,b)\,(x,\neg b)$ from anywhere inside the word. The group operations are: identity $=$ the class of the empty word; product $=$ the class of the concatenation of representatives; inverse $=$ the class of the reversed word with every boolean flipped. When $\alpha$ is empty, $F(\alpha)$ is the trivial group. The statement quantifies over this case too.

Subgroup of finite index. For a subgroup $H \le G$, the index of $H$ is defined as the number of left cosets $gH$ of $H$ in $G$, encoded as a natural number by the convention that the number is $0$ when the set of cosets is infinite (the encoding sends any infinite cardinality to $0$). "$H$ has finite index" is then, by definition, the assertion that this natural number is nonzero. Since the set of cosets is nonempty (it contains $H$ itself), this is exactly the assertion that $G/H$ is a finite set, i.e. that $H$ has finite index in the ordinary sense.

Normal subgroup. $N \le G$ is normal when $g n g^{-1} \in N$ for every $n \in N$ and every $g \in G$.

The family being intersected. The intersection is indexed by the collection of all pairs (subgroup $N$ of $F(\alpha)$, proof that $N$ is normal and of finite index); two such data with the same underlying subgroup are the same index. So it is literally the intersection of the underlying sets of all finite-index normal subgroups. This family is never empty: the whole group $F(\alpha)$ is normal in itself and has index $1$, so the intersection is over a nonempty family and equals the set of elements lying in every finite-index normal subgroup.

The right-hand side. $\{1\}$ is the trivial subgroup, whose underlying set is exactly $\{g : g = 1\}$. The equality is an equality of subgroups of $F(\alpha)$, which (subgroups being determined by their underlying sets) is the same as equality of the two underlying sets.

Edge cases the quantifier includes

• $\alpha = \varnothing$: $F(\alpha)$ is trivial, every subgroup is $\{1\}$, and the claim reduces to $\{1\} = \{1\}$.
• $\alpha$ a singleton: $F(\alpha)$ is infinite cyclic; the claim is that the intersection of all finite-index subgroups of $\mathbb{Z}$ (all subgroups are normal) is trivial.
• $\alpha$ of arbitrary infinite cardinality: no finiteness or countability assumption is made on $\alpha$; the claim is asserted for free groups of every rank.