What is the purpose of this axiom in the definition of a uniform space?

I started reading about uniform spaces the other day, and in an attempt to familiarize myself with the notion I attempted to work through some examples based on the entourage definition of a uniformity. I was able to understand and verify the uniformity on a metric space in which $\{\{(x,\ y)\ |\ d(x,\ y)<a\}\ |\ a\in\mathbb{R}_{>0}\}$ is a fundamental system of entourages.

However, something still doesn't sit well with me, and that's the 4th axiom wikipedia gives:

If $U\in\Phi$, then there is some $V\in\Phi$ for which $(x,\ y)\in V$ and $(y,\ z)\in V$ implies $(x,\ z)\in U$.

I recognize that this implies that for all $U\in\Phi$, there is $V\in\Phi$ for which $V\circ V\subseteq U$ where $\circ$ is the binary relation composition operator. My question is: why?

Why is it necessary to include this in the definition of a uniform space? As far as I've read, it appears to me uniform spaces were invented to generalize the notion of uniform continuity between spaces: a map between uniform spaces is uniformly continuous if the image of an entourage in the domain space is an entourage in the codomain space. This also allows us to generalize the notion of Cauchy sequences to Cauchy nets.

I understand that a net $(x_\alpha)$ is a Cauchy net if for every entourage $U$ there is $\lambda$ such that whenever $\alpha,\ \beta > \lambda$, $(x_\alpha,\ x_\beta)\in U$.

Now, I can understand how the notion of a Cauchy net generalizes the notion of a Cauchy sequence (under the condition that an entourage is taken to be a binary relation of "closeness"), but I haven't worked through the details of why the above definition of a uniformly continuous map preserves Cauchy nets (though it doesn't sound too far off from the usual proof that uniformly continuous functions on $\mathbb{R}$ preserve Cauchy sequences).

I read in this answer that the 4th axiom is essentially inspired by the triangle inequality, but that doesn't quite make sense to me either. I suppose it could be that axiom 4 is meant to generalize the notion of a metric space, in which uniform continuity can be defined (and in fact, the notion of a metric is required in the usual definition of uniform continuity), but I still don't quite see how omitting the 4th axiom would break the definition of uniform continuity.

So, what part of all of this, the generalization of Cauchy sequences and uniform continuity, is the 4th axiom required for? Why would omitting the 4th axiom be a problem, and why does the 4th axiom "fix" the problem, and properly generalize uniform continuity?

The purpose of the axiom is to make sure that this common proof motif from metric spaces will generalize to uniform spaces:

Let $x$ (with certain properties) be given. The enemy chooses an $\varepsilon>0$ and we must then find a $z$ (with certain other properties) within $\varepsilon$ of $x$.

We can do that by first finding an $y$ (with yet other properties) within $\varepsilon/2$ of $x$ and then finding a $z$ with the desired property within $\varepsilon/2$ of $y$. Then, due to the triangle inequality, that $z$ will be within $\varepsilon$ of $x$.

The task is now reduced to proving separately that (1) there is an $y$ near every $x$, and (2) there is a $z$ near every $y$.

In a uniform space, the enemy choosing a $\varepsilon$ corresponds to choosing an $U$, and splitting the $\varepsilon$ into two $\frac\varepsilon2$s corresponds to picking a $V$ by the axiom.

With metric spaces, we sometimes need to prove nearness in a chain of more than two steps, for example splitting $\varepsilon$ into seven $\frac\varepsilon7$s. This would be modeled in a uniform space by several successive uses of the axiom.

Sometimes it is convenient that the pieces you split the $\varepsilon$ into are equal, which is why the axiom gives you a single $V$ such that $V\circ V\subseteq U$, and not just $V$ and $W$ such that $V\circ W\subseteq U$ (which would have been enough to enable the simple case I've sketched above). But the difference is not really important; the $V\circ W$ version implies the $V\circ V$ version because entourages are preserved by intersection.

• That certainly makes sense, and explains my confusion about the axiom generalizing the triangle inequality. So, just to check, it's not necessarily that the 4th axiom is required for uniform continuity to make sense, it's just that since uniform spaces are also used to generalize metric spaces, the 4th axiom helps make proofs like that work? Is there any reason the axiom is required for uniform continuity? – user3002473 Jan 7 '17 at 18:10
• @user3002473: That depends on what you consider "makes sense" to be. You could omit the axiom and then "uniform continuity" would still define something, but you probably wouldn't be able to prove most of the useful theorems about it you're trying to generalize from metric spaces. Essentially, without this axiom all you would have is an arbitrary filter on $\hat X=\{U\subseteq X\mid \#U=2\}$, without any dependence on the inner structure of $\hat X$. – Henning Makholm Jan 7 '17 at 20:31

From a purely intuitive standpoint, I believe the axiom can be interpreted as implying that there are arbitrarily small entourages. That is, makes sense if we view an entourage $$U$$ as a mapping taking any point $$x$$ to a neighborhood $$U[x]$$ (with the different neighborhoods having the same 'size').

In particular, this allows us to find a entourage $$V$$ of at most half the 'radius' (imagining an open ball), since $$V \circ V$$ is twice the 'radius' of $$V$$. Since entourages are closed under taking supersets (which allows us to increase the 'radius'), we get that there are entourages of any size.

Why does this matter? Well, consider a case where this isn't true: if $$(X, d)$$ is a metric space, we could let $$U_{r}[x] = \{ y \in X : 1 < d(x,y) < r \} \cup \{x\}$$ for each $$r > 1$$. Then I believe the collection of $$U_{r}$$ generates a collection of sets satisfying every axiom but the fourth. What topology does it induce? Well, it will be hard for the sets $$U_{r}[x]$$ to be neighborhoods; you would hope that it gives the metric topology, but it can't since there is no way to represent distances less than 1 (let alone arbitrarily small distances).

I am relatively certain that this does not actually induce a valid topology, in which case you have a problem. If uniform continuity were defined on such a space, you could have it being uniformly continuous yet not continuous in any meaningful sense. More generally, if uniform spaces specialize topological spaces, then every entourage must generate a topology.