Can we define a norm on $\Bbb{R^\omega}$ in a basis free way? Let $\Bbb{R^\omega}=\{(x_n)_{n\in \mathbb{N}}: x_n \in \Bbb{R}\}$.
Then, $(\Bbb{R^\omega}, +, \cdot) $ is a linear space.
I know , if $(x_n) $ are $p$- summable, then we can define norm , $\ell_p$-norm ($1\le p<\infty $) on $\Bbb{R^\omega}$. And if $(x_n) 's$ are bounded we can define supremum norm, $\ell_{\infty}$ on $\Bbb{R^\omega}$.
The best thing I can do for general $\Bbb{R^\omega}$ (no special assumption on sequences) is to define a metric on $\Bbb{R^\omega}$ by
$$d(x, y) =\sum_{j\in\mathbb{N}}{(a_j)} \frac{|x_j -y_j|}{1+|x_j -y_j|}$$
where $(a_j) _{j\in\mathbb{N}}$ is any convergent series of positive reals. I can show that the metric isn't induced by a norm on $\Bbb{R^\omega}$. But by checking a particular metric on $\Bbb{R^\omega}$ , doesn't gives us an opportunity to make sure that the linear space $\Bbb{R^\omega}$ is not a normed space.
I also know that the existence of Hamel basis of a linear space implies the linear space is a normed space. Again to prove existence of Hamel basis we need Zorn's lemma, an equivalent version of AC.

Question: Can we define a norm in a basis-free way on $\Bbb{R^\omega}$ to make it a normed space?

 A: *

*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 $n\in\mathbb{N}$.

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*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.

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

I don't know.
