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](http://books.google.com/books?id=bQwhdmL6IjUC&pg=PA142&lpg=PA142&dq=if+a+uniformity+U+has+a+countable+fundamental+system+of+entourages,+then+there+is+a+pseudo+metric&source=bl&ots=eKHTzHWIG5&sig=H_H3KeyN-y_WcHhnxNMqShu5jf0&hl=en&sa=X&ei=aokuUcmFL9DitQbTuYDYAg&ved=0CD0Q6AEwAQ#v=onepage&q=if%2520a%2520uniformity%2520U%2520has%2520a%2520countable%2520fundamental%2520system%2520of%2520entourages%252C%2520then%2520there%2520is%2520a%2520pseudo%2520metric&f=false). I can't see the proof page.


even a sketch of proof is welcome.
thanks.



A uniformity with a countable base is a pseudometric uniformity.
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.


