As to the right translations, the result is clear, since the relation $yx^{-1} \in V$ is equivalent to $(ya)(xa)^{-1} \in V$ [in other words, the mapping $(x, y) \to (xa, ya)$ leaves $V_d$ fixed]. For the left translations, the result follows from $(GV_{\text{III}})$; for $yx^{-1} \in V$ if and only if $(ay)(ax)^{-1} \in aV a^{-1}$; hence $x \to ax$ is uniformly continuous on $G_d$.

Similarly, the right and left translations are isomorphisms of the left uniformity onto itself.

Every inner automorphism $x \to axa^{-1}$ of $G$ is therefore an automorphism for the group structure of $G$, for the topology of $G$, and for both the uniformities of $G$.

This is an immediate consequence of Definition 1.

The reader should beware of supposing that the mapping $(x, y) \to xy$ of the uniform space $G_d \times G_d$ into the uniform space $G_d$ is in general uniformly continuous. Similarly, the symmetry $x \to x^{-1}$, considered as a mapping of $G_d$ onto $G_d$, is not in general uniformly continuous (see Exercises 3 and 4).

For if $V'$ is a neighbourhood of the identity element in $G'$, and $V = \overline{f}^{-1}(V')$, then the relation $yx^{-1} \in V$ implies

Suppose for example that $V$ is complete with respect to the right uniformity, and let $\mathfrak{F}$ be a Cauchy filter on $G_d$; then $\mathfrak{F}$ contains a $V_d$-small set $M$, and if $x_1 \in M$ we therefore have $M \subset Vx_1$. Hence the trace of $\mathfrak{F}$ on the complete subspace $Vx_1$ of $G_d$ is a Cauchy filter, which converges to a point $x_0$; since $x_0$ is a cluster point of $\mathfrak{F}$, it is a limit of $\mathfrak{F}$ (Chapter II, § 3, no. 2, Corollary 2 to Proposition 5).

For every compact space is complete with respect to its unique uniformity (Chapter II, § 4, no. 1, Theorem 1).

For every complete subspace of a Hausdorff uniform space is closed (Chapter II, , no. 4, Proposition 8)

$\bar{u}$ is uniformly continuous with respect to the right uniformities of $H_1$ and $H_2$ (no. 1, Proposition 3), hence admits a unique extension to a mapping $\bar{u}$ of $G_1$ into $G_2$ which is uniformly continuous with respect to the right uniformities of these groups (Chapter II, § 3, no. 6, Theorem 2). Moreover, by virtue of the principle of extension of identities (Chapter I, § 8, no. 1, Corollary 1 to Proposition 2), $\bar{u}$ is a homomorphism of $G_1$ into $G_2$, whence the first assertion of the Proposition. To prove the second assertion, it is enough to consider the isomorphism $v$ of $H_2$ onto $H_1$, which is the inverse of $u$, and its extension $\bar{v}$ to a continuous homomorphism of $G_2$ into $G_1$; by reason of the uniqueness of the extension, $\bar{v} \circ \bar{u}$ and $\bar{u} \circ \bar{v}$ are the identity mappings of $G_1$ and $G_2$ respectively, and therefore (Set Theory, R, § 2, no. 12) $\bar{u}$ is bijective.

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:

Since $\hat{G}$ is regular, every neighbourhood of the identity element in $\hat{G}$ contains the closure $V$ of an open neighbourhood $U$ of $e$ in $\hat{G}$, and $V$ is also the closure of the trace of $U$ on $G$.

Let $G$ be a group which is not necessarily Hausdorff; let $N = \overline{\{e\}}$, and let $G' = G/N$ be the Hausdorff group associated with $G$ (§ 2, no. 6). If $G'$ has a completion $\hat{G}'$, this completion is called the Hausdorff completion of $G$ and is denoted by $\hat{G}$; $\hat{G}'_d$ (resp. $\hat{G}'_s$) is then the Hausdorff completion (Chapter II, § 3, no. 7) of the uniform space $G_d$ (resp. $G_s$).

Since the kernel of $u$ is closed and contains $e$, it contains $N$, and hence $u$ can be written as $u = w \circ \psi$, where $w$ is a continuous homomorphism of $G'$ into $H$; now apply Proposition 5 of no. 3 to $w$.

The uniform continuity of $x^{-1}$ follows from Proposition 2 of no. 1, and that of $xy$ from Proposition 3 of no. 1, since $(x, y) \to xy$ is a continuous homomorphism of $G \times G$ into $G$. If $G$ is Hausdorff, it satisfies the condition of Theorem 1 of no. 4 (as does every Hausdorff group whose left and right uniformities coincide); moreover the functions $xy$ and $yx$ are equal on $\hat{G} \times \hat{G}$ by the principle of extension of identities; hence the second part of the theorem, by considering in the general case the Hausdorff group associated with $G$.

It follows in particular from this theorem that if $f$ and $g$ are two uniformly continuous mappings of a uniform space $X$ into a commutative group $G$, written additively, then the functions $-f$ and $f + g$ are uniformly continuous.

Suppose that $G$ is written additively. Let $\mathfrak{U}_1$ be the uniformity on $G$ corresponding (no. 1) to the topology $\mathfrak{T}_1$: it will suffice to show that if $\mathfrak{F}$ and $\mathfrak{F}'$ are two minimal Cauchy filters (Chapter II, § 3, no. 2) with respect to $\mathfrak{U}_1$, which converge in $G_2$ to the same point $a$, then $\mathfrak{F} = \mathfrak{F}'$ (Chapter II, § 3, no. 7). For this, it is enough to show that $\mathfrak{F} \cap \mathfrak{F}'$ is a Cauchy filter with respect to $\mathfrak{U}_1$. Let $V$ be a neighbourhood of $o$ in $G$ with respect to $\mathfrak{T}_1$, such that $V$ is closed in $\mathfrak{T}_2$, and let $W$ be a symmetric neighbourhood of $o$ in $G$ with respect to $\mathfrak{T}_1$, such that $W + W \subset V$. By hypothesis, there is a $W_d$-small set $M$ (resp. $M'$) in $\mathfrak{F}$ (resp. $\mathfrak{F}'$); if $x \in M$ and $y \in M$ we have $y - x \in W$, i.e. $y \in x + W$. If $\overline{W}$ and $\overline{V}$ are the closures of $W$ and $V$ in $G_2$, it follows that $y \in x + \overline{W}$, and therefore, since $a$ is in the closure of $M$, that $a \in x + \overline{W}$ for each $x \in M$. Similarly, $a \in x' + \overline{W}$ for each $x' \in M'$, and hence $x - x' \in \overline{W} + \overline{W}$; but since $(x, y) \to x + y$ is a continuous mapping of $G_2 \times G_2$ into $G_2$, we have $\overline{W} + \overline{W} \subset \overline{W} + \overline{W} \subset \overline{V}$. It follows that if $x \in M$ and $x' \in M'$, then $x - x' \in \overline{V} \cap G = V$, because $V$ is closed in $\mathfrak{T}_2$; and this completes the proof.

If $A_1$ is the closure of $A$ in $G_1$, then $f(A_1)$ is contained in the closure of $A$ in $G_2$, which by hypothesis is equal to $A$. Since $f(A) = A$ by definition and $f$ is injective, we have $A_1 = A$.

The sets of $\mathfrak{B}$ are closed in the topology $\mathfrak{T}_2$, hence complete for the uniformity $U_1$ by Corollary 1; the result therefore follows from Proposition 4 of no. 3.