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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?

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    $\begingroup$ Concerning your first questeion: You are looking for a norm that induces the same topology as $d$? Or just any norm? $\endgroup$
    – Gerd
    Dec 15, 2021 at 10:22
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    $\begingroup$ And in addition you want to avoid a Hamel base? As you write you can define norms by means of a Hamel base. $\endgroup$
    – Gerd
    Dec 15, 2021 at 10:27
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    $\begingroup$ 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. $\endgroup$
    – Gerd
    Dec 15, 2021 at 10:53
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    $\begingroup$ @Gerd: But isn't that exactly not what the question is about? $\endgroup$
    – Asaf Karagila
    Dec 15, 2021 at 18:31
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    $\begingroup$ @DanielWainfleet No, $\|x\|=d(x,0)$ fails badly the homogeneity $\|tx\|=|t| \|x\|$. $\endgroup$
    – Jochen
    Dec 20, 2021 at 13:21

1 Answer 1

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  • 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.

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  • $\begingroup$ Very impressive answer! $\endgroup$
    – Jochen
    Jan 5, 2022 at 9:53
  • $\begingroup$ 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 $\|ae_N\|\le 2$. $\endgroup$
    – Edward H
    Jan 17, 2022 at 2:02

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