Let B be a reflexive and infinite dimensional real Banach space-which could be Hilbert space l^2- and let B be endowed with the weak topology. Although this topology is regular and Hausdorff, it is not metrizable. But is it close enough to being metrizable to enable B to satisfy the axioms for a Moore space? My motive for asking this question is to obtain a clearer understanding of the relationships that exist between various topologies which are closely connected with and closely resemble metrizable ones-even though they may not themselves be metrizable.
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asked Jul 4, 2014 at 18:59
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4$\begingroup$ You might want to include the definition of "Moore space" into your question. $\endgroup$– André HenriquesCommented Jul 4, 2014 at 20:26
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$\begingroup$ A Moore space is a regular, Hausdorff developable topological space $\endgroup$– Garabed GulbenkianCommented Jul 5, 2014 at 20:14
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$\begingroup$ I see, so your question is, essentially: are Banach spaces with the weak topology regular and developable? I'm pretty sure that they are regular, so the question is really: are they developable? $\endgroup$– André HenriquesCommented Jul 5, 2014 at 21:36
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$\begingroup$ If $B$ is separable, we can just take $C_n=\{ x + B_{1/n}(0): x\in B\}$ as the required coverings (as in item (1) of the Wikipedia article on Moore spaces; $B_{1/n}$ of course refers to a metrization of the unit ball). $\endgroup$– Christian RemlingCommented Jul 5, 2014 at 23:57
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$\begingroup$ @Christian Remling: Many thanks for your response. Since the space l^2 is separable, it seems that if we take B to be l^2, then the anmswer to my question is "YES". $\endgroup$– Garabed GulbenkianCommented Jul 6, 2014 at 18:50
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