0
$\begingroup$

Let $(X, \mathcal{U})$ be an uniform space. It is known that every entourage $U\in\mathcal{U}$ is a neighborhood of $\Delta_X$, but the converse is not true, in general.

What can say about closed neighborhood of $\Delta_X$? Is it true that for a closed neighborhood $D\neq \Delta_X$ of $\Delta_X$, there is $U\in\mathcal{U}$ with $U\subseteq D$?

Thanks a lot.

$\endgroup$
1
$\begingroup$

No, that's not true in general. Consider $\mathbb{R}$ with its standard uniform structure. Let$$D^\star=\left\{(x,y)\in\mathbb{R}^2\,\middle|\,-\frac1{1+x^2}\leqslant y\leqslant\frac1{1+x^2}\right\}$$and let $D$ be what you obtain when you apply to $D^\star$ a rotation of $\frac\pi4$ radians around the origin. Then $D$ is a closed neighborhood of $\Delta_{\mathbb R}$, but it contains no entourage.

$\endgroup$
1
$\begingroup$

Show that in a separated uniform space: If we have a neighbourhood $U$ of $\Delta_X$ that is no entourage, then $\overline{U}$ is a closed neighbourhood of $\Delta_X$ that contains no entourage.

$\endgroup$

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.