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 $D\circ D \subseteq U$?

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    $\begingroup$ There are counterexamples. Look in good books on uniform spaces... $\endgroup$ – 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 $(x,z)\in U_t\times U_t \subset U$.

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