I am trying to prove an equivalent weak version of Baire's category theorem which states that $:$

Every complete metric space is a set of second category or non-meagre.

I am trying to prove it using another version of Baire's category theorem which states that

Let $(X,d)$ be a metric space. Let $\{U_n \}_{n \geq 1}$ be countable collection of open dense subsets of $X$. Then $$\cap_{n=1}^{\infty} U_n$$ is dense in $X$.

Now suppose that $X$ is complete. If $X$ was meagre then $$X= \cup_{n=1}^{\infty} V_n$$ where each $V_n$ in the union is a nowhere dense set. So $$\cap_{n=1}^{\infty} (X \setminus V_n)= \emptyset.$$ Now complement of a nowhere dense set is dense. So if nowhere dense sets are closed then we get a contradiction by the above theorem for otherwise $\emptyset$ becomes dense in $X$ which is an impossibility. That will prove the weak equivalent version of Baire's category theorem which I have stated at the beginning.

Now my question is $:$

Is every nowhere dense set closed?

If it is so then why? Please help me in this regard.

Thank you in advance.

  • 1
    $\begingroup$ Can't you remove a point from, say, the Cantor set to obtain a nowhere dense set that is not closed? $\endgroup$ – Bungo Sep 1 '18 at 6:57
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    $\begingroup$ Even more simple: the set $S = \{1/n\mid n\geqslant 1\}$ is nowhere dense, but not closed: $0\in\overline S\setminus S$. $\endgroup$ – Fimpellizieri Sep 1 '18 at 7:00
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    $\begingroup$ I mean: a nowhere dense set is a set such that its closure has empty interior. If they were all closed, what point would there be in keeping that terminological detour? "If and only if closed with empty interior" would be like lemma 1. $\endgroup$ – Saucy O'Path Sep 1 '18 at 7:01
  • $\begingroup$ So what should be my argument to prove the required theorem? $\endgroup$ – Arnab Chatterjee. Sep 1 '18 at 7:02
  • $\begingroup$ A nowhere dense set is not necessarily closed, but the closure of a nowhere dense set is still nowhere dense, and is of course closed. Does that help? $\endgroup$ – bof Sep 1 '18 at 7:04

Hint: To fix your argument, consider $\overline{V_n}$ rather than $V_n$.

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  • $\begingroup$ I have already understood it. See my above comment @Fimpellizieri. Thanks for your answer. $\endgroup$ – Arnab Chatterjee. Sep 1 '18 at 7:15

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