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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 $x\in M$.

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    $\begingroup$ 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. $\endgroup$ Commented May 31, 2022 at 22:10

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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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  • $\begingroup$ 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? $\endgroup$
    – fsp-b
    Commented Jun 1, 2022 at 21:48
  • $\begingroup$ (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?) $\endgroup$
    – fsp-b
    Commented Jun 1, 2022 at 21:57
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    $\begingroup$ @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. $\endgroup$ Commented Jun 2, 2022 at 3:45
  • $\begingroup$ 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? $\endgroup$ Commented Jun 2, 2022 at 9:04
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    $\begingroup$ @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 $M_1\times M_2$. $\endgroup$ Commented Jun 2, 2022 at 9:15

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