A norm in a linear space induces the topology defined by the metric \begin{equation} d(x,y)=\|x-y\| \end{equation} Consider a product of normed linear spaces $\Pi_{j\in J} V_j$. One can define the product topology as for general products of topological spaces, namely the topology generated by the basis of $\Pi_{j\in J} A_j$ where $A_j$ is an open set in $V_j$, and $A_j=V_j$ for all but a finite number of $j$. My question is, is there a norm that induces this product topology?

The definition of direct product, which I denote by $\Pi^{'}_{j\in J}V_j$, inspires my question. The norm in the direct product is defined as \begin{equation} \|x\|_\infty = {\rm sup}_{j\in J} \|P_j(x)\| \end{equation} where $P_j$ is the projection to the j-th component. By this definition the direct product is a proper subset of the Cartesian product, for example, in the product of countable $\mathbb{R}$'s, $(1,2,3,\dots)$ is not a member of the direct product. I noticed that the topology induced by the norm $\|\ \|_\infty$ does not give the direct product the subspace topology inherited from the product topology of the Cartesian product. Consider the unit ball $\{x\in \Pi^{'}_{j\in J}V_j,\|x\|_\infty <1\}$. If we have the subspace topology, then we will have $B\cap \Pi^{'}_{j\in J}V_j$ is contained in the unit ball for some $B$ in the basis of $\Pi_{j\in J}V_j$. A member of the basis must have an "unconstrained" component, say $A_i=V_i$. The vector with a large i-th component but all other components vanishing is clearly in the intersection $B\cap \Pi^{'}_{j\in J}V_j$, but not in the unit ball.

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    $\begingroup$ No, if $J$ is uncountable it will not even be metrizable (since the topology is not first countable), and even in the countable case, it will not be normable by this result: math.stackexchange.com/a/168037/1210477 $\endgroup$
    – M W
    Nov 20 at 3:15
  • $\begingroup$ Let me clarify in the previous comment by "countable" I meant "countably infinite". In the finite case the norm $\|x\|_\infty$ you have defined will indeed induce the product topology, as will $\|x\|_p:=\left(\sum_{j\in J}\|P_j(x)\|^p\right)^{\frac{1}{p}}$ for any $p\in [1,\infty)$. $\endgroup$
    – M W
    Nov 20 at 3:43


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