Uniform spaces, Dugundji chapter 9, section 11

Question

I think I'm encountering a lot of errors in the section of Dugundji about uniform spaces. Let me write out my concerns.

We now use a uniformity to derive a topology. Any uniformity $\mathscr{F}$ in $Y$ gives a topology $\mathscr{T}(\mathscr{F})$ in $Y$ by taking the family $\{V[y] : V\in\mathscr{F}, y\in Y\}$ as basis. With this topology in $Y$, it is easy to verify that in the space $Y\times Y$, the family $\{\text{Int}\ V: V\in\mathscr{F}\}$ is a uniformity equivalent to $\mathscr{F}$ and inducing the same topology.

There is two concerns of mine while reading this. First off, the fact that it's a basis is very suspicious.

Second concern is pretty apparent. Take $\mathscr{F} = \{V(r): r>0\}$ where $V(r) = \{(x, y)\in\mathbb{R}^2: |x-y|<r\}$ and $V(r)[x]$ are clearly just open intervals, so the topology here is the Euclidean topology.

But if we take $W(r) = \{(x, y) : |x-y|\leq r\}$ instead, $W(r)[x]$ are closed intervals, which not only don't form a basis, since $[0, 1]\cap [1, 2] = \{1\}$ for example, but there is no closed interval which would be a subset of $\{1\}$ (degenerated intervals are not taken here), but even if we accept this as a subbasis, it would be discrete!

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For people that don't have this book, here is some of the notation and definitions:

$\Delta := \{(x, x) : x\in Y\}$, $V[x] := \{y : (y, x)\in V\}$ (this was also corrected)

Definition. A uniformity in a set $Y$ is a family $\mathscr{F}$ of subsets of $Y\times Y$ such that

1. $\Delta\subseteq V$ for $V\in\mathscr{F}$,
2. For any $V_1, V_2\in\mathscr{F}$ there is $W\in\mathscr{F}$ such that $W\subseteq V_1\cap V_2$,
3. For any $V\in \mathscr{F}$ there is $W\in\mathscr{F}$ such that $W\circ W^{-1} \subseteq V$.

I really don't know what to do, I've read so much of this book but now there's so much errors.

Maybe I misunderstood something? I really have no idea. This concerns me so much.

Answer 1

Let $W_r=\left\{\langle x,y\rangle\in\Bbb R^2:|x-y|\le r\right\}$, and let $\mathscr{W}=\{W_r:r>0\}$. For each $x\in\Bbb R$ let $\mathscr{W}(x)=\{W[x]:W\in\mathscr{W}\}$. Then $\bigcup_{x\in\Bbb R}\mathscr{W}(x)$ is a nbhd base for the Euclidean topology in the sense that $U\subseteq\Bbb R$ is open the Euclidean topology iff for each $x\in U$ there is a $W[x]\in\mathscr{W}(x)$ such that $W[x]\subseteq U$.

In other words, Dugundji is using basis in a slightly non-standard fashion here. If $\langle X,\tau\rangle$ is a space, he’s thinking here of a family $\mathscr{B}=\bigcup_{x\in X}\mathscr{B}_x$, where each $\mathscr{B}_x$ is a nbhd base at $x$. Note that the members of $\mathscr{B}$ do not themselves have to be open: saying that $\mathscr{B}_x$ is a nbhd base at $x$ is just saying that a set $A$ is a (not necessarily open) nbhd of $x$ iff it contains some member of $\mathscr{B}_x$.