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I'm largely following Ioan M. James' book for this. Some definitions for a uniform space $X$:

For any entourage $U$ of $X$ and any $x \in X$, the neighbourhood induced by $U$ is the set $$ U[x] = \{\, y \in X \mid (x,y) \in U\, \}. $$

A subset $S \subseteq X$ is said to be open if for every $x \in S$, there exists an entourage $U$ of $X$ such that we have $x \in U[x] \subseteq S$. The uniform topology on $X$ is the collection of subsets defined by $$ \mathcal{T} = \{\, S \subseteq X \mid S \text{ is open in } X\, \}. $$

I'm fairly certain that induced neighbourhoods may not themselves be open in general. Thus I'm really confused by a proof to a proposition provided in the book.

Proposition 8.11. Let $X$ be a uniform space. For each symmetric entourage $D$ of $X$ and each subset $M$ of $X \times X$ the subset $D \circ M \circ D$ is a neighbourhood of $M$ in the topological product $X \times X$.

Proof. Let $D$ be a symmetric entourage. Then $(x,y) \in D \circ M \circ D$ if and only if there exists $(\xi ,\eta) \in M$ such that $(x, \xi) \in D$ and $(\eta, y) \in D$, in other words, such that $x \in D[\xi]$ and $y \in D[\eta]$, or again such that $$(x,y) \in D[\xi] \times D[\eta]. $$ Since $D[\xi] \times D[\eta]$ is a neighbourhood of $(\xi, \eta)$ in the topological product $X \times X$, this proves the assertion.


Edited for clarification.

By the product topology on $X \times X$, for $D[\xi] \times D[\eta]$ to be a neighbourhood of $(\xi, \eta)$, we require that $D[\xi]$ and $D[\eta]$ are neighbourhoods of $\xi$ and $\eta$ respectively.

A set $A$ is a neighbourhood of $x$ if there exists an open set $O_x$ such that we have $x \in O_x \subseteq A$, right? However, I can't figure out the corresponding open sets for $D[\xi]$ and $D[\eta]$. Are uniform neighbourhoods always neighbourhoods in the uniform topology?

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You’re right that $D[x]$ need not be an open neighbourhood of $x$. As a simple example of this, note that $D=\{(u,v) \in \mathbb{R}^2: |u-v| \le \varepsilon\}$ is an entourage for $\mathbb{R}$ in the standard uniformity induced by the metric $d(x,y)=|x-y|$. And $D[x]= \{y\in \mathbb{R}: |x-y| \le \varepsilon\} = [x-\varepsilon, x+\varepsilon]$ is not an open neighbourhood of $x$. This is basically due to the fact that entourages are closed upwards for inclusion, the basic entourages induced by a metric are usally set to be of the form $D=\{(x,y): d(x,y) < r\}$ and these induce just the usual (open) balls as neighbourhoods.

As to the product $X \times X$, where we see $X$ just as a topological space (as James also does). This has a base of open sets of the form $U \times V$ where $U,V$ are open subsets of $X$, as is well-known. This implies that a set of the form $N_x \times N_y$ is a neighbourhood of $(x,y)$ in $X \times X$ iff $N_x$ is a neighbourhood of $x$ and $N_y$ is a neighbourhood of $y$:

Proof If $N_x \times N_y$ is of that form, then there is an open set $O_x$ of $X$ such that $x \in O_x \subseteq N_x$ and an open set $O_y$ of $X$ such that $y \in O_y \subseteq N_y$. This is just the definition of $N_x$ resp. $N_y$ being a neighbourhood of $x$ resp. $y$ in a topological space. But then of course $O_x \times O_y$ is product open in $X \times X$ and $(x,y) \in O_x \times O_y \subseteq N_x \times N_y$, so again by the definition of neighbourhood we have that $N_x \times N_y$ is a neighbourhood of $(x,y)$ in $X \times X$. On the other hand, if we know $N_x \times N_y$ is a neighbourhood of $(X,y)$ in $X \times X$ there must be a product open set $O$ such that $(x,y) \in O \subseteq N_x \times N_y$. So by the definition of a base there exists a basic open set $U \times V$ in $X \times X$ (so $U$ and $V$ are open sets of $X$) such that $$(x,y) \in U \times V \subseteq O \subseteq N_x \times N_y$$ It follows easily that then $x \in U \subseteq N_x$ and $y \in V \subseteq N_y$, showing that indeed $N_x$ and $N_y$ are neighbourhoods of $x$ resp. $y$ and we are done.

Another easy observation which holds for all topological spaces: if $A \subseteq X$ and for each $x \in A$ we have a neighbourhood $N_x$ of $x$ in $X$, then $\bigcup_x N_x$ is a neighbourhood of $A$ in $X$. This is immediate from the definition of neighbourhoods and the fact that open sets are closed under all unions.

Now, what James has shown in the quoted computation is in fact that if $D$ is an entourage and $M \subseteq X \times X$:

$$D \circ M \circ D = \bigcup\{D[\xi] \times D[[\eta]: (\xi, \eta) \in M \}$$

which by all the previous remarks is just a union of product neighbourhoods of all points of $M$ and so a neighbourhood of $M$ in the product.

Hope this clarifies the matter somewhat. It’s mostly observations on neighbourhoods in general that are needed here, plus knowing what basic open sets in the product are.

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  • $\begingroup$ Thanks for yet another answer. I realised I wasn't being very clear in my question. I know that $D[\xi]$ is a uniform neighbourhood of $\xi$, but is it a neighbourhood of $\xi$ in the sense that there exists an open set $O_x$ such that $x \in O_x \subseteq D[\xi]$? $\endgroup$ – jessica Jan 11 at 3:24
  • $\begingroup$ @jessica By definition the $D[x]$ are a neighbourhood base, so yes. Open sets are defined using neighbourhood bases. $\endgroup$ – Henno Brandsma Jan 11 at 3:30

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