# Product topology from two premetric spaces induced by sum of premetrics?

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 $$x\in M$$.

• This is unimportant, but $d = d_1 + d_2$ can be seen as wrong, since the domain of the right side is not $M_1 \times M_2 \times M_1 \times M_2$. There is a lack of projections. Commented May 31, 2022 at 22:10

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 $$d_1+d_2$$.
• Thank you, @Taras! Do you know if the question can be answered positively if the product $M_1\times M_2$ is assumed to be sequential? Commented Jun 1, 2022 at 21:48
• (Also (excuses for my naivety but I'm far from being a topologist): If you say that ''the product $M_1\times M_2$ is not sequential'' then you do mean that the product topology on $M_1\times M_2$ is not sequential, right?) Commented Jun 1, 2022 at 21:57
• @rmcerafl Yes, I mean that the product topology on $M_1\times M_2$ can be non-sequential. If the product topology is sequential, then I do not know the answer. Commented Jun 2, 2022 at 3:45
• The elements of $M_1$ are probably $1/n+i/mn$ with $n\in\mathbb N$ and $m\in\mathbb N$? What is a sequentially closed subset of $M_1\times M_2$ which is not closed? Commented Jun 2, 2022 at 9:04
• @JochenWengenroth Of course, you are right about $m$. The sequentially closed subset of $M_1\times M_2$ which is not closed is the diagonal'' $\{(z,z): z\in M_1\setminus\{0\}\}$ of $M_1\times M_2$. Commented Jun 2, 2022 at 9:15