#pseudometric spaces

I never liked the first definition I saw of an ultraproduct of metric spaces:

Definition. Let \(\mathcal{U}\) be an ultrafilter on \(I\), and let \((X_i, d_i)\) be a sequence of uniformly bounded metric spaces.

Define \(d\) on \(\prod_{i \in I} X\) as follows: \(d\left((x_i)_{i \in I}, (y_i)_{i \in I}\right)\) is the ultralimit \(\lim_{i \to \mathcal{U}} d_i(x_i, y_i)\).

Note that \(d\) is a pseudometric on \(\prod_{i \in I} X_i\).

The ultraproduct of the uniformly bounded metric spaces \(X_i\) with respect to \(\mathcal{U}\) is the metric space \(\prod_{i \to \mathcal{U}} (X_i, d_i)\) obtained by quotienting out the distance-0 balls from \(\left(\prod_{i \to I} X_i, d \right)\).

Ultraproducts of sets are filtered colimits of infinite products.

Definition. Let \(\mathcal{U}\) be an ultrafilter on \(I\), and let \((A_i)_{i \in I}\) be an \(I\)-indexed sequence of sets.

The ultraproduct \(\prod_{i \to \mathcal{U}} A_i\) is defined to be the filtered colimit of the following diagram: the objects are products \(\prod_{i \in P} A_i\) as \(P\) ranges over the sets in the ultrafilter \(\mathcal{U}\), and the transition maps are the truncation maps

\[\prod_{i \in P} A_i \to \prod_{i \in P \cap P'} A_i.\]

In a category with products and filtered colimits, we call the result of this kind of construction a categorical ultraproduct.

In this post, we’re going to show that the above definition can be redeemed: while metric spaces do not form a nice category, pseudometric spaces do—in a way, this is because pseudometric spaces are geometrically axiomatizable (see e.g. the discussion in D1.1 of the Elephant), while metric spaces are not—and so modulo quotienting out the distance-0 balls, an ultraproduct of uniformly bounded metric spaces in the sense of the above definition is the same thing as their categorical ultraproduct in the category of pseudometric spaces.

Precisely put, our goal in this post will be to prove the following

Proposition. Let \((X_i)_{i \in I}\) be a uniformly bounded family of metric spaces. Then the metric space ultraproduct \(\prod_{\mathcal{U}} X_i\) in the sense of the first definition coincides with the quotient of \[\left(\operatorname{\underset{\longrightarrow}{\lim}}_{P \in \mathcal{U}} \left(\prod_{i \in P} (X_i, d_i)\right)\right)\] by distance-\(0\) balls, where the filtered colimit above is computed in the category \(\mathbf{PseudoMet}\) of pseudometric spaces with morphisms the weakly contractive maps.

A product of uniformly bounded metric spaces \(\prod_{i \in I} (X_i, d_i)\) is given a pseudometric \(d\) by \(d\left((x_i)_{i \in I}, (y_i)_{i \in I}\right) \overset{\operatorname{df}}{=} \sup_{i \in I} d_i(x_i, y_i)\).

Let’s figure out how to compute filtered colimits.

Lemma. Let \(\left(X_i, f_{ij} : X_i \to X_j \right)\) be a filtered diagram of pseudometric spaces. Then \(\operatorname{\underset{\longrightarrow}{\lim}} \left(X_i, f_{ij} : X_i \to X_j \right)\) exists, and is given by \[\left( \operatorname{\underset{\longrightarrow}{\lim}} X_i, \overline{d} \right),\] where \(\operatorname{\underset{\longrightarrow}{\lim}} X_i\) is the colimit taken in the \(\mathbf{Set}\), and where \[\overline{d}([x], [y]) \overset{\operatorname{df}}{=} \inf_i d_i(x_i, y_i),\] where \([x]\) and \([y]\) are germs and the infimum is computed only considering those \(X_i\) which have representatives \(x_i\), \(y_i\) for both \([x]\) and \([y]\).

Proof. First, we verify that \(\overline{d}\) defined above is a pseudometric. Symmetry and positivity are clear, so we check the triangle identity. We need to verify \[\inf_i d_i(x_i, y_i) \overset{?}{\leq} \inf_j d_j(x_j, z_j) + \inf_k d_k (z_k, y_k).\]

Now, if this was not true, then \[\inf_i d_i(x_i, y_i) > \inf_j d_j(x_j, z_j) + \inf_k d_k (z_k, y_k),\] so in particular there exist \(j\) and \(k\) such that \[d_j(x_j, z_j) + d_k(z_k, y_k) < \inf_i d_i(x_i, y_i).\] Since we are in a filtered diagram, \(X_j\) and \(X_k\) have an amalamgation \(X_j, X_k \rightrightarrows X_{\ell}\). Since the maps are contractive, \[d_{\ell}(x_{\ell}, y_{\ell}) \leq d_{\ell}(x_{\ell}, z_{\ell}) + d_{\ell}(z_{\ell}, y_{\ell}) \leq d_j(x_j, z_j) + d_k(z_{k}, y_{k}) < \inf_i d_i(x_i, y_i),\] a contradiction.

Now, we convince ourselves that \(\left( \operatorname{\underset{\longrightarrow}{\lim}} X_i, \overline{d} \right)\) is truly a colimit for \(\left(X_i, f_{ij} : X_i \to X_j \right)\). If \((Y, d_Y)\) is a cocone to this diagram, then there is a unique map of sets \(c : \operatorname{\underset{\longrightarrow}{\lim}} X_i \to Y\), and \(c\) is contractive because there is a contractive map \(c_i : X_i \to Y\) for each \(i\) which factors as a function through \(c\), so for any pair \((x_i, y_i)\),

\[\forall j \in J, d_Y(c_i(x_i, y_i)) \leq d_j(x_j, y_j) \hspace{1mm}\implies\hspace{1mm} d_Y(c([x_i], [y_i])) \leq \inf_i d_i (x_i, y_i),\]

proving the lemma. \(\square\)

Proof of the Proposition. First, let us compare the pseudometric spaces \[\prod_{i \in I} (X_i, d_i) \text{  vs  } \prod_{i \to \mathcal{U}} (X_i, d_i).\] There is a natural quotient map \(\prod X_i \to \prod_{i \to \mathcal{U}} X_i\). Is this contractive? Unraveling definitions, we see that we need to check the inequality \[\overline{d}([x_i], [y_i]) = \inf_P \left(\sup_{i \in P} d_i(x_i, y_i) \right) \overset{?}{\leq} \lim_{i \to \mathcal{U}} d_i(x_i, y_i) \overset{\operatorname{df}}{=} a \in \mathbb{R}.\] By definition of ultralimit, we have that for each \(\epsilon\), the \(\epsilon\)-ball \(B_{\epsilon}(a)\) around \(a\) contains \(\mathcal{U}\)-many \(d_i(x_i, y_i)\). Sending \(\epsilon \to 0\), we conclude that the inequality is true.

Actually, we can conclude more. If \(P\) indexes a set \(S \overset{\operatorname{df}}{=} \{d_i(x_i, y_i)\}_{i \in P}\) which witnesses this inequality, we can intersect \(P\) with smaller and smaller \(P'\). These intersections will be nonempty, which means that \(\sup S = a\). So the weak inequality can actually be improved to equality.

Therefore, the quotient map \(\prod X_i \to \prod_{i \to \mathcal{U}} X_i\) is isometric, and so any failure of injectivity must be because points distance \(0\) apart are being glued together. So there is a factorization