Any uniformity on a set $X$ is generated by a family of pseudometrics on $X$. So if $(X,\mathcal D)$ is a uniform space there's a set $P$ of pseudometrics on $X$ with


$$\mathcal D=\left< \bigcup\_{d\in P} \mathcal D\_d\right> \tag1 $$


where the angles show the filter generated by the subbace $\bigcup\_{d\in P} \mathcal D\_d$; where each $\mathcal D\_d$ is the uniformity on $X$ with base
$$\lbrace U\_d(r) \mid r>0 \rbrace$$
in which
$$U\_d(r)=\lbrace (x,y)\in X^2\mid d(x,y)\le r\rbrace$$


$(1)$ immediately implies that $(X,\mathcal D)$ can be embedded uniformly in a product of pseudometric spaces.


Now suppose $(X,\mathcal D)$ is Hausdorff. [Isbell](http://books.google.com/books?id=2llOQvY39GcC&pg=PA18&lpg=PA18&dq=every+uniform+space+has+completion&source=bl&ots=MDTtnTKzdr&sig=LVvEFaOcw9fAYjJqvY8jLUBgoxs&hl=en&sa=X&ei=xugPUfeCLcqZtAbz9IHABA&ved=0CHwQ6AEwCA#v=onepage&q&f=true) says $(X,\mathcal D)$ can be embedded in a product of metric spaces. (The proof given uses covering definition of uniform space, while I'm only familiar with diagonal definition).




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My question is:


Is any Hausdorff uniformity on a set $X$ generated by a family of metrics on $X$? Is there a couterexample?


I think none of the pseudometrics that generate a Hausdorff uniformity need to be a metric. Because:
$$\bigcap\_{d\in P}\bigcap\_{n\in \Bbb N}U\_d(\frac{1}{n})=\Delta\_X \text{ (diagonal)}$$
but this intersection need not be countable and so there may be no metrics in the set $P$ (of pseudometrics that generate the Hausdorff uniformity $\mathcal D$). So uniformities generated by metrics need not be all Hausdorff uniformities.



Uniformities generated by metrics.
Here is a result that completely characterizes the uniformities that are generated by metrics as opposed to pseudometrics.


$\mathbf{Theorem}$ Let $(X,\mathcal{U})$ be a uniform space. Then the following are equivalent.


1. There are $R\_{n}\in\mathcal{U}$ for all natural numbers $n$ such that $\bigcap\_{n}R\_{n}:=1\_{X}=\{(x,x)|x\in X\}$.
2. there is a non-empty set of metrics $\mathcal{D}$ such that $R\in\mathcal{U}$ if and only if there is some $d\in\mathcal{D}$ and some $\epsilon>0$ with $U\_{d}(\epsilon)\subseteq R$.


$\mathbf{Proof}$
$2\rightarrow 1$ Assume that $\mathcal{D}$ is a set of metrics such that $R\in\mathcal{U}$ if and only if there is some $d\in\mathcal{D},\epsilon>0$ with $U\_{d}(\epsilon)\subseteq R$. Then $1\_{X}=\bigcap\_{n}U\_{d}(\frac{1}{n})$ and $U\_{d}(\frac{1}{n})\in\mathcal{U}$ for all $n$.


$1\rightarrow 2$ Assume that there is a sequence $(R\_n)\_n$ where $R\_{n}\in\mathcal{U}$ for all $n$ and such that $\bigcap\_n R\_n=1\_X$. Let $V$ be the set of all sequences $S=(S\_{n})\_{n}$ such that $\bigcap\_{n}S\_{n}=1\_{X}$, $S\_{n}=S\_{n}^{-1}:=\{(y,x)|(x,y)\in S\_{n}\}$ for all $n$, and where $S\_{n+1}\circ S\_{n+1}:=\{(x,z)|\exists y,(x,y),(y,z)\in S\_{n+1}\}\subseteq S\_{n}$. Then by [this question](https://mathoverflow.net/questions/123160/a-uniformity-with-a-countable-base-is-a-pseudometric-uniformity), for all $S\in V$ there is a pseudometric $d\_{S}$ such that $S\_{n}\subseteq R$ for some $n$ if and only if there is some $\epsilon>0$ where $U\_{d\_{S}}(\epsilon)\subseteq R$. Furthermore, the pseudometric $d\_{S}$ is clearly a metric. I claim that the system of metrics $\{d\_{S}|S\in V\}$ generates the uniformity $\mathcal{U}$. If $S=(S\_{n})\_{n}\in V$ and $\epsilon>0$, then $S\_{n}\subseteq U\_{d\_{S}}(\epsilon)$ for some $n$, so since $S\_{n}\in\mathcal{U}$, we have $U\_{d\_{S}}(\epsilon)\in\mathcal{U}$. However, if $R\in\mathcal{U}$, then since $\bigcap\_{n}R\_{n}=1\_{X}$, there is some $S=(S\_{n})\_{n}\in V$ with $S\_{n}\subseteq R\_{n}$ for all $n$ and where $S\_{0}\subseteq R$. Then there is some $\epsilon>0$ where $U\_{d\_{S}}(\epsilon)\subseteq S\_{0}\subseteq R$. 


$\mathbf{QED}$


Of course, there are many uniformities in which $\bigcap\_{n}R\_{n}\neq 1\_{X}$ for any choice of $R\_{n}\in\mathcal{U}$ for all $n$. For instance, if $(X,\mathcal{U})$ is generated by a system of metrics, then every point in $X$ is a $G\_{\delta}$-set. In fact, there are even uniformities $(X,\mathcal{U})$ such that the intersection of countably many entourages is an entourage (I am currently writing a paper with a result about those uniformities). For example, if $X$ is a set with $|X|>2^{\aleph\_{0}}$ and $\mathcal{E}$ is the set of all equivalence relations that partition $X$ into at most $2^{\aleph\_{0}}$ pieces, then $\mathcal{E}$ generates a uniformity on $X$ such that the intersection of countably many entourages is an entourage.


