I'm having trouble constructing the Hausdorff completion of a uniform space $(X,U)$ using pseudometrics. I know that every uniformity on a space $X$ is made by pseudometrics. Here is my idea:

Let $(X,U)$ be a uniformity generated by pseudometrics $(d_i: X \rightarrow\mathbb{R}_{\geq 0})_{i \in I}$. Define 'Generalized' Cauchy Sequences (GCS) as followed: $(x_n)_{n \in \mathbb{N}}$ is a GCS iff $$\forall \varepsilon > 0 \forall K \subset I \text{ finite }\forall i \in K \exists n_0 \forall p,q \geq n_0 : d_i(x_p,x_q)<\varepsilon$$ Let $\mathcal{C}$ be the set of all GCS and define a equivalence relation as followed: $$ (x_n)_{n \in \mathbb{N}}R(y_n)_{n \in \mathbb{N}} \Leftrightarrow \forall \varepsilon > 0 \forall K \subset I \text{ finite }\forall i \in K \exists n_0 \forall n \geq n_0 : d_i(x_n,y_q)<\varepsilon$$

Let $X'$ be the set $\mathcal{C}/R$. Define pseudometrics $d'_i((x_n)_{n \in \mathbb{N}},(y_n)_{n \in \mathbb{N}})$ as $\lim_{n \rightarrow \infty} d_i(x_n,y_n)$. Let $k: X \rightarrow X'$ be the canonical function.

Now, I can prove that $X'$ is Hausdorff (thanks to the equivalence relation), $k(X)$ is dense is $X'$ (every point in $X'$ is the limit of images of points in $X$), but I'm having trouble proving that $X'$ is complete.

Basically, I have 2 questions: is what I'm doing correct (and if so, how do I proceed) or else, what am I doing wrong here (and how to correct it).

As always, any help would be appreciated.


Notice first that you’ve not gained anything by looking at finite $K\subseteq I$: your definition of GCS is equivalent to saying that the sequence is $d_i$-Cauchy for every $i\in I$. The real problem, though, is that in general sequences don’t suffice in uniform spaces: you have to look at nets or filters. Thus, you can’t expect that adding points for sequences that ought to converge will be enough. Let me suggest a different approach (which is the standard one).

For $i\in I$ let $\langle X_i,d_i\rangle$ be the pseudometric space with underlying set $X$ and pseudometric $d_i$. For $x,y\in X$ define $x\stackrel{i}\sim y$ iff $d_i(x,y)=0$, let $Y_i$ be the quotient $X_i/\stackrel{i}\sim$, let $q_i$ be the quotient map, and let $\rho_i$ be the metric on $Y_i$ induced by $d_i$: $$\rho_i\big(q_i(x),q_i(y)\big)=d_i(x,y)$$ for any $q_i(x),q_i(y)\in Y_i$. Let $Y=\prod_{i\in I}Y_i$. Then the map

$$e:X\to Y:x\mapsto\langle q_i(x):i\in I\rangle$$

is a uniform embedding. (This takes a bit of argument, but it’s fairly straightforward.)

Now let $\hat Y_i$ be the metric completion of $Y_i$ for each $i\in I$. Show that $\prod_{i\in I}\hat Y_i$ is a complete uniform space, and conclude that the closure of $e[X]$ in $\prod_{i\in I}\hat Y_i$ is a Hausdorff completion of $X$.

  • 1
    $\begingroup$ @MinimusHeximus: A uniformly continuous embedding. $\endgroup$ – Brian M. Scott Jun 3 '13 at 6:35

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