I've seen in various textbooks and notes that if $X$ is paracompact, then the collection of all the neighborhoods of the diagonal is a uniformity.

I am trying to show that this uniformity is complete using Cauchy filters. So far, I let $\mathfrak{F}$ be a filter on $X$ that does not converge. By definition, saying $\mathfrak{F}$ does not converge to any point is to say that for any $A \subset X$, there is a open neighborhood $O_{A}$ of $A$ which is not an element of $\mathfrak{F}$.

With that said, the set $\alpha = \{ X \setminus \overline{F} : F \in \mathfrak{F} \}$ is an open cover of $X$. This is where I'm stuck. Why does it follow from here that $\mathfrak{F}$ is not Cauchy?

Can anyone help?

  • $\begingroup$ If paracompactness includes the Hausdorff condition you can use characterization given in Kelley, General Topology, Theorem 5.28 (d): for every open cover there exists a neighborhood $N$ of the diagonal such that the cover $\{N(x)\}_{x \in X}$ is a refinement of $\alpha$. Then $\mathfrak{F}$ can't contain any $N$-small set, so it isn't Cauchy. $\endgroup$ – t.b. Jun 4 '12 at 16:49

Let $\mathscr{U}=\{X\setminus\operatorname{cl}F:F\in\mathfrak{F}\}$. Let $\mathscr{V}$ be a locally finite open refinement of $\mathscr{U}$. A paracompact Hausdorff space is normal, so $X$ has an open cover $\mathscr{W}=\{W_V:V\in\mathscr{V}\}$ such that for each $V\in\mathscr{V}$, $\operatorname{cl}W_V\subseteq V$; clearly $\mathscr{W}$ is locally finite. (I’m assuming that your definition of paracompactness, unlike mine, includes Hausdorffness; otherwise, you need to add that hypothesis.)

Consider any $x\in X$. If $x\in\operatorname{cl}W_V\in\mathscr{W}$, then $x\in V\in\mathscr{V}$, and $\mathscr{V}$ is point-finite, so $$G_x=\bigcap_{x\in\operatorname{cl}W_V}V$$ is open. $\mathscr{W}$ is locally finite and hence closure-preserving, so $$H_x=\bigcup_{x\notin\operatorname{cl}W_V}\operatorname{cl}W_V$$ is closed. Thus, $N_x=G_x\setminus H_x$ is an open neighborhood of $x$. Let $\mathscr{N}=\{N_x:x\in X\}$; $\mathscr{N}$ is an open cover of $X$.

Fix $x\in X$; there is some $V\in\mathscr{V}$ such that $x\in\operatorname{cl}W_V$, and I claim that $\operatorname{st}(x,\mathscr{N})\subseteq V$. To see this, suppose that $x\in N_y\in\mathscr{N}$; then $x\notin H_y$, so $y\in\operatorname{cl}W_V$, and therefore $N_y\subseteq G_y\subseteq V$. Since $N_y$ was an arbitrary element of $\mathscr{N}$ containing $x$, it follows that $\operatorname{st}(x,\mathscr{N})\subseteq V$. (In other words, $\mathscr{N}$ is a barycentric open refinement of $\mathscr{V}$ and hence also of $\mathscr{U}$.)

Now let $$D=\bigcup_{x\in X}(N_x\times N_x)\;;$$ clearly $D$ is an open neighborhood of the diagonal. Let $F\in\mathfrak{F}$ be arbitrary, and suppose that $F\times F\subseteq D$. Fix $x\in F$. Then for each $y\in F$, $\langle x,y\rangle\in F\times F\subseteq D$, so there is some $z\in X$ such that $x,y\in N_z$. Thus, $F\subseteq\operatorname{st}(x,\mathscr{N})\subseteq V$ for some $V\in\mathscr{V}$. But $\mathscr{V}$ refines $\mathscr{U}$, so $V$ (and hence $F$) is disjoint from some member of the filter $\mathfrak{F}$. This is impossible, so for all $F\in\mathfrak{F}$ we must have $F\times F\nsubseteq D$, and $\mathfrak{F}$ is therefore not Cauchy.

  • $\begingroup$ In case you missed this question on complete uniformizability of metric spaces: It is very easy from what you have done here (I haven't seen a direct way of proving that result without making a detour via paracompactness of metric spaces). $\endgroup$ – t.b. Jun 5 '12 at 5:45
  • $\begingroup$ @t.b.: I saw your comment on it and completely forgotten about it. Thanks. $\endgroup$ – Brian M. Scott Jun 5 '12 at 5:54

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