# Are the sections of entourages in a uniform space open?

Wikipedia's article on uniform spaces defines the following.

• A nonempty family $\mathcal{U}$ of subsets $U \subseteq X \times X$ is a uniform structure if it satisfies the following axioms:

1. If $U \in \mathcal{U}$, then $\Delta \subseteq U$, where $\Delta = \{ (x,x) : x \in X \}$ is the diagonal on $X \times X$.
2. If $U \in \mathcal{U}$ and $U \subseteq V$ for $V \subseteq X \times X$, then $V \in \mathcal{U}$.
3. If $U \in \mathcal{U}$ and $V \in \mathcal{U}$, then $U \cap V \in \mathcal{U}$.
4. If $U \in \mathcal{U}$, then there is $V \in \mathcal{U}$ such that $V \circ V \subseteq U$, where $V \circ V$ denotes the composites of $V$ with itself. (The composite of two subsets $V$ and $U$ of $X \times X$ is defined by $V \circ U = \{ (x,z) : (\exists y \in X)((x,y) \in U \wedge (y,z) \in V ) \}$.)
5. If $U \in \mathcal{U}$, then $U^{-1} \in \mathcal{U}$, where $U^{-1} = \{ (y,x) : (x,y) \in \mathcal{U} \}$.

... The elements $U$ of $\mathcal{U}$ are called entourages....

• Every uniform space $X$ becomes a topological space by defining a subset $O$ of $X$ to be open if and only if for every $x$ in $O$ there exists an entourage $V$ such that $V[x] = \{ y : (x,y) \in V \}$ is a subset of $O$.

I am having trouble showing that the sets $V[x]$ themselves are open using this definition (this is claimed in the following sentence). In a metric space, it is clear what to do: if $z \in B_{\varepsilon}(x)$, then there is some smaller $\delta < \varepsilon$ with $z \in B_{\delta}(x)$ and then $B_{\varepsilon - \delta}(z) \subseteq B_{\varepsilon}(x).$

In a uniform space, it's claimed that "this can be proved with a recursive use of the existence of a "half-size" entourage". I guess this is meant as an adaptation of the metric space argument but it isn't clear to me that this works. If $X$ is not Hausdorff, then I can't see any axiom that says that any $(x,y) \in U$ must always lie in a smaller entourage $(x,y) \in V \subseteq U$; and even if $X$ is Hausdorff I don't see how we can "take the difference" in entourages and look at $B_{\varepsilon-\delta}.$

• Reference: General Topology by R. Engelking. ..One of the 8 chapters is entirely on Uniform Spaces. – DanielWainfleet Nov 9 '15 at 1:21
• @user254665 Thanks but I'm still not convinced. I don't see anywhere where Engelking shows that $V[x]$ is open, and for example, in Corollary 8.1.3 Engelking refers specifically to the interior of $V[x]$ (which he calls $B(x,V)$) rather than $V[x]$ itself. I am concerned because this conflicts with Bourbaki who defines the topology on $X$ to have the sets $V[x]$ as a basis. – user288288 Nov 9 '15 at 1:35
• I will take a look at it. I have a copy but it's been a while.... – DanielWainfleet Nov 9 '15 at 1:42
• Please read carefully what the next sentence from the Wikipedia article states, and also look at the article on neighbourhoods. – user642796 Nov 9 '15 at 5:12
• @ArthurFischer Thanks, I see that I've been misinterpreting the term "neighborhood base" and "neighborhood filter". I guess it's not claimed there that $U[x]$ is open after all. – user288288 Nov 9 '15 at 5:24

The sets of the type $V[x]$ do not have to be open at all.
Take for example $\mathbb{R}$ with the standard metric and the induced uniform structure.
Then for every $\epsilon > 0$

$$V_\epsilon = \bigcup \limits_{x \in \mathbb{R}} \{ x \} \times B_\epsilon(x)$$

is an entourage.
However by the second axiom $W = V_\epsilon \cup \{0,2\epsilon\}$ is also an entourage.
But now we have

$$W[0] = V_\epsilon[0] \cup \{2 \epsilon\} = B_\epsilon(0) \cup \{2 \epsilon \} = (-\epsilon, \epsilon) \cup \{2 \epsilon \}$$

which is certainly not open.
Note however that its interior is nonempty!

• Sorry for piggybacking onto this post. I'm wondering about something. If $U$ is an entourage of $X$ and $x \in X$ is any point, is $U[x]$ at least a neighbourhood of $x$? (i.e. for every $y \in U[x]$ does there exist an open set that contains $y$ and is a subset of $U[x]$?) – jessica Jan 10 at 8:03