Question(s) about uniform spaces.

I was reviewing questions and notes related to uniform spaces and came across this interesting statement: Every metric space is homeomorphic, as a topological space, to a complete uniform space.

It seems pretty staightforward, but I am having trouble proving it.

Also, there was a follow-up question that seemed intesting: Is it true that every metric space is homeomorphic to a complete metric space?

Can anyone help? Thank you!

Solutions Collecting From Web of "Question(s) about uniform spaces."

As t.b. noted in the comments, every metric space is paracompact, so an even stronger result is:

Theorem. Every paracompact Hausdorff space is completely uniformizable.

Let $\langle X,\tau\rangle$ be a paracompact Hausdorff space. The first step of the proof is to show that the collection $\mathfrak{N}$ of all open nbhds of the diagonal of $X\times X$ is a base for a diagonal uniformity on $X$. It’s clear that $\mathfrak{N}$ satisfies most of the required conditions; the only one that isn’t immediately clear is that for each open nbhd $N$ of the diagonal there is an open nbhd $M$ such that $M\circ M\subseteq N$, i.e., such that $\langle x,z\rangle\in N$ whenever $\langle x,y\rangle,\langle y,z\rangle\in M$.

To see this, let $N\in\mathfrak{N}$. Let $\mathscr{U}=\{U\in\tau\setminus\{\varnothing\}:U\times U\subseteq N\}$; $\mathscr{U}$ is an open cover of $X$. In this answer I showed how to find an open barycentric refinement $\mathscr{V}$ of $\mathscr{U}$, i.e., a refinement with the property that for each $x\in X$ there is a $U\in\mathscr{U}$ such that $\operatorname{st}(x,\mathscr{V})\subseteq U$. Now repeat the process to get a barycentric open refinement $\mathscr{W}$ of $\mathscr{V}$.

Fix $W_0\in\mathscr{W}$ and $x_0\in W_0$ arbitrarily. If $W\in\mathscr{W}$ and $W_0\cap W\ne\varnothing$, we may pick an $x\in W_0\cap W$. $\mathscr{W}$ is a barycentric refinement of $\mathscr{V}$, so there is some $V_W\in\mathscr{V}$ such that $W_0\cup W\subseteq\operatorname{st}(x,\mathscr{W})\subseteq V_W$. Moreover, $x_0\in W_0$, so $W\subseteq V_W\subseteq\operatorname{st}(x_0,\mathscr{V})$. And $\mathscr{V}$ is a barycentric refinement of $\mathscr{U}$, so there is a $U\in\mathscr{U}$ such that $\operatorname{st}(W_0,\mathscr{W})=\bigcup\{W\in\mathscr{W}:W_0\cap W\ne\varnothing\}\subseteq\operatorname{st}(x_0,\mathscr{V})\subseteq U$. In other words, $\mathscr{W}$ is an open star refinement of $\mathscr{U}$.

Let $M=\bigcup\{W\times W:W\in\mathscr{W}\}$; clearly $M$ is an open nbhd of the diagonal. Suppose that $\langle x,y\rangle,\langle y,z\rangle\in M$ for some $x,y,z\in X$. Then there are $W_0,W_1\in\mathscr{W}$ such that $x,y\in W_0$ and $y,z\in W_1$, and since $W_0\cap W_1\ne\varnothing$, there is a $U\in\mathscr{U}$ such that $W_0\cup W_1\subseteq U$. Then $x,z\in U$, so $\langle x,z\rangle\in U\times U\subseteq N$, as desired.

In the answer to which I linked above I showed that $\mathfrak{N}$ is complete, so it only remains to verify that it generates the topology $\tau$. As usual, for each $N\in\mathfrak{N}$ and $x\in X$ let $N[x]=\{y\in X:\langle x,y\rangle\in N\}$, and let $\mathscr{N}=\{N[x]:N\in\mathfrak{N}\text{ and }x\in X\}$; $\mathscr{N}$ is a base for the topology $\tau_\mathfrak{N}$ generated by $\mathfrak{N}$. The members of $\mathfrak{N}$ are open in $X\times X$, so $\mathscr{N}\subseteq\tau$. On the other hand, suppose that $\varnothing\ne V\in\tau$, and let $x\in V$. $X$ is $T_3$, so there is an open set $U$ such that $x\in U\subseteq\operatorname{cl}U\subseteq V$; let $W=X\setminus\operatorname{cl}U$. Then $\{U,W\}$ is an open cover of $X$, so $N=(U\times U)\cup(W\times W)\in\mathfrak{N}$. Clearly $x\in N[x]=U\subseteq V$, so $V\in\tau_\mathfrak{N}$. Thus, $\tau_\mathfrak{N}=\tau$, and the proof is complete.