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The following question came up during a reading of Rudin's functional analysis. I have not been able to find any information through searching online, but I apologise if the answer is obvious, or the question is trivial.

Consider $\Phi$ to be the space of sequences that have finitely many non-zero terms. The space is not closed in $\ell_1$, therefore $\ell_1/\Phi$ with the quotient topology is not Hausdorff, and so it cannot be metrizable. However, does there exist a metric on $\ell_1/\Phi$ that gives rise to a non-trivial topology? Furthermore, is $\ell_1/\Phi$ normable?

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You should explain what you mean by non-trivial. If you do not specify a way the topology is to be related to the topology of $\ell^1$ and to the quotient map, then of course there are a lot of non-trivial toplogies on this quotient set -- just find a interesting topological space with the same cardinal. –  Benoît Kloeckner Sep 9 '11 at 19:40
    
What Benoit said. Do you want this non-trivial metric-induced topology to be weaker than the quotient one? stronger? –  Yemon Choi Sep 9 '11 at 20:10
    
Crossposted to MSE: math.stackexchange.com/questions/63179/… where it has received answers. I therefore vote to close here –  Yemon Choi Sep 28 '11 at 21:40

1 Answer 1

Thank you for your answers, and please excuse my lack of proficiency in the subject. With that been said, I am still confused how there could be such choice in topology. For example, how can there be a weaker-than-the-quotient metrizable topology, wouldn't this make the quotient topology Hausdorff? Further, is there a metrizable topology for any cardinal number space which is not the trivial one or the topology of all subsets?

What about the existence of a norm on the space, regardless of the quotient topology? Is it constructable?

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Please add your comments to the original question, rather than adding them as an "answer" –  Yemon Choi Sep 9 '11 at 23:22
    
It is hard to answer your further questions without you making more precise what you mean by "a metric on $\ell^1/\Phi$". You seem to be assuming that such a metric is related in some way to the given quotient topology, but you do not say so explicitly. –  Yemon Choi Sep 9 '11 at 23:23
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Thank you again for your response. I now realize that the original question was very poorly defined. What I was trying to do originally was to see if a seminorm exists on \ell_1 that vanishes on \Phi and therefore gives rise to a norm on the \ell_1/\Phi. Such a norm cannot be continuous with respect to the \ell_1 norm, but the point was to see if it would be possible to explicitly construct a norm on the quotient space(and not use an axiom of choice type of argument). Asking about metrizability was pointless. To the moderators, I apologize for posting a comment earlier as an answer. –  Ivan Sep 10 '11 at 0:44
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Ivan: thanks for clarifying. I suggest that you edit your original question, and add this more precise version to it - people don't always read through the comments. –  Yemon Choi Sep 10 '11 at 1:21
    
Ivan, you may need to register an account and have it merged with the unregistered one you used post the question. Just flag your post for moderator attention when you've registered. –  j.c. Sep 29 '11 at 3:58

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