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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 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).


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.

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2 Answers 2

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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, 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.

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From your remarks one can see that the core of your question lies in the case of a product of metric spaces. But a large cardinal product, say of the real line, cannot have a structure generated by metrics. This follows from the simple fact that a continuous pseudometric on such a product can only depend on countably many components and so cannot be a metric.

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