# Is a nowhere dense set closed?

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?

• Even more simple: the set $S = \{1/n\mid n\geqslant 1\}$ is nowhere dense, but not closed: $0\in\overline S\setminus S$. – Fimpellizieri Sep 1 '18 at 7:00
Hint: To fix your argument, consider $\overline{V_n}$ rather than $V_n$.