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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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    $\begingroup$ 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. $\endgroup$
    – Benoît Kloeckner
    Commented Sep 9, 2011 at 19:40
  • $\begingroup$ What Benoit said. Do you want this non-trivial metric-induced topology to be weaker than the quotient one? stronger? $\endgroup$
    – Yemon Choi
    Commented Sep 9, 2011 at 20:10
  • $\begingroup$ Crossposted to MSE: math.stackexchange.com/questions/63179/… where it has received answers. I therefore vote to close here $\endgroup$
    – Yemon Choi
    Commented Sep 28, 2011 at 21:40

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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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  • $\begingroup$ Please add your comments to the original question, rather than adding them as an "answer" $\endgroup$
    – Yemon Choi
    Commented Sep 9, 2011 at 23:22
  • $\begingroup$ 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. $\endgroup$
    – Yemon Choi
    Commented Sep 9, 2011 at 23:23
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    $\begingroup$ 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. $\endgroup$
    – Ivan
    Commented Sep 10, 2011 at 0:44
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    $\begingroup$ 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. $\endgroup$
    – Yemon Choi
    Commented Sep 10, 2011 at 1:21
  • $\begingroup$ 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. $\endgroup$
    – j.c.
    Commented Sep 29, 2011 at 3:58

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