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I have been reading about uniform spaces and topological groups. There does not look to be a lot of literature on the topic, much less accesible literature, and the books that I have been reading do not mention any examples of uniform spaces, other than metric spaces and topological groups. There is another which I came up with, which is $\mathbb R$ with uniform structure containing $U_r = \lbrace (x,y) \in \mathbb R ^2 \ | \ x = y$ or exactly one of $x,y$ irrational and $|x-y| < r \rbrace$ but this example is a bit trivial since the interior of each $U_r[x]$ is just the point $x$. Can anyone give me some more examples?

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  • $\begingroup$ Completely regular spaces are exactly the uniform spaces. $\endgroup$ – Camilo Arosemena-Serrato Oct 8 '13 at 23:16
  • $\begingroup$ Correction: $\:$ If Dependent Choice then completely regular spaces are exactly the uniformizable spaces. $\hspace{.36 in}$ $\endgroup$ – user57159 Oct 8 '13 at 23:22
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    $\begingroup$ @Ricky: That’s just hyper-pedantry: in this, as in most contexts, full $\mathsf{AC}$ is the default assumption. The real correction is that the completely regular spaces are precisely the uniformizable spaces; they don’t become uniform spaces until they’re given a uniform structure. It seems reasonably clear that Daron is looking for examples of uniform structures. I’ll see if I can come up with a nice example or two. $\endgroup$ – Brian M. Scott Oct 9 '13 at 10:50
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For now, at least, I’m going to give only one example, because the first order of business is to explain why your sets

$$U_r=\Delta\cup\{\langle x,y\rangle\in\Bbb R^2:|x-y|<r\text{ and exactly one of }x\text{ and }y\text{ is irrational}\}$$

for $r>0$ do not give you a uniformity, where $\Delta$ is the diagonal in $\Bbb R^2$.

$\mathscr{U}=\{U_r:r>0\}$ is not a uniformity: it doesn’t satisfy the condition that if $U\subseteq V\subseteq\Bbb R^2$, and $U\in\mathscr{U}$, then $V\in\mathscr{U}$. It also fails to satisfy the condition that if $U\in\mathscr{U}$, there is a $V\in\mathscr{U}$ such that $V\circ V\subseteq U$. If $\langle x,y\rangle,\langle y,z\rangle\in V\in\mathscr{U}$, then $x$ and $z$ are either both rational or both irrational; if $x\ne z$, then $\langle x,z\rangle\in(V\circ V)\setminus U$. In other words, $\mathscr{U}$ fails to satisfy conditions $(2)$ and $(4)$ of this definition.

Failure to satisfy $(2)$ isn’t catastrophic: we could just replace $\mathscr{U}$ by $$\mathscr{V}=\{V\subseteq\Bbb R^2:V\supseteq U_r\text{ for some }r>0\}\;,$$ taking $\mathscr{U}$ as a base for the uniformity rather than the uniformity itself. However, $\mathscr{V}$ still isn’t a uniformity: it still fails to satisfy $(4)$, since there is no $V\in\mathscr{V}$ such that $V\circ V\subseteq U_1$, for instance.

I can come up with a uniformity on $\Bbb R$ that is somewhat similar to your attempt. For $r>0$ let

$$U_r=\{\langle x,y\rangle\in\Bbb R^2:x=y\in\Bbb R\setminus\Bbb Q\text{ or }|x-y|<r\}\;,$$

and let $$\mathscr{U}=\{U\subseteq\Bbb R^2:U\supseteq U_r\text{ for some }r>0\}\;.$$ This $\mathscr{U}$ is a uniformity. Most of the conditions are very easy to check. For $(4)$, let $U\in\mathscr{U}$. There is an $r>0$ such that $U_r\subseteq U$, and it’s easy to check that $U_{r/2}\circ U_{r/2}\subseteq U_r\subseteq U$.

Now let’s see what topology $\mathscr{U}$ generates on $\Bbb R$. Call this topology $\tau$, and for each $U\in\mathscr{U}$ and $x\in\Bbb R$ let $$U[x]=\{y\in\Bbb R:\langle x,y\rangle\in U\}\;.$$ Suppose first that $x$ is irrational. Then $U_r[x]=\{x\}$ for each $r>0$, so $\{x\}$ is a nbhd of $x$. Thus, $x$ is in the interior of $\{x\}$, so $\operatorname{int}_\tau\{x\}=\{x\}$, and $\{x\}$ is $\tau$-open: each irrational is an isolated point of $\langle\Bbb R,\tau\rangle$.

Note that I use the term neighborhood (nbhd) in the broad sense, not requiring it to be open: a set $N$ is a nbhd of $x$ if $x$ is in the interior of $N$.

Now suppose that $x$ is rational. Then $$U_r[x]=\{y\in\Bbb R:|x-y|<r\}=(x-r,x+r)$$ for each $r>0$. Since $\{U_r:r>0\}$ is a base for the uniformity $\mathscr{U}$, $\{U_r[x]:r>0\}$ is a local base for the topology $\tau$ at the point $x$, so $\{(x-r,x+r):r>0\}$ is a local base at each rational.

Without using uniformities we can describe the space $\langle\Bbb R,\tau\rangle$ by saying that it’s obtained from the usual topology on $\Bbb R$ by isolating each irrational. The resulting space is a fairly important example in topology; it’s called the Michael line, and you can read about it in Dan Ma’s Topology Blog. Note that this is an example of a uniform space that is not metrizable and cannot be turned into a topological group.

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It is sometimes useful to know that for every compact Hausdorff space $X$ there is precisely one uniform structure whose uniform topology is the topologyof $X$. The entourages are just the neighborhoods of the diagonal in $X \times X$.

This is consonant with the fact that on compact Hausdorff spaces, every continuous function to a uniform space is uniformly continuous (I'm going on memory for this last bit).

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