Proposition 6

Let $\mathfrak{F}$ and $\mathfrak{G}$ be two Cauchy filters on $G_d$. Then the image of the filter $\mathfrak{F} \times \mathfrak{G}$ under the mapping $(x, y) \to xy$ is a Cauchy filter base on $G_d$.

Let us evaluate the "proximity" of $xy$ and $x'y'$ in $G_d$; in other words, let us form the product $(x'y')(xy)^{-1} = x'y'y^{-1}x^{-1}$. For each $a \in G$, we can also write $(x'y')(xy)^{-1} = (x'a^{-1})(ay'y^{-1}a^{-1})(ax^{-1})$. We shall see that by suitable choice of $a$, each of the three factors of this product is very small whenever the pairs $(x, y)$ and $(x', y')$ belong to a sufficiently small set of $\mathfrak{F} \times \mathfrak{G}$. Let $V$ be any neighbourhood of $e$ in $G$; then there is a $V_d$-small set $A \in \mathfrak{F}$. Choose $a$ in $A$, then if $x$ and $x'$ are any two points of $A$, we have $x'a^{-1} \in V$ and $ax^{-1} \in V$. On the other hand, the relation $ay'y^{-1}a^{-1} \in V$ is equivalent to

and since $W$ is a neighbourhood of $e$, there is a $W_d$-small set $B \in \mathfrak{G}$. Hence for all $(x, y)$ and $(x', y')$ in $A \times B$, we have $(x'y')(xy)^{-1} \in V^3$, and this completes the proof.

In order that $x^{-1}$ can be extended by continuity to $\hat{G}_d$ it is necessary and sufficient that the image, under the symmetry $x \to x^{-1}$, of a Cauchy filter on $G_d$ is a Cauchy filter on $G_d$ (Chapter II, § 3, no. 6, Proposition 11). There are examples of topological groups in which this condition is not satisfied (cf. Chapter X, § 3, Exercise 16); we shall suppose that it is satisfied for the remainder of this proof.

2) The extended functions $xy$ and $x^{-1}$ define a group structure on $\hat{G}_d$. For if we apply the principle of extension of identities (Chapter I, § 8, no. 1, Proposition 2, Corollary 1) to the functions $x(yz)$ and $(xy)z$, defined on $\hat{G}_d \times \hat{G}_d \times \hat{G}_d$ and equal on the dense subspace $G_d \times G_d \times G_d$, we see that the law of composition $(x, y) \to xy$ is associative on $\hat{G}_d$. For the same reason, the functions $x, ex, xe$ are identical on $\hat{G}_d$, and the functions $e, xx^{-1}, x^{-1}x$ are identical on $\hat{G}_d$.

3) The topological group $\hat{G}_d$ is complete. Let $\mathfrak{U}_d$ be its right uniformity, and let $\mathfrak{U}$ be the uniformity on $\hat{G}_d$ obtained by completing the right uniformity of $G$. Then $\mathfrak{U}$ and $\mathfrak{U}_d$ induce the same uniformity on $G$, and therefore every Cauchy filter base $\mathfrak{B}$ on $G$ with respect to $\mathfrak{U}_d$ is also a Cauchy filter base with respect to $\mathfrak{U}$. Now $\mathfrak{B}$ converges in $\hat{G}_d$, since $\mathfrak{U}$ is a complete uniformity; and since $\mathfrak{U}$ and $\mathfrak{U}_d$ induce the same topology on $\hat{G}_d$, it follows (Chapter II, , no. 4, Proposition 9) that $\mathfrak{U}_d$ is a complete uniformity. This conclusion also shows that $\mathfrak{U}$ and $\mathfrak{U}_d$ coincide (Chapter II, § 3, no. 6, Corollary to Theorem 2)

4) Uniqueness. This follows from Proposition 5 of no. 3.

To sum up, we have proved the following theorem: