Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

up vote 2 down vote accepted

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

share|improve this answer
    
what is 'basic'? are you assuming convexity? –  Tom Artiom Fiodorov Jan 23 '12 at 23:38
1  
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
3  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.