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  1. #1
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    How can I show

    That subset of a set of 1st category (can write as countable union of nowhere dense sets) is also 1st category? Intuitively it is clear, but I am not sure how to argue formally

    Thank you for helping
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  2. #2
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    Quote Originally Posted by MMath09 View Post
    That subset of a set of 1st category (can write as countable union of nowhere dense sets) is also 1st category?
    If X is of category I then X = \bigcup\limits_{n \in \mathbb{N}} {A_n } where each {A_n } is nowhere dense.


    B \subseteq X\; \Rightarrow \;B \subseteq \bigcup\limits_{n \in \mathbb{N}} {A_n } \; \Rightarrow \;\bigcup\limits_{n \in \mathbb{N}} {\left( {B \cup A_n } \right)}
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  3. #3
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    Quote Originally Posted by Plato View Post
    If X is of category I then X = \bigcup\limits_{n \in \mathbb{N}} {A_n } where each {A_n } is nowhere dense.


    B \subseteq X\; \Rightarrow \;B \subseteq \bigcup\limits_{n \in \mathbb{N}} {A_n } \; \Rightarrow \;\bigcup\limits_{n \in \mathbb{N}} {\left( {B \cup A_n } \right)}
    Is this last bit supposed to be X = \;\bigcup\limits_{n \in \mathbb{N}} {\left( {B \cup A_n } \right)} ? So you write the set X as the union of the set X and the subset B

    So if B was made up of somewhere dense sets or an uncountable number of sets then you could write X like that too & that would be a contradiction

    OK thanks
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