# norm on product of normed linear spaces which induces the product topology

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.

• 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
– M W
Nov 20 at 3:15
• 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 $p\in [1,\infty)$.
– M W
Nov 20 at 3:43