Quotient of \ell_1 by space of finite sequences

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?

-
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 2011 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 2011 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 2011 at 21:40

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?

-
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 2011 at 23:23