TITLE: Product topology from two premetric spaces induced by sum of premetrics?
QUESTION [5 upvotes]: For metric spaces $(M_1, d_1)$ and $(M_2, d_2)$, it is an exercise that the product topology on $M_1\times M_2$ is induced by the metric $d((x_1, y_1), (x_2, y_2)) =d_1(x_1, x_2) + d_2(y_1, y_2)$.
Do you know if this statement generalises to premetric spaces?
Here, we call $(M,\tilde{d})$ a premetric space if $M$ is a set and $\tilde{d}:M\times M\rightarrow[0,\infty)$ is such that $\tilde{d}(x,x)=0$ for all $x\in M$.

REPLY [5 votes]: The answer to this question is negative.
Consider the subspace $M_1=\{0\}\cup\{\frac 1n+\tfrac{i}{nm}:n,m\in\mathbb N\}$ of the complex plane and the space $M_2=M_1\cup\{\frac1n:n\in\mathbb N\}$ endowed with the symmetric
$$d_2(x,y)=\begin{cases}|x-y| &\mbox{if $0\notin \{x,y\}$ or $x,y\in\mathbb R$ or $x=y$};\\
1&\mbox{otherwise}
\end{cases}
$$
It can be shown that the product $M_1\times M_2$ is not sequential, so its topology cannot be generated by a premetric, in particular, it is not generated by the symmetric $d_1+d_2$.
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