I am stuck trying to prove why a net in a product of uniform spaces is Cauchy if and only if every projection of it is a Cauchy net.

I assume, analogously to the fact that continuous uniformity implies Cauchy uniformity with respect to sequences, that it has to do with the projections being uniformly continuous. However, when it comes to the details I can´t really prove any of the two implications because I don´t really understand how the Cauchy net works in the product space.

Any kind of help is welcome, thank you for your time!


Suppose we have a product $X=\prod_{i \in I} X_i$ of uniform spaces $(X_i, \mathcal{U}_i)$, with projections $\pi_i: X \to X_i$. (I'm assuming you're using entourages as the type of uniform space, so each $\mathcal{U}_i$ is a collection of subsets of $X_i \times X_i$.) Recall that the collection $$\mathcal{S}_p=\bigcup_{i \in I}\{(x,y) \in X^2: (\pi_i(x), \pi_i(y)) \in U: U \in \mathcal{U}_i\}$$ is a subbase for a (product) uniformity $\mathcal{U}_p$ on $X$, which is the minimal one which makes all $\pi_i$ uniformly continuous.

Some more definitions and facts: a net $(x_n)$, $n \in D$, where $D$ is a set directed by a relation $\ge$, in a uniform space $(X,\mathcal{U})$, is called Cauchy, iff

$$\forall U \in \mathcal{U}: \exists n_0 \in D: \forall n,n' \in D: (n \ge n_0 \land n' \ge n_0) \to (x_n, x_{n'}) \in U$$ and if $\mathcal{S}$ is a subbase for $\mathcal{U}$, it's easy to check from the definition of a subbase and the fact that $D$ is a directed set, that this is equivalent to

$$\forall U \in \mathcal{S}: \exists n_0 \in D: \forall n,n' \in D: (n \ge n_0 \land n' \ge n_0) \to (x_n, x_{n'}) \in U$$

so that we only have to check the property for subbase elements.

Another basic fact: if $(X,\mathcal{U}) \to (Y, \mathcal{V})$ is uniformly continuous, and $(x_n), n \in D$ is Cauchy in $X$ then $f(x_n), n \in D$ is Cauchy in $Y$.

Now to the main question:

A net $(x_n), n \in D$ is Cauchy in $(\prod_i X_i, \mathcal{U}_p)$ iff for all $i \in I$ the net $(\pi_i(x_n)), n \in D$ is Cauchy in $(X_i, \mathcal{U}_i)$.

The left to right implication is an immediate consequence of the fact that the uniformly continuous $\pi_i$ all preserve Cauchy sequences.

The right to left implication follows from the subbase-checking criterion, using the subbase $\mathcal{S}_p$ as defined in the beginning: let $$O=\{(x,y) \in X^2: \pi_i(x), \pi_i(y)) \in U\}$$ be an element of $\mathcal{S}_p$, for some fixed $i \in I$ and $U \in \mathcal{U}_i$. Because $\pi_i(x_n)$ is a Cauchy net, and $U \in \mathcal{U}_i$, there is some $n_0\in D$ such that $$\forall n,m \in D: \text{ if } n \ge n_0 \text{ and } m \ge n_0, (\pi_i(x_n), \pi_i(x_m)) \in U$$ or by definition of $O$:

$$\forall n,m \in D: \text{ if } n \ge n_0 \text{ and } m \ge n_0, (x_n, x_m) \in O$$

which says that $(x_n)$ fulfills the Cauchy definition for $O$.

So basically it all follows from the definitions, plus the subbase Cauchy criterion.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.