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I need a proof for this proposition:

If a uniformity $\mathfrak U$ on $X$ has a countable fundamental system of entourages, then it can be defined by a pseudometric on $X$.

which is the proposition 2 on page 142 of this book. I can't see the proof page.

even a sketch of proof is welcome. thanks.

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This is not really "research level" but I coincidentally worked through this proof earlier today. In the future, you may want to consider math.stackexchange.com for arguments that can easily be found in books you don't happen to have access to. People there also tend to give more detailed answers than here. –  François G. Dorais Feb 28 '13 at 0:08
    
@François G. Dorais: Although it is not (or may not be) research level, finding an answer to it is very hard by searching. or getting an answer in math.EX. However thanks for your answer. –  user31813 Feb 28 '13 at 5:03
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1 Answer 1

up vote 5 down vote accepted

First, find a fundamental system of entourages $U_0 \supseteq U_1 \supseteq \cdots$ such that

  1. $U_0 = X \times X$
  2. $(x_0,x_1) \in U_i$ iff $(x_1,x_0) \in U_i$.
  3. If $(x_0,x_1),(x_1,x_2),(x_2,x_3) \in U_{i+1}$ then $(x_0,x_3) \in U_i$.

Define $e(x_0,x_1) = \inf\lbrace 2^{-i} : (x_0,x_1) \in U_i\rbrace$. This is a well-defined nonnegative number since $(x_0,x_1) \in U_0$. The function $e$ is symmetric by property 2 and, of course, $e(x,x) = 0$. However, it does not necessarily satisfy the triangle inequality just yet.

By induction on $n$, we show that $$\sum_{i=0}^{n-1} e(z_i,z_{i+1}) \geq \frac{e(z_0,z_n)}{2}$$ for all sequences $z_0,\dots,z_n$. This is clear when $n = 1$. Suppose the result is true for all $n \lt m$. Given $z_0,\dots,z_m$, set $s = \sum_{i=0}^{m-1} e(z_i,z_{i+1})$. We may assume $0 \lt s \lt \frac12$, otherwise the result is trivial. Let $n \lt m$ be such that $$\sum_{i=0}^{n-1} e(z_i,z_{i+1}) \leq \frac{s}{2} \lt \sum_{i=0}^n e(z_i,z_{i+1}).$$ By the induction hypothesis, $e(z_0,z_n) \leq s$ and also $e(z_{n+1},z_m) \leq s$ since $$\sum_{i=n+1}^{m-1} e(z_i,z_{i+1}) = s - \sum_{i=0}^{n} e(z_i,z_{i+1}) \leq \frac{s}{2}.$$ Clearly, $e(z_n,z_{n+1}) \leq s$ too. Let $k$ be such that $2^{-k-1} \leq s \lt 2^{-k}$. Then $(z_0,z_n),(z_n,z_{n+1}),(z_{n+1},z_m) \in U_{k+1}$ which implies that $(z_0,z_m) \in U_k$ and hence $e(z_0,z_m) \leq 2^{-k} \leq 2s$, as required.

If we define $$d(x,y) = \inf \sum_{i=0}^{n-1} e(z_i,z_{i+1})$$ where the infimum is taken over all finite sequences $z_0,\dots,z_n$ with $z_0 = x$ and $z_n = y$, then $d$ is a pseudometric and since we always have $$\frac{e(x,y)}{2} \leq d(x,y) \leq e(x,y),$$ this pseudometric generates the correct topology.

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Do you know where this argument comes from? It's an exact duplicate of the argument used in current textbooks to prove the Birkhoff-Kakutani theorem (that every Hausdorff top. group with a countable basis of neighborhoods of 1 is metrizable), and now I wonder where this argument appeared first... –  Julien Melleray Feb 28 '13 at 14:00
    
Since topological groups are uniform, the result for topological groups is a special case. That special case may have come first since I believe that uniform spaces started as the common part of topological groups and metric spaces before taking a life of its own. –  François G. Dorais Feb 28 '13 at 15:10
    
Good point. Birkhoff-Kakutani. I encountered it a few years ago. thanks all. –  user31813 Feb 28 '13 at 16:25
    
Ah, I did not know that topological groups played a part in the advent of uniform spaces - that makes sense, as witnessed by the above proof... –  Julien Melleray Feb 28 '13 at 21:37
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François, uniform spaces have some (pre)history independent of topological groups (see the chapter by Bentley, Herrlich, and Husek in "Handbook of the History of General Topology, Vol. 2"). In particular your proof above is originally from a paper by Alexandroff and Urysohn (C. R. Acad. Sci., Paris 177 (1923), 1274-1276) –  Sergey Melikhov Mar 1 '13 at 23:49
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