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I learnt from a paper that "Let cov(K) be the least cardinal k such that a perfect Polish space can be expressed as a union of k meager sets. (It does not matter which perfect Polish space is used to define cov(K) ) " . I don't know why cov(K)'s are equal for different perfect Polish spaces. I want to use this result in my own paper, but I fail to find any books contain this result, could anybody kindly to help me ?

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This should be in standard texts on descriptive set theory, like Moschovakis's "Descriptive Set Theory" or Kechris's "Classical Descriptive Set Theory". The basic idea is that any two perfect Polish spaces become homeomorphic after you remove (at most) countably many points. Countably many points constitute a meager set and therefore don't affect cov(K).

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    $\begingroup$ Not points' but nowhere dense sets' I think:after you remove the boundaries of the memebrs of some countable base what is left is zero-dimenional and Polish and perfect and nowhere locally compact, hence homeomorphic to the space of irrationals. NB the plane minus a countable set and the line minus a countable set are not homeomorphic: the former is connectedm the latter is not. $\endgroup$
    – KP Hart
    Sep 15, 2010 at 17:33
  • $\begingroup$ Thanks for your help, i have understood the union of the boundaries of the memebrs of some countable base is meager, and i have found Alexanderov-Urysohn theorem which tell us baire space is unique up to zero-dimenional and Polish and perfect and nowhere locally compact, but I don't know why does a perfect polish space became a baire space with the union of boundaries removed ? $\endgroup$
    – sonicyouth
    Sep 16, 2010 at 4:42
  • $\begingroup$ The starting space is Polish after removing those boundaries you have a G_delta-subset of that space, now apply the theorem that a G_delta-subset of a completely metrizable space is completely metrizable (using some other metric of course). $\endgroup$
    – KP Hart
    Sep 16, 2010 at 6:52
  • $\begingroup$ Oh, and to avoid confusion: Baire space' (capitalized) is the topological product N^N, whereas a baire space' is a space that is not meager in itelf, It seems to me that you confused the two but I might be wrong. $\endgroup$
    – KP Hart
    Sep 16, 2010 at 6:58
  • $\begingroup$ oh, it is my fault, the "b" should be capitalized. thanks again, i think i have understood. $\endgroup$
    – sonicyouth
    Sep 16, 2010 at 7:07
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Andreas gave an argument while I was typing, so let me give a different approach: The proof that the covering number is the least $\kappa$ for which ${\sf MA}_\kappa({\rm countable})$ fails is soft enough that it can be easily adapted to work for any Polish space you begin with. Details of this equivalence are, for example, in the excellent ``Set theory: on the structure of the real line'' by Tomek Bartoszynski and Haim Judah, that you will want to have as a reference anyway.

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  • $\begingroup$ Thank you! I have found the charactorization of cov(M) for R^1 in that book, but it is not mentioned cov(M) for general perfect polish spaces, do you mean the proof of the charactorizations also relevant to any perfect polish spaces ? $\endgroup$
    – sonicyouth
    Sep 16, 2010 at 4:39

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