# Question about neighborhood of diagonal of a topological space

Let $$X$$ be a topological space but it is not uniform space and $$U$$ be a neighborhood of diagonal $$X$$, $$\Delta_X$$. Is there a neighborhood $$D(\neq \Delta_X)$$ of $$\Delta_X$$ such that $$D\circ D \subseteq U$$?

• There are counterexamples. Look in good books on uniform spaces... – Henno Brandsma Mar 17 at 22:23

The answer is positive provided $$X$$ is a $$T_1$$ paracompact space. In order to show this we shall use definitions and notations from pp. 302-304 of Ryszard Engelking’s “General Topology” (2nd ed., Heldermann, Berlin, 1989), see below.
Since $$U$$ is a neighborhood of the diagonal $$\Delta_X$$, for each $$x\in X$$ we can pick its open neighborhood $$U_x$$ such that $$U_x\times U_x\subset U$$. By Theorem 5.1.12.ii, an open cover $$\{U_x:x\in X\}$$ of the space $$X$$ has and open barycentric refinement $$\mathcal V$$. Put $$D=\bigcup \{V\times V:V\in \mathcal V\}$$. Then $$D$$ is a neighborhood of $$\Delta_X$$. We claim that $$D\circ D\subset U$$. Indeed, let $$(x,z)\in D\circ D$$ be an arbitrary point. There exists a point $$y\in X$$ such that points $$(x,y)$$ and $$(y,z)$$ belong to $$D$$. Therefore there exist elements $$V_x$$ and $$V_y$$ of $$\mathcal V$$ such that $$\{x,y\}\subset V_x$$ and $$\{y,z\}\subset V_z$$. Then $$\{x,z\}\subset V_x\cup V_z\subset \operatorname{St}(y,\mathcal V)\subset U_t$$ for some $$t\in X$$. Then $$(x,z)\in U_t\times U_t \subset U$$.