Let $(X, \mathcal{U})$ be a compact uniform space which is not metrizable and and $\{U_i\}_{i=0}^{\infty}$ be a countable set of $U_i\in \mathcal{U}$ with $U_{i-1}\subseteq U_i$ and $U_1\subseteq V$ that $V\in \mathcal{U}$ and $V\circ V\circ V\subseteq U$ for $U\in \mathcal U$. In Proof of Theorem 1 in https://projecteuclid.org/download/pdf_1/euclid.pjm/1103038056, author claimed that since $\{U_i\}_{i=1}^{\infty}$ is not a base for the uniformity $\mathcal{U}$, there is $W\in \mathcal{U}$ with $W\subseteq U$ such that for every $i>0$, $U_i\cap comp W\neq \emptyset$,

Q. Proof of $U_i\cap comp W\neq \emptyset$ is not clear for me, ( The notation $camp W$ is not define in this paper and it is not clear for me)

can someone help me please



$\operatorname{comp}W$ is just the complement of $W$ in $X$, which I'll denote by $W^c$ instead.

Recall that $\{U_i\}_{i=1}^\infty$ being a base (for entourages inside $U$) means:

$$\forall W \in \mathcal{U}( W \subset U) \implies (\exists i: U_i \subseteq W)\text{.}$$

So not being a base means, applying standard logic:

$$\exists W \in \mathcal{U}: (W \subset U)\land (\forall i: U_i \nsubseteq W)$$

and the last negation of inclusion can be written as $U_i \cap W^c \neq \emptyset$ (not all members of $U_i$ are in $W$, so some of them are in $W^c$...). So indeed:

$$\exists W \in \mathcal{U}: (W \subset U)\land (\forall i: U_i \cap W^c \neq \emptyset)$$

So the negation of being a base inside $U$ gives us the promised $W$.

(If the $(U_i)$ would be a base for entourages inside $U$, then for any $V \in \mathcal{U}$, $V \cap U \in \mathcal{U}$ would contain some $U_i$ so $V$ would contain a $U_i$ as well, and $\mathcal{U}$ would have a countable base and thus be metrisable, which it is not. So the negation holds.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.