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Claim: $$R^\infty \text{ is not a Banach space when equipped with its natural product topology}$$ I need help proving this 'obvious' claim. I just got acquainted with a definition of a product topology and the concept does not seem to be easy to work with. How would I even go about showing whether $R^\infty$ is metrizable?

Edit: Intuitively, I know that $$d(x,y)=\sum_{j=1}^{\infty}\frac{1}{2^j} \frac{|x_j-y_j|}{1+|x_j-y_j|}$$ should prove that $R^{\infty}$ is metrizable.

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Off-hand, I'm not sure about the question you ask. But are you also asking for a proof of the claim, or are you fine with that? – user16299 Jan 23 '12 at 23:24
I am asking for a proof of a claim. Sorry for omitting it, I will edit – Tom Artiom Fiodorov Jan 23 '12 at 23:27

1 Answer 1

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Hint: any basic neighbourhood of $0$ in the product topology will contain a straight line through $0$.

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what is 'basic'? are you assuming convexity? – Tom Artiom Fiodorov Jan 23 '12 at 23:38
A basic neighbourhood of $0$ in the product topology of ${\mathbb R}^\infty$ is $\prod_{j \in N} U_j$ where all but finitely many $U_j$ are $\mathbb R$ and the others are arbitrary neighbourhoods of $0$ in $\mathbb R$. Convexity is not relevant here. – Robert Israel Jan 24 '12 at 0:20
That's pretty much the standard definition of the product topology. If you prefer to use your metric, the hint still works: show that $\{x: d(x,0) < \varepsilon\}$ contains a straight line through $0$. – Robert Israel Jan 24 '12 at 0:24
I was just wondering whether basic neighbourhood was different to a neighbourhood. I understand why it has a line segment, thank you. Should I try to show that $R^{\infty}$ is not complete or that there's no norm? – Tom Artiom Fiodorov Jan 24 '12 at 0:46
Show that for every norm, the unit ball does not contain a straight line through $\mathbf{0}$. $\hspace{0.8 in}$ – Ricky Demer Jan 24 '12 at 1:29

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