Can we define any metric on $\Bbb{R^\omega}$ so that it represents a norm?

Solution 1:

• Is there a norm that induces the product topology on $\mathbb{R}^\omega$?

No. If so, there would be an bounded open neighbourhood of $0$ (i.e. there exists an open set $U\ni 0$ such that for every open set $V\ni 0$ we have $U\subset nV$ for some $n\in\mathbb{N}$). Any open set $U$ in $\mathbb{R}^\omega$ constrains only finitely many coordinates, so a chosen open set $V$ that projects to $(-1,1)$ on a different coordinate cannot satisfy $U\subset nV$ for any $n\in\mathbb{N}$.

• Is there a norm on the $\mathbb{R}$-vector space $\mathbb{R}^\omega$, in $\mathsf{ZF}$?

We work in $\mathsf{ZF}+\mathsf{DC}$, so the Baire category theorem holds. Let $\tau$ be the product topology on $\mathbb{R}^\omega$. Suppose $\mathbb{R}^\omega$ also has a norm $\|\cdot\|$ (inducing another topology). Let $B$ be the closed unit ball with respect to the norm, and let $U$ be an open set with respect to $\tau$. We denote $e_n\in\mathbb{R}^\omega$ the element with $1$ at $n$th coordinate and $0$ at other coordinates.

Claim. $B\mathop{\triangle}U$ is not $\tau$-meagre.

Proof. If $U=\varnothing$ then the claim follows from the Baire category theorem as $\mathbb{R}^\omega=\bigcup_{n\in\mathbb{N}}nB$. If $U\ne\varnothing$ let $x\in U$. Let $a_n$ be sufficiently large that $x+a_ne_n\not\in B$, then $x+a_ne_n\to x\in U$ pointwise, so for sufficiently large $N$ we have $y=x+a_Ne_N\in U\setminus B$. Now for arbitrary $z\in\mathbb{R}^\omega$ we have $y+z/n\to y\in U\setminus B$ both pointwise and with respect to $\|\cdot\|$, so for sufficiently large $M$ we have $y+z/M\in U\setminus B$. That is, $\mathbb{R}^\omega= \bigcup_{n\in\mathbb{N}}n((U\setminus B)-y)$. Again, the claim follows from the Baire category theorem. $\square$

However, in the Solovay model of $\mathsf{ZF}+\mathsf{DC}$ for every subset $B$ of a complete separable metric space there exists an open set $U$ such that $B\mathop{\triangle}U$ is meagre. So in this model $\mathbb{R}^\omega$ has no norms.

• Is it true that if $\mathbb{R}^\omega$ has a norm then it has a basis"?

I don't know.