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Let $\mathbb{R}^\omega$ be the countably infinite product of $\mathbb{R}$ with itself in the product topology. $\mathbb{R}^\omega$ is metrizable but the metric doesn't arise from a norm, so a natural analogue to a unit sphere is the quotient of $\mathbb{R}^\omega - \{0\}$ under the action of $(0, \infty)$ (that is, we say two elements of $\mathbb{R}^\omega$ are equivalent if they are positive multiples of one another).

Denote the quotient space $\mathbb{R}^\omega - \{0\} \: / \: \mathbb{R}^+$. Is $\mathbb{R}^\omega - \{0\} \: / \: \mathbb{R}^+$ itself metrizable?

So far I've shown the quotient map $q: \mathbb{R}^\omega \rightarrow \mathbb{R}^\omega - \{0\} \: / \: \mathbb{R}^+$ is open. Since $\mathbb{R}^\omega$ is second countable, the quotient space is also second countable, so I was hoping it would be easy to prove the quotient space is regular and then apply the Urysohn Metrization Theorem. Unfortunately I don't think the quotient map is closed, and showing regularity by hand seems a bit difficult, so I'm stuck. Any help or advice would be appreciated, and let me know if more details would be helpful.

Dan Asimov
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Compact_Ball
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1 Answers1

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Let us call your quotient space $S$. I claim that $S$ is in fact not regular. For instance, let $x\in\mathbb{R}^\omega$ be the point whose first coordinate is $1$ and all other coordinates are $0$, and let $U$ be the image in $S$ of the set of all points with positive first coordinate. Then $U$ is a neighborhood of $q(x)$ in $S$, but I claim there is no neighborhood $V$ of $q(x)$ such that $\overline{V}\subseteq U$. Indeed, for any such $V$, there would exist $n$ such that $q^{-1}(V)$ contains all points of $\mathbb{R}^\omega-\{0\}$ whose first $n$ coordinates agree with those of $x$. In particular, if $y_N$ has first coordinate $1$, $(n+1)$st coordinate $N$, and all other coordinates $0$, then $y_N\in q^{-1}(V)$, and hence $y_N/N\in q^{-1}(V)$ as well. As $N\to\infty$, $y_N/N$ converges to the point $y$ whose $(n+1)$st coordinate is $1$ and all other coordinates are $0$. Thus $q(y)\in\overline{V}$, but $q(y)\not\in U$.

Eric Wofsey
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