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Given the space $\mathbb N^ \mathbb N$ with the topology generated by basis sets of the form:

$$[V,n] = \{x \in \mathbb N^ \mathbb N ; V \text{ is an n prefix of x}\}$$

I can see that this space is separable.

My question is: is it metrisable? or even completely metrisable?

Thank you!

Martin Sleziak
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topsi
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    This is completely metrizable, and in fact it plays an important role in the study of descriptive set theory in complete separable metric spaces. It's often called Baire Space. – Dave L. Renfro Jan 27 '14 at 17:35
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    I think that this was asked before on the site... Perhaps under a guise of "$\Bbb{N^N}$ under the product topology ...", in which case all you have to do is show that this topology is in fact the product topology. – Asaf Karagila Jan 27 '14 at 17:49
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    The countable product of metric spaces is always metrizable. – Michael Greinecker Jan 27 '14 at 17:50
  • ok. Thank you all for your clues. i'll check it out now. Thanks!! – topsi Jan 27 '14 at 18:16
  • Maybe a duplicate of some of the following questions? http://math.stackexchange.com/questions/250966/proof-that-omega-omega-is-completely-metrizable-and-second-countable, http://math.stackexchange.com/questions/586469/prove-that-baire-space-omega-omega-is-completely-metrizable, http://math.stackexchange.com/questions/49859/basic-questions-about-mathbbz-mathbbn-with-the-product-topology (The first one seems to be the closets, since the topology is defined using finite sequences, similarly as in the OP.) – Martin Sleziak Jan 29 '14 at 07:33

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