For metric spaces $(M\_1, d\_1)$ and $(M\_2, d\_2)$, it is [an exercise](https://math.stackexchange.com/questions/781734/proving-a-metric-induces-the-product-topology) 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 $x\in M$.



Product topology from two premetric spaces induced by sum of premetrics?
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 $d\_1+d\_2$.


