From Wikipedia:

A uniform space $(X, Φ) $is a set $X$ equipped with a nonempty family $Φ$ of subsets of the Cartesian product $X × X$ ($Φ$ is called the uniform structure or uniformity of $X$ and its elements entourages) that satisfies the following axioms:

  1. if $U$ is in $Φ$, then $U$ contains the diagonal $Δ = \{ (x, x) : x ∈ X \}$.
  2. if $U$ is in $Φ$ and $V$ is a subset of $X × X$ which contains $U$, then$V$ is in $Φ$
  3. if $U$ and $V$ are in $Φ$, then $U ∩ V$ is in $Φ$
  4. if $U$ is in $Φ$, then there exists $V$ in $Φ$ such that, whenever $(x, y)$ and $(y, z)$ are in $V$, then $(x, z)$ is in $U$.
  5. if $U$ is in $Φ$, then $U^{-1} = \{ (y, x) : (x, y) \in U \}$ is also in $Φ$.

I was wondering

A. Are 2 and 3 equivalent to that $Φ$ is a filter on $X \times X$?

B. Is $Δ = \{ (x, x) : x ∈ X \}$ also in $Φ$?

C. How shall I understand 4?

It reminds me of the triangle inequality in the definition of a metric. Are they really related?

Is 4 equivalent to "For any $V$ is in $Φ$, and for any $(x, y)$ and $(y, z)$ in $V$, there exists $U$ in $Φ$ such that $(x, z)$ is in $U$?"

D. in 5, What does it mean that it doesn't require $U \equiv U^{-1}$? To distinguish distances with different "directions"/"orientations"?

Thanks and regards!

  • $\begingroup$ I re-labeled your questions to avoid confusion. B. Usually $\Delta$ is not in $\Phi$. If it is, this would be the discrete uniform structure. $\endgroup$ – GEdgar Feb 5 '12 at 14:09
  • $\begingroup$ In general, to understand uniform structure, take a metric space and examine the corresponding uniform structure. Indeed, 4 is what we get from the triangle inequality. And 5 corresponds to $d(x,y) = d(y,x)$. $\endgroup$ – GEdgar Feb 5 '12 at 14:13
  • $\begingroup$ The other formative example of uniform space was topological group. So next examine some non-metrizable topological group and see what its (say, right) uniform structure is. $\endgroup$ – GEdgar Feb 5 '12 at 14:14
  • $\begingroup$ @GEdgar: Thanks! About your first comment,from wiki, "the discrete uniformity on X is defined by letting every superset of the diagonal {(x,x) : x is in X} in X × X be an entourage". Isn't this equivalent to 1? In other words, the uniform structure defined in my post is actually discrete uniform structure. What is going wrong here? $\endgroup$ – Tim Feb 5 '12 at 14:24
  • $\begingroup$ For the discrete uniform structure you take all supersets of $\Delta$. In general (as your 1 says) you take some supersets of $\Delta$, probably not all of them. Correspondingly in a metric space: every ball about $x$ contains $x$, but usually $\{x\}$ itself is not a ball about $x$. $\endgroup$ – GEdgar Feb 5 '12 at 14:41

Yes, 2) and 3) just say it is a filter of $X \times X$, all of whose elements contain the diagonal $\Delta$ (axiom 1). This is also often said in the definition (but here they don't want to depend on readers already knowing what a filter is).

The diagonal itself can be in $\Phi$, but then any subset that contains the diagonal will be too (by 2) and this is called the discrete uniformity. This is just $$\Phi = \{ U \subset X \times X \mid \Delta \subset U \},$$ and it will turn out that it induces on $X$ the discrete topology.

4 is indeed inspired by the triangle inequality, and 5 by the symmetry axiom for a metric ($d(x,y) = d(y,x)$). This is clear if you look at the uniformity induced by a metric $d$, then the uniformity is generated (as a filter base) by the sets $$U_r = \{ (x,y) \in X \times X \mid d(x,y) < r \}$$

If take $U = U_r$ for 4), then we take $V = U_{\frac{r}{2}}$, and note that if $(x,y), (y,z) \in V$, then $d(x,z) <= d(x,y) + d(y, z) < \frac{r}{2} + \frac{r}{2} = r$ so that $(x,z) \in U_r$, as required.

It's not equivalent to your formulation. For any $V$ that contains the diagonal we can take $$V \circ V = \{ (x,z) \in X \times X \mid \exists y : (x,y) \in V \land (y,z) \in V \}$$ and then $V \subset V \circ V$, so the latter is always in $\Phi$ when $V$ is. So your condition (for any 2 pairs $(x,y), (y,z)$ from $V$, the "composition pair" is in some $U$) is always satisfied from 1 + 2 alone, so adds nothing new.

Note that using the discrete metric $d(x,y) = 1$ when $x \neq y$, $U_1 = \Delta$, making the connection with the discrete uniformity again.

Another example worth looking into is the case of topological groups (a group with a topology that makes the group operations multiplication and inversion continuous), where there are 2 natural uniformities: for any neighbourhood $U$ of the identity $e$ we define one uniformity as generated by all subsets of the form $$\{ (x,y) \in G \times G \mid x\ast y^{-1} \in U\}$$ and the other by all sets of the form $$(x,y) \in G \times G \mid y \ast x^{-1} \in U \}.$$ The axiom 4 then comes down to the fact that for any neighbourhood $U$ of $e$ there is a neighbourhood $V$ of $e$ such that (e.g.) $V \ast V^{-1} \subset U$. which follows from the continuity of the group operations, if you think about it.

So it's similar to the triangle equality in spirit, and for uniformities that are induced by metrics, it is proved from it, but it's not always of that origin. It's a very important property of uniformities, and allows us to construct continuous functions for the topology induced by a uniformity, similarly to the proof of the Urysohn lemma for normal spaces. It makes such topological spaces completely regular.

  • $\begingroup$ Thanks! (1) I was wondering if on a set, the discrete uniformity is the unique uniformity that induces the discrete topology? (2) Similarly, on a set, is the discrete metric the unique metric that induced the discrete uniformity/topology? $\endgroup$ – Tim Feb 6 '12 at 2:44
  • $\begingroup$ There are many metrics that will work in general, even for non-discrete, we could use scaling and cutting-off etc. to get alternative equivalent metrics. For uniformities I'm not sure. $\endgroup$ – Henno Brandsma Feb 6 '12 at 6:05
  • $\begingroup$ An example: for $X = \{ \frac{1}{n} \mid n=1,2,\ldots \}$ in the inherited metric from $\mathbb{R}$, both the metric and the uniformity induce the discrete topology, but the metric is not the (standard) discrete one, as arbitarily small values occur, and the uniformity does not contain the diagonal $\Delta$, so it's not the maximal uniformity, called the discrete uniformity. $\endgroup$ – Henno Brandsma Feb 6 '12 at 13:30

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