Bourbaki

Theorem 1

Tag 0BNF, permanent. Cite it as [Bourbaki, Tag 0BNF].

General Topology, chapter III, § 6. Topological groups with operators; topological rings division rings and fields

Let E, F, G be three complete Hausdorff commutative groups; let A be a dense subgroup of E and let B be a dense subgroup of F. If f is a Z\mathbf{Z} continuous Z\mathbf{Z}-bilinear (*) mapping of A×BA \times B into G, then f can be extended by continuity to a continuous Z\mathbf{Z}-bilinear mapping of E×FE \times F into G.

Let (x0,y0)(x_0, y_0) be an arbitrary point of E×FE \times F, and let U,B\mathfrak{U}, \mathfrak{B} be the traces on A, B respectively of the neighbourhood filters of x0,y0x_0, y_0 (U,B\mathfrak{U}, \mathfrak{B} are filters, by hypothesis). To show that f can be extended by continuity, it is enough to show that f(U×B)f(\mathfrak{U} \times \mathfrak{B}) is a Cauchy filter-base on G (Chapter II, § 3, no. 6, Proposition 11). Consider the identity

f(x′,y′)−f(x,y)=f(x′−x,y1)+f(x1,y′−y)+f(x′−x,y′−y1). f(x', y') - f(x, y) = f(x' - x, y_1) + f(x_1, y' - y) + f(x' - x, y' - y_1).

We shall show that by taking (x,y)(x, y) and (x′,y′)(x', y') in a sufficiently small set of U×B\mathfrak{U} \times \mathfrak{B}, and by choosing x1x_1 and y1y_1 suitably, we can make each of the terms on the right-hand side very small. Let W be any neighbourhood of 0 in G; since f is continuous at (0,0)∈A×B(0, 0) \in A \times B, there is a set U∈BU \in \mathfrak{B} and a set V∈BV \in \mathfrak{B} such that f(x′−x,y′−y)∈Wf(x' - x, y' - y) \in W whenever x∈U,x′∈U,y∈V,y′∈Vx \in U, x' \in U, y \in V, y' \in V. Take a point x1∈Ux_1 \in U and a point y1∈Vy_1 \in V; then for all x,x′x, x' in U and all y,y′y, y' in V we shall have

f(x′−x,y′−y1)+f(x−x1,y′−y)∈W+W. f(x' - x, y' - y_1) + f(x - x_1, y' - y) \in W + W.

On the other hand, the partial mapping x→f(x,y1)x \to f(x, y_1) is continuous on A; hence there is a set U′⊂UU' \subset U, belonging to U\mathfrak{U}, and such that, whenever xx and x′x' belong to U′U', we have f(x′−x,y1)∈Wf(x' - x, y_1) \in W. Likewise there exists V′⊂VV' \subset V belonging to B\mathfrak{B} such that, whenever yy and y′y' belong to V′V', we have f(x1,y′−y)∈Wf(x_1, y' - y) \in W. Consequently, if (x,y)(x, y) and (x′,y′)(x', y') are any two points of U′×V′U' \times V', then

f(x′,y′)−f(x,y)∈W+W+W+W; f(x', y') - f(x, y) \in W + W + W + W;

this proves the existence of the extension f‾\overline{f} of f. The fact that f‾\overline{f} is Z\mathbf{Z}-bilinear is an immediate consequence of the principle of extension of identities (Chapter I, § 8, no. 1, Corollary 1 to Proposition 2).

Q.E.D.

In the application of this theorem to a Hausdorff topological ring A, we take E, F and G to be A^\hat{A}, A and B to be the ring A, and f to be

(*) ff is said to be Z\mathbf{Z}-bilinear if, for all elements x,x′x, x' of A and all elements y,y′∈By, y' \in B we have

f(x+x′,y)=f(x,y)+f(x′,y) f(x + x', y) = f(x, y) + f(x', y)
and
f(x,y+y′)=f(x,y)+f(x,y′). f(x, y + y') = f(x, y) + f(x, y').

the Z\mathbf{Z}-bilinear mapping (x,y)→xy(x, y) \to xy, which by hypothesis is continuous. We denote again by xyxy the value of the extended function on A^×A^\hat{A} \times \hat{A}; this function is a law of composition on A^\hat{A}, and to say that it is Z\mathbf{Z}-bilinear means that it is distributive on both sides with respect to addition; and it is also associative, by the principle of extension of identities. Consequently:

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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