Bourbaki

Proposition 2

Tag 0C1H, permanent. Cite it as [Bourbaki, Tag 0C1H].

General Topology, chapter IX, § 1. Generation of a uniformity by a family of pseudometrics; uniformizable spaces

If a uniformity U\mathscr{U} on XX has a countable fundamental system of entourages, then there is a pseudometric ff on XX such that U\mathscr{U} is identical with the uniformity defined by ff.

Let (Vn)(V_n) be a countable fundamental system of entourages of U\mathscr{U}. Define inductively a sequence (Un)(U_n) of symmetric entourages of U\mathscr{U} such that U1⊂V1U_1 \subset V_1 and

⋃n+13⊂Un∩Vnforn≥1. \bigcup_{n+1}^3 \subset U_n \cap V_n \quad \text{for} \quad n \geq 1.
Clearly (Un)(U_n) is another fundamental system of entourages of U\mathscr{U}, and we have in particular ⋃n+13⊂Un\bigcup_{n+1}^3 \subset U_n for n≥1n \geq 1. We define a real-valued function gg on X×XX \times X as follows: g(x,y)=0g(x, y) = 0 if (x,y)∈Un(x, y) \in U_n for all nn; g(x,y)=2−kg(x, y) = 2^{-k} if (x,y)∈Un(x, y) \in U_n for 1≤n≤k1 \leq n \leq k, but (x,y)∉Uk+1(x, y) \notin U_{k+1}; g(x,y)=1g(x, y) = 1 if (x,y)∉U1(x, y) \notin U_1. The function gg is symmetric and positive, and we have g(x,x)=0g(x, x) = 0 for all x∈Xx \in X. Put
f(x,y)=inf⁡∑i=0p−1g(zi,zi+1), f(x, y) = \inf \sum_{i=0}^{p-1} g(z_i, z_{i+1}),
the greatest lower bound being taken over the set of all finite sequences (zi)0≤i≤p(z_i)_{0 \leq i \leq p} (pp arbitrary) such that z0=xz_0 = x and zp=yz_p = y. We shall show that ff is a pseudometric which satisfies the inequalities
12g(x,y)≤f(x,y)≤g(x,y). \frac{1}{2} g(x, y) \leq f(x, y) \leq g(x, y).
It follows immediately from the definition that ff is symmetric and positive and satisfies the triangle inequality. Also it is clear that f(x,y)≤g(x,y)f(x, y) \leq g(x, y), hence f(x,x)=0f(x, x) = 0 for all x∈Xx \in X, and therefore ff is a pseudometric. To prove the left-hand half of the inequalities (2), let us show by induction on pp that, for every finite sequence (zi)0≤i≤p(z_i)_{0 \leq i \leq p} of p+1p + 1 points of XX such that z0=xz_0 = x and zp=yz_p = y, we have
∑i=0p−1g(zi,zi+1)≥12g(x,y). \sum_{i=0}^{p-1} g(z_i, z_{i+1}) \geq \frac{1}{2} g(x, y).
This is clear if p=1p = 1. Put a=∑i=0p−1g(zi,zi+1)a = \sum_{i=0}^{p-1} g(z_i, z_{i+1}); the inequality (3) is true if a≥1/2a \geq 1/2, because g(x,y)≤1g(x, y) \leq 1. Suppose then that a<1/2a < 1/2, and let hh be the largest of the indices qq such that
∑i<qg(zi,zi+1)≤a2; \sum_{i < q} g(z_i, z_{i+1}) \leq \frac{a}{2};
we have then ∑i<hg(zi,zi+1)≤a/2\sum_{i < h} g(z_i, z_{i+1}) \leq a/2 and ∑i<h+1g(zi,zi+1)>a/2\sum_{i < h+1} g(z_i, z_{i+1}) > a/2, whence
∑i>hg(zi,zi+1)≤a2. \sum_{i > h} g(z_i, z_{i+1}) \leq \frac{a}{2}.

By the inductive hypothesis we have g(x,zh)≤ag(x, z_h) \leq a and g(zh+1,y)≤ag(z_{h+1}, y) \leq a; on the other hand it is clear that g(zh,zh+1)≤ag(z_h, z_{h+1}) \leq a. Let kk be the smallest integer >0> 0 such that 2−k≤a2^{-k} \leq a; then k≥2k \geq 2, and (x,zh)∈Uk,(zh,zh+1)∈Uk,(zh+1,y)∈Uk(x, z_h) \in U_k, (z_h, z_{h+1}) \in U_k, (z_{h+1}, y) \in U_k by the definition of gg; hence (x,y)∈Uk⊂Uk−13(x, y) \in U_k \subset U_{k-1}^3, which implies that g(x,y)≤21−k≤2ag(x, y) \leq 2^{1-k} \leq 2a.

Hence the inequalities (2) are proved; they show that, for each a>0a > 0, the set f−1([0,a])f^{-1}([0, a]) contains UkU_k for each index kk such that 2−k<a2^{-k} < a, and conversely that each UkU_k contains the set f−1([0,2−k−1])f^{-1}([0, 2^{-k-1}]); hence the sets f−1([0,a])f^{-1}([0, a]) form a fundamental system of entourages of the structure U\mathscr{U}.

Q.E.D.

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

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Cites

Nothing.

Transcribed from the printed volume. Not translated.