# 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?

• Concerning your first questeion: You are looking for a norm that induces the same topology as $d$? Or just any norm?
– Gerd
Commented Dec 15, 2021 at 10:22
• And in addition you want to avoid a Hamel base? As you write you can define norms by means of a Hamel base.
– Gerd
Commented Dec 15, 2021 at 10:27
• I posted an answer how to get norms form a Hamel base. A norm on $\mathbb{R}^\omega$ without Hamel base seems difficult to me. Maybe someone else has an idea.
– Gerd
Commented Dec 15, 2021 at 10:53
• @Gerd: But isn't that exactly not what the question is about? Commented Dec 15, 2021 at 18:31
• @DanielWainfleet No, $\|x\|=d(x,0)$ fails badly the homogeneity $\|tx\|=|t| \|x\|$. Commented Dec 20, 2021 at 13:21

• 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.

• Very impressive answer! Commented Jan 5, 2022 at 9:53
• Alternative proof: $\mathbb{R}^\omega$ is a Polish group, so $B+B$ contains a basic open neighborhood $U\ni 0$ of zero, but that means for some $N\in\mathbb{N}$ and all $a\in\mathbb{R}$ we have $ae_N\in U$ or $\|ae_N\|\le 2$. Commented Jan 17, 2022 at 2:02