TITLE: Product norm on infinite product space
QUESTION [7 upvotes]: Today I proved that if $V$ is a normed space with norm $\|\cdot\|$ then I can define a norm on $V \times V$ that induces the same topology as the product topology as follows: $\| (v,w) \|_{V \times V} = \|v\| + \|w\|$.
I think I can do the same for an infinite product $V^{\mathbb N}$ by defining $\|(v_n)\|_{\mathbb N} = \sum_{n=0}^\infty \frac{1}{2^n} \|v_n\|$ and I proved it using the proof of the case $V \times V$ and changing some minor things.
Can you confirm that this result is correct? Thanks.

REPLY [15 votes]: Proposition. If $X_i, i \in \mathbb{N}$ is any sequence of nonzero topological vector spaces, then the product topology on $X = \prod_i X_i$ is not normable.
Proof.  Suppose to the contrary there is a norm $\|\cdot\|$ on $X$ which induces the product topology.  Let $B \subset X$ be the open unit ball of $\|\cdot\|$.  $B$ is open in the product topology and contains 0, so we can find a basic open neighborhood of 0 which is contained in $B$.  That is, there is a set of the form $U = U_1 \times U_2 \times \dots \times U_n \times X_{n+1} \times \cdots$, where $U_i \subset X_i$ is open and nonempty, such that $0 \in U \subset B$.  Choose any nonzero $x_{n+1} \in X_{n+1}$ and set $x = (0,\dots, 0, x_{n+1}, 0, \dots)$. Then for any $t \in \mathbb{R}$ we have $tx \in U \subset B$.  If we take $t = 2/\|x\|$, we have $\|tx\| = 2$ which is absurd since $B$ is the unit ball.
In short, the problem is that every open set of the product topology contains a line, and balls of a norm do not.