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This is not a homework assignment, its a question from a topology test which im trying to solve to prepare for my own.

I dont know how to solve it, so anything helpful will be appreciated.

Let $(X,d)$ be a complete metric space and $\{G_i : i=1,2,\ldots\}$ a countable collection of open subsets such that $B=\bigcap G_i$ is not empty. Let $\{A_j : j=1,2,\ldots\}$ be a countable collection of nowhere dense sets in $X$.

a. Prove $B\subset \overline{B\setminus\bigcup A_j}$

b. Show by counter example that the condition of completeness on $(X,d)$ is necessary.

I know that a complete metric space is a Baire space, so any intersection of open dense sets is dense, also $X\setminus\overline{A_i}$ is dense in X, but i dont quite see how to apply it. Is $B\setminus\overline{A_i}$ dense in B?

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    What have you tried? Where did you get stuck? Please don't just ask us to do your homework for you. – 2017-02-06
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    Questions posted here should not be phrased in language suitable for assigning homework. It can make people wonder if you merely copied a question without even understanding what it says, so that no actual question is present in your mind. It you need help understanding what a question says, that appears in assigned homework, you can ask about that. – 2017-02-06
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    Yeah i was in the process of editing it with proper mathjax and content while you guys started yelling at me... Its my first time trying the mathjax so it didnt work properly – 2017-02-06
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    And Noah im not asking you to solve anything, a hint, a direction to look in or any number of helpful tips would have been appreciated. – 2017-02-06
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    Aren't the $G_i$ supposed to be dense too (so open and dense)? This would make more sense in the context of Baire's theorem. – 2017-02-06
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    I checked the original about 5 times, this is exactly how it was phrased in the test... – 2017-02-07
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    How would i proceed if $G_i$ were indeed dense? – 2017-02-08

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The claim is false - take $ X = \mathbb R $, $ G_i = (-1/i, 1/i) $ and $ A_k = \{ 0 \} $ for all $ k $. Then,

$$ B = \bigcap G_i = \{ 0 \} $$ $$ C = B - \bigcup A_j = B - \{ 0 \} = \emptyset $$

so that $ \bar{C} = \emptyset $, and $ B $ is not a subset of $ \bar C $.

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    Wow thanks, thats very odd... – 2017-02-06