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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. And even stronger question is, is $\ell_1/\Phi$ normable?

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2 Answers 2

This is simply an expansion of Robert Israel’s answer.

For $n\in\omega$ let $e_n \in \ell_1$ be defined by $e_n(n) = 1$ and $e_n(k) = 0$ if $k \ne n$. Let $E = \{e_n:n\in\omega\}$, and note that $\Phi = \operatorname{span}E$. Extend $E$ to an ordinary basis $B$ for $\ell_1$ as a vector space. Let $B_0 = B \setminus E$. If $b_1,\dots,b_n \in B_0$ and $\alpha_1,\dots,\alpha_n \in \mathbb{R}$ are such that $\sum_{k=1}^n \alpha_k x_k \in \Phi$, then $$\sum_{k=1}^n \alpha_k b_k \in \operatorname{span}B_0 \cap \operatorname{span}E = \{0\},$$ so $\alpha_1 = \dots = \alpha_n = 0$. It follows that if $\varphi$ is the natural projection from $\ell_1$ to $\ell_1/\Phi$, $\varphi[B_0]$ is a vector space basis for $\ell_1/\Phi$. It also follows that $\varphi \upharpoonright B_0$ is one-to-one, so $\vert \varphi[B_0] \vert = \vert B_0 \vert = \vert B \vert$. (The last equality follows from the countability of $E$ and the fact that $B_0 = B \setminus E$ is infinite.) Thus, $\ell_1$ and $\ell_1/\Phi$ are real vector spaces of the same dimension and must therefore be isomorphic as vector spaces. Let $h:\ell_1/\Phi \to \ell_1$ be a vector space isomorphism.

Define a norm on $\ell_1/\Phi$ by setting $\|x\| = \|h(x)\|_1$, where $\|\cdot\|_1$ is the ordinary $\ell_1$ norm; $h$ is then an isomorphism of normed linear spaces, and of course $\|\cdot\|$ induces not just a Hausdorff, but a metric topology on $\ell_1/\Phi$. But as Robert said, this construction depends heavily on the axiom of choice.

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As vector spaces, $\ell_1/\Phi$ and $\ell_1$ (or, for that matter, any other separable Banach space) are isomorphic, so, if you don't ask for any particular relation between the quotient topology of $\ell_1/\Phi$ and that given by your new norm, the answer is yes. However, I doubt that such a norm would be constructible - this relies strongly on the Axiom of Choice.

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Thank you for you answer, however, in what topology is \ell_1/\Phi separable? It cannot be in the quotient topology, since it is not Hausdorff while a normed space is. Am I missing something? –  user15806 Sep 9 '11 at 21:57
    
@Ivan: Although $\ell_1/\Phi$ is not Hausdorff, it is separable in the quotient topology since continuous images of separable spaces are separable. –  LostInMath Sep 10 '11 at 0:16

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