We recently need the following result about product space in an applied analysis paper:


Let $(M,d)$ be a metric space, and $\bigotimes^{k+1}M = \overbrace{M \times M \times \dots \times M}^{(k+1)-{\rm times}}$ be equipped with the product topology.
Let $U \subset \bigotimes^{k+1}M$ be a neighborhood of the hyperdiagonal
$ \{ (x,x,\dots,x) : x\in M\}$. When $k\geq 2$, there is a smaller open neighborhood
$V' \subset U$ of the hyperdiagonal
of the form
$
V' = \{(x\_0,x\_1,\dots,x\_k) \,:\, (x\_i,x\_{i+1}) \in V \,, 0 \leq i < k\}
$
for $V \subset M \times M$ an open neighborhood of the diagonal of $M\times M$.


The proof (see Appendix B of <http://www.math.drexel.edu/~tyu/Papers/SE7.pdf>) relies on a metric; we wondered if any form of this result is written down somewhere, and if the metrizability of the space $M$ is required for the result to hold true. Thanks.



Neighborhoods of the hyperdiagonal in a product space
The result does not hold for general topological spaces. 


For a counterexample, equip $M = \{a, b, c\}$ with the topology whose open sets are $\emptyset$, $M$, $\{a, b\}$, $\{b\}$ and $\{b, c\}$. 
The set $U= \{a,b\}^3\cup \{b,c\}^3\subseteq M^3$ is an open neighborhood of the ternary diagonal.


Now let $V\subseteq M^2$ be any open neighborhood of the binary diagonal. Since (i) $V$ contains $(a,a)$, (ii) $V$ is open in the product topology, and (iii) $\{a,b\}$ is the smallest open set of $M$ containing $a$, it follows that $\{a,b\}^2\subseteq V$. Similarly $\{b,c\}^2\subseteq V$. But now $V$ is already too big for our purposes. If $V'$ is defined to be
$\{(x\_0,x\_1,x\_2) \,:\, (x\_i,x\_{i+1}) \in V \,, 0 \leq i \lt 2\},
$
then $(a,b,c)\in V'-U$. That is, there is no neighborhood $V$ of the binary diagonal that induces a neighborhood $V'$ of the ternary diagonal such that $V'\subseteq U$.


