If $\mathfrak{U}$ is a not necessarily separated uniform structure for some set $X$, then an equivalence relation $R$ can be introduced on $X$ by letting $x R y$ provided $(x,y)\in U$ for every $U\in \mathfrak{U}$. Let $\pi: X\rightarrow X/R$ denote the quotient map and put $(Y,\mathfrak{W})= (X/R,\mathfrak{U}/R)$.

$(Y,\mathfrak{W})$ is called he separated uniform space associated with $(X,\mathfrak{U})$.

I am trying to show that $(Y,\mathfrak{W})$ is a hausdorff uniform space.

$\mathfrak{L}=\{ (\pi\times\pi)(M):M\in \mathfrak{U}\}$ is a basis for a uniformity ($\mathfrak{W}$) on $X/R$.

how can we show that: for every $(\pi\times\pi)(M)\in \mathfrak{L}$; $(\pi\times\pi)(N)\circ(\pi\times\pi)(N) \subset (\pi\times\pi)(M)$ for some $(\pi\times\pi)(N) \in \mathfrak{L}$.

and why $(\pi\times\pi)^{-1}((\pi\times\pi)(M))= R\circ M\circ R$; for every $M\in \mathfrak{U}$ ?

$(Y,\mathfrak{W})$ Used for construct the hausdorff completion of a hausdorff uniform space $(X,\mathfrak{U})$.


1 Answer 1


Let $(X,{\frak U})$ be a uniform space. We set $R = \bigcap {\frak U}$. It is a standard exercise to show that $R$ is an equivalence relation.

So we look at the following set: $${\frak U}/R := \{A\subseteq (X/R)\times (X/R): \exists M\in{\frak U}\big((\pi\times\pi)(M) \subseteq A\big)\}.$$

First we need to verify that $(X/R,{\frak U}/R)$ is a uniform space. By $[x]_R=\pi(x)$ for $x\in X$ we denote the equivalence class (= unique member of $X/R$) that contains $x$.

  1. Let $x\in X$, let $A\in {\frak U}/R$ and show that $([x]_R,[x]_R)\in A$. By definition of ${\frak U}/R$ we have $(\pi\times\pi)(M)\subseteq A$ for some $M\in{\frak U}$, so $(x,x)\in M$ which implies $(\pi(x), \pi(x)) = ([x]_R, [x]_R)\in (\pi\times\pi)(M)\subseteq A$.
  2. It's easy to see that ${\frak B} := \{(\pi\times\pi)(M): M\in {\frak U}\}$ is a filter base for ${\frak U}$ and therefore ${\frak U}/R$ is a filter.
  3. Let $A\in {\frak U}/R$ and show that $A^{-1}\in{\frak U}/R$. Let $M\in{\frak U}$ such that $(\pi\times\pi)(M)\subseteq A$. An easy exercise shows that $$\big((\pi\times\pi)(M)\big)^{-1} = (\pi\times\pi)(M^{-1}).$$ So we have $(\pi\times\pi)(M^{-1}) = \big((\pi\times\pi)(M)\big)^{-1} \subseteq A^{-1}$, which together with $M^{-1}\in{\frak U}$ directly implies $A^{-1}\in {\frak U}/R$ by definition of ${\frak U}/R$.
  4. Last we need to show the following: If $A\in {\frak U}/R$ there is $B\in {\frak U}/R$ such that $B\circ B \subseteq A$. Suppose $(\pi\times\pi)(M)\subseteq A$ for $M\in {\frak U}$. Pick $N\in{\frak U}$ with $N\circ N \subseteq M$. We want to prove that $$(\pi\times\pi)(N) \circ (\pi\times\pi)(N) \subseteq (\pi\times\pi)(M).$$ Recall that $$(\pi\times\pi)(N) = \{(\pi(x_1), \pi(x_2)): (x_1,x_2) \in N\}.$$ Let $(a,b), (b,c) \in (\pi\times\pi)(N)$. We want to show that $(a,c)\in (\pi\times\pi)(M)$. There are $x_1, x_2, x_3\in X$ such that $\pi(x_1) = a, \pi(x_2) = b, \pi(x_3) = c$ and $(x_1, x_2), (x_2,x_3) \in N$. So by assumption $(x_1, x_3) \in M$ because $N\circ N\subseteq M$, so $(\pi(x_1), \pi(x_3)) = (a,c) \in (\pi\times\pi)(M)$.
  • 1
    $\begingroup$ I really appreciate your concern. $\endgroup$
    – M. A.
    Dec 9, 2015 at 19:42
  • $\begingroup$ In fact, I am stuck with the following. There will be $(x_1, x_2), (x_3, x_4)$ such that $(a,b) = ( \pi(x_1), \pi(x_2))$ and $(b,c) = (\pi(x_3), \pi(x_4))$ and $(x_2,x_3) \in R$. In that case the composition of the above elements of $N$ is not possible. We must prove that for any entourage $N$ if $(x_1, x_2) \in N$ and $(x_2,x_3) \in R$, then $(x_1, x_3) \in N$ $\endgroup$ Jun 1, 2020 at 18:04

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