a) If the stabilizer of each point of $X_\alpha$ is a compact subgroup of $G_\alpha$ for each $\alpha \in I$, then the stabilizer of each point $x = (x_\alpha)$ of $X$ is a compact subgroup of $G$; the orbit of $x$ (with respect to $G$) is canonically homeomorphic to the inverse limit of the orbits of the $x_\alpha$ (with respect to $G_\alpha$); and the canonical mapping $h : X/G \to \varprojlim X_\alpha/G_\alpha$ is injective.

b) If, for each $\alpha \in I$, every orbit of a point of $X_\alpha$ (with respect to $G_\alpha$) is compact, then every orbit of a point of $X$ (with respect to $G$) is relatively compact, and $h$ is surjective. If also $h$ is bijective, then every orbit of a point of $X$ is compact.

Let $x = (x_\alpha) \in X$, and for each $\alpha \in I$ let $X'_\alpha = G_\alpha \cdot x_\alpha$ be the orbit of $x_\alpha$. If $\alpha \leq \beta$, it follows from (i) and the relation $g_{\alpha\beta}(x_\beta) = x_\alpha$ that $g_{\alpha\beta}(X'_\beta) \subset X'_\alpha$, and hence that $(X'_\alpha)$ is an inverse system of subsets of the $X_\alpha$. For each $\alpha \in I$ let $u_\alpha : G_\alpha \to X'_\alpha$ be the continuous mapping $s_\alpha \to s_\alpha \cdot x_\alpha$; the $u_\alpha$ form an inverse system of mappings, and $u = \varprojlim u_\alpha$ is the continuous mapping $s \to s.x$ of $G$ into the subspace $X' = \varprojlim X'_\alpha$ of $X$. The hypothesis of a) implies that $\overline{u}_\alpha(y_\alpha)$ is compact for each $y_\alpha \in X'_\alpha$. Since also $u_\alpha$ is surjective, the conditions of Chapter I, § 9, no. 6, Proposition 8, Corollary 2 are satisfied, and the first two assertions of a) are therefore established. Hence, if $x = (x_\alpha)$ and $y = (y_\alpha)$ are such that $x_\alpha$ and $y_\alpha$ belong to the same orbit with respect to $G_\alpha$ for each $\alpha \in I$, then $x$ and $y$ belong to the same orbit with respect to $G$, and therefore $h$ is injective.

Likewise, the hypothesis of b) implies that the inverse system of canonical mappings $v_\alpha : X_\alpha \to X_\alpha/G_\alpha$ satisfies the conditions of Chapter I, § 9, no. 6, Proposition 8, Corollary 2, and therefore its inverse limit $v = \varprojlim v_\alpha : X \to \varprojlim X_\alpha/G_\alpha$ is surjective, and the inverse image under v of every point of $\varprojlim X_\alpha / G_\alpha$ is compact. Since v factorizes into

where $\psi$ is the canonical mapping, the assertions of b) follow.

For the hypotheses of a) and b) are satisfied, since every closed subgroup of $G_\alpha$ is compact and $u_\alpha : s_\alpha \to s_\alpha \cdot x_\alpha$ is a continuous mapping of a compact space into a Hausdorff space.

For hypothesis a) is satisfied, and $X'_\alpha = X_\alpha$ for each $\alpha$.

The fact that $h$ is bijective follows from Corollary 1, applied by replacing $X_\alpha$ by $G_\alpha$, and $G_\alpha$ by $K_\alpha$ operating by right translations (§ 2, no. 5). Let $\varphi$ be the canonical mapping $G \to G/K$, and for each $\alpha$ let $f_\alpha$ be the canonical mapping $G \to G_\alpha$. If, for each $\alpha$, $V_\alpha$ runs through a fundamental system of open neighbourhoods of the identity element $e_\alpha$ of $G_\alpha$, then the sets $V = \overline{f}_\alpha^{-1}(V_\alpha)$ ($\alpha$ and $V_\alpha$ variable) form a fundamental system of neighbourhoods of the identity element $e$ in $G$ (Chapter I, § 4, no. 4, Proposition 9), and the sets $\varphi(V.K)$ form a fundamental system of neighbourhoods of $\varphi(e)$ in $G/K$. We have to show that the image of $\varphi(V.K)$ under $h$ contains a neighbourhood of $h(\varphi(e))$, that is that there exists $\beta \geq \alpha$ and a neighbourhood $W_\beta$ of $e_\beta$ in $G_\beta$ such that $\overline{f}_\beta^{-1}(W_\beta . K_\beta) \subset V.K$. Now, the relation $x \in V.K$ is equivalent to the existence of $y$ in $K$ such that $f_\alpha(xy^{-1}) \in V_\alpha$, i.e. $f_\alpha(x) \in V_\alpha \cdot f_\alpha(K)$; so that $V.K = \overline{f}_\alpha^{-1}(V_\alpha \cdot f_\alpha(K))$. Let $U_\alpha = V_\alpha \cdot f_\alpha(K)$; we shall see that there exists $\beta \geq \alpha$ such that if we put $U_\beta = \overline{f}_{\alpha \beta}^{-1}(U_\alpha)$, we have $K_\beta \subset U_\beta$; it will then follow that there is a neighbourhood $W_\beta$ of $e_\beta$ in $G_\beta$ such that $W_\beta . K_\beta \subset U_\beta$ (Chapter II, § 4, no. 3, Corollary to Proposition 4), and this will establish the desired relation

We proceed by reductio ad absurdum: for each $\beta \geq \alpha$, let $M_\beta$ denote $K_\beta \cap G U_\beta$; since $f_{\beta \gamma}^{-1}(U_\beta) = U_\gamma$ if $\alpha \leq \beta \leq \gamma$, the $M_\beta$ form an inverse system of compact subsets of the $G_\beta$ (for $\beta \geq \alpha$). If they were all non-empty, then their inverse limit $M$ would also be non-empty (Chapter I, § 9, no. 6, Proposition 8). It is clear that $M \subset K$ and $f_\alpha(M) \subset M_\alpha$; but this is absurd since $f_\alpha(K) \subset U_\alpha$, and the proof is therefore complete.

(AP) For each $\alpha \in I$, $H_\alpha$ is closed in $G$ and every neighbourhood of the identity element $e$ in $G$ contains one of the $H_\alpha$ (in other words, the filter base formed by the $H_\alpha$ converges to $e$).

Then the mapping $i : G \to \tilde{G} = \varprojlim G / H_\alpha$ is a strict morphism of $G$ onto $i(G)$; $\tilde{G}$ is Hausdorff and $i(G)$ is dense in $\tilde{G}$; and finally the kernel of $i$ is the closure of $\{e\}$ in $G$. If in addition one of the $H_\alpha$ is complete, then $i$ is surjective.

Clearly the $G_\alpha = G / H_\alpha$ are Hausdorff (§ 2, no. 5, Proposition 13), hence so is $\tilde{G}$ (since it is a subspace of $\prod_{\alpha \in I} G_\alpha$). The kernel $H$ of $i$ is the intersection of the $H_\alpha$ and is therefore a closed subgroup of $G$. Since each neighbourhood of $e$ contains some $H_\alpha$, it contains $H$ and therefore [§ 3, no. 1, formula (1)] $H$ is the closure of $\{e\}$. Let us next show that $i(G)$ is dense in $\tilde{G}$. Let $f_\alpha$ be the canonical mapping $\tilde{G} \to G_\alpha$, which is the restriction to $\tilde{G}$ of the projection $pr_\alpha$; $\varphi_\alpha = f_\alpha \circ i$ is the canonical mapping $G \to G/H_\alpha$. If $U$ is any non-empty open set of $\tilde{G}$, then there is an index $\alpha \in I$ and a non-empty open set $U_\alpha$ in $G_\alpha$ such that $\overline{f}_\alpha^{-1}(U_\alpha) \subset U$ (Chapter I, § 4, no. 4, Proposition 9); therefore

but since $\varphi_\alpha$ is surjective, $\overline{i}^{-1}(U)$ is not empty, and therefore $i(G) \cap U \neq \emptyset$.

To see that $i$ is a strict morphism of $G$ onto $i(G)$, consider a neighbourhood $V$ of $e$ in $G$. There is a neighbourhood $W$ of $e$ in $G$ such that $W^2 \subset V$, and an index $\alpha \in I$ such that $H_\alpha \subset W$; it follows that $V$ contains $WH_\alpha = \overline{\varphi}_\alpha(\varphi_\alpha(W)) = \overline{i}^{-1}(\overline{f}_\alpha^{-1}(\varphi_\alpha(W)))$. Since $\overline{f}_\alpha(\varphi_\alpha(W))$ is a neighbourhood of the identity element in $\tilde{G}$, the result follows.

Finally, suppose that there is an index $\gamma \in I$ such that $H_\gamma$ is complete. To show that $i$ is surjective it is enough to prove that every family $(T_\alpha)_{\alpha \in I} \in \tilde{G}$ has a non-empty intersection. Since $T_\gamma$ is obtained by translation from $H_\gamma$, it is a complete subspace of $G$ (with respect to both the right and the left uniformities). Moreover, since every neighbourhood $U$ of $e$ in $G$ contains one of the $H_\alpha$, the corresponding set $T_\alpha$ is $U_{d-}$ (or $U_{s-}$) small, and hence the set of $T_\alpha$ contained in $T_\gamma$ is a Cauchy filter base; therefore it converges in $T_\gamma$, and since the $T_\alpha$ are closed in $G$ (since they are translations of the $H_\alpha$), their intersection is not empty.

Q.E.D.

For $\tilde{G}$ is then complete (no. 2), and Proposition 2 shows that $i(G)$ is isomorphic with the Hausdorff group associated with $G$; since it is dense in $\tilde{G}$, the corollary follows ($§ 3$, no. 3, Proposition 5). In particular:

The subgroup $H_\alpha$ of $G$ is open and therefore also closed ($\S$ 2, no. 1, Corollary to Proposition 4), and $G/H_\alpha$ is discrete ($\S$ 2, no. 6, Proposition 18); hence the conditions of Corollary 1 are satisfied.

For the remainder of this sub-section we shall assume that $G$ is Hausdorff and that $(H_\alpha)$ is a family of compact normal subgroups of $G$, which is directed with respect to the relation $H_\alpha \supset H_\beta$ and which satisfies the condition (AP); by virtue of Proposition 2, the mapping

is then an isomorphism of topological groups which permits us to identify $G$ and $\tilde{G}$. We denote by $f_\alpha$ the canonical mapping $G \to G/H_\alpha$.

For $E$ is the intersection of the sets $EV$ where $V$ runs through the neighbourhood filter of $e$ [$\S$ 3, no. 1, formula (1)], and every neighbourhood of $e$ contains an $H_\alpha$; whence the result, since $E \subset EH_\alpha$.

a) Since $H_\alpha$ is compact, $LH_\alpha$ is closed in $G$ ($\S$ 4, no. 1, Proposition 1, Corollary 1) and therefore $L_\alpha$ is closed in $G_\alpha$. Since $i$ identifies the topological groups $G$ and $\varprojlim G_\alpha$, and since $\varprojlim L_\alpha$ may be identified with a (topological) subgroup of $\varprojlim G_\alpha$, $i$ identifies the subgroup $\bigcap_\alpha LH_\alpha$ of $G$ with $\varprojlim L_\alpha$, and to prove the first assertion it is enough to remark that $L = \bigcap_\alpha LH_\alpha$ by Lemma 1. On the other hand, if $L$ is normal, then for each $\alpha \in I$ the mapping $f'_\alpha : G/L \to G_\alpha/L_\alpha$ induced by $f_\alpha$ is a surjective strict morphism ($\S$ 2, no. 8, Remark 3), whose kernel is the compact normal subgroup $H_\alpha L/L$ of $G/L$, the canonical image of the compact subgroup $H_\alpha$ of $G$. Since the subgroups $H_\alpha L/L$ of $G/L$ satisfy condition (AP) (§ 2, no. 8, Proposition 24), and since $G/L$ is Hausdorff, the last assertion of a) is a consequence of Proposition 2.

b) Let $f'_{\alpha \beta}$ be the restriction of $f_{\alpha \beta}$ to $L_\beta$ ($\alpha \leq \beta$). Then $(L_\alpha, f'_{\alpha \beta})$ is an inverse system of topological groups, whose inverse limit $L$ can be identified with the subgroup $G \cap \prod_\alpha L_\alpha$ of $G$. By hypothesis, $f'_{\alpha \beta}$ is surjective and its kernel is the compact subgroup $f_\beta(H_\alpha) \cap L_\beta$ of $L_\beta$; consequently (Chapter I, § 9, no. 6, Corollary 1 to Proposition 8) we have $L_\alpha = f_\alpha(L)$ for all $\alpha \in I$. If $L'$ is another closed subgroup of $G$ such that $f_\alpha(L') = L_\alpha$ for all $\alpha \in I$, then $L'H_\alpha = \overline{f'}_\alpha(L_\alpha)$, whence (Lemma 1) $L' = \bigcap_\alpha L'H_\alpha = \bigcap_\alpha \overline{f'}_\alpha(L_\alpha) = L$. Finally, the last assertion of b) follows from the formula $L = \bigcap_\alpha \overline{f'}_\alpha(L_\alpha)$, since the $\overline{f'}_\alpha(L_\alpha)$ are now normal subgroups of $G$.

The proposition is a consequence of the following lemma:

Once this lemma is established, we shall have $f_\alpha(C) = C_\alpha$ for each $\alpha \in I$, and since $C$ is a closed subgroup of $G$ (§ 2, no. 2, Proposition 7) it suffices to apply Proposition 3 a).

To prove the lemma, notice first that if $C'$ is the component of the identity element $e'$ in $G'$, then $\varphi(C) \subset C'$ since $\varphi(C)$ is connected. Suppose that $\varphi(C) \neq C'$. Since $C$ is a closed normal subgroup of $G$ (§ 2, no. 2, Proposition 7), $\varphi(C)$ is a normal subgroup of $G'$; if $\psi$ is the canonical mapping $G' \to G'/\varphi(C)$, then $\psi(C')$ is connected and does not consist of the identity element alone; hence the identity component of $G'/\varphi(C)$ does not consist of the identity element alone. But $G'/\varphi(C)$ is isomorphic to $(G/H)/(HC/H)$, hence to $G/HC$, and consequently also to $(G/C)/(HC/C)$ (§ 2, no. 7, Corollary to Proposition 22, and Proposition 20). Now, $G/C$ is Hausdorff and totally disconnected (Chapter I, § 11, no. 5, Proposition 9), and $HC/C$, being the canonical image of the compact normal subgroup $H$ of $G$, is a compact subgroup of $G/C$. It is therefore sufficient to prove Lemma 2 under the additional hypothesis that $G$ is totally disconnected, i.e. $C = \{e\}$.

Suppose then that $C' \neq \{e'\}$; replacing $G$ by its subgroup $\overline{f'}(C')$, which is totally disconnected and contains $H$, we may suppose that $G'$ is connected and does not consist of a single point.

Let $\mathfrak{M}$ be the set of closed subgroups $L$ of $G$ such that $LH = G$. We shall show that the set $\mathfrak{M}$, ordered by the relation $\supset$, is inductive. Indeed, if $\mathfrak{T}$ is a linearly ordered subset of $\mathfrak{M}$, then for each $x \in G$ the set of sets $xH \cap L$ with $L \in \mathfrak{T}$ is a filter base formed of closed sets in the compact space $xH$; hence the intersection of these sets is not empty, which shows that the intersection of the subgroups $L \in \mathfrak{T}$ belongs to $\mathfrak{M}$. Applying Zorn's lemma, we see therefore that $\mathfrak{M}$ has a minimal element $L_0$. Since $H$ is compact, $G/H = L_0H/H$ is isomorphic to $L_0/(L_0 \cap H)$ (§ 4, no. 1, Proposition 1, Corollary 3); since $L_0$ is totally disconnected and $L_0 \cap H$ is compact, we see that we may replace $G$ by $L_0$; in other words, we may suppose in addition that there is no closed subgroup $L \neq G$ such that $LH = G$.

Now let $F$ be the intersection of the neighbourhoods of the identity element which are both open and closed in $G$, and let us show that $F$ is a closed subgroup of $G$. Clearly $F$ is closed in $G$; hence it is enough to show that $F^{-1}F \subset F$. But if $x \in F$ and if $V$ is an open and closed neighbourhood of $e$ in $G$, then so is $xV$; for otherwise $e$ would belong to the complement $W$ of $xV$ in $G$, which is again open and closed, and we should have $x \notin W$; hence $x \notin F$, contrary to hypothesis. It follows that $xF$, the intersection of the sets $xV$ as $V$ runs through the open and closed neighbourhoods of $e$ in $G$, contains $F$; in other words, $x^{-1}F \subset F$, which proves our assertion. Since $G$ is totally disconnected and $G \neq \{e\}$, we have $F \neq G$. But if $V$ is an open and closed neighbourhood of $e$ in $G$, then $VH$ is also open and closed in $G$ (§ 4, no. 1, Corollary 1 to Proposition 1), hence $\varphi(V)$ is both open and closed in $G/H$, and this implies that $\varphi(V) = G/H$ by virtue of the hypothesis. We shall conclude from this that $FH = G$, which will give us the desired contradiction and complete the proof of the lemma. Indeed, for each $x \in G$, $xH$ meets every neighbourhood $V$ of $e$ which is both open and closed, hence also meets the intersection $F$ of these neighbourhoods, since the sets $Vx \cap H$ form a filter base consisting of closed sets in the compact space $xH$.

Q.E.D.