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I was reading a textbook and saw an alternative formulation of nowhere dense. I am not sure how to prove this alternate formulation below:

The Normal Nowhere Dense Statement:

Let $X$ be a metric space. A subset $A ⊆ X$ is called nowhere dense in $X$ if the interior of the closure of $A$ is empty, i.e. $(\overline{A})^{\circ} = ∅$. Otherwise put, $A$ is nowhere dense iff it is contained in a closed set with empty interior.

Alternate Formulation:

"Passing to complements, we can say equivalently that $A$ is nowhere dense iff its complement contains a dense open set."

Does anyone know how I can prove this? It seems rather painfully straightforward but I am not sure how to show it exactly. Thank you!

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  • $\begingroup$ Which direction do you find difficult: From "normal" to "alternate" or vice versa? $\endgroup$ Jun 10, 2014 at 20:46
  • $\begingroup$ From normal to alternate. I just cant get the intuitive feel. Thanks! $\endgroup$
    – user123276
    Jun 10, 2014 at 21:03
  • $\begingroup$ Using another definition of nowhere dense, the proof will be much easier, almost obvious. The definition: a set X is called nowhere dense iff each nonempty open set contains a nonempty open subset disjoint from X. $\endgroup$
    – Michael
    Aug 24, 2022 at 12:06
  • $\begingroup$ @Michael Sorry to bother up so late but May I know the proof using the definition that you mentioned ? $\endgroup$ Oct 18, 2023 at 12:49
  • $\begingroup$ @user-492177 X is nowhere dense <=> each nonempty open set contains a nonempty open subset disjoint from X <=> each nonempty open set contains a nonempty open subset in the complement of X <=> Let B be the union of all such open sets in the last clause, then B is a dense open set contained in the complement of X. $\endgroup$
    – Michael
    Oct 25, 2023 at 12:29

1 Answer 1

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First, you should know that, for any $B\subseteq X$, $X\setminus\overline{B}=(X\setminus B)^\circ$ and that $X\setminus B^\circ=\overline{X\setminus B}$. Now

\begin{align*} A\text{ nowhere dense }&\iff\left(\overline{A}\right)^\circ=\varnothing\\ &\iff X\setminus(\overline{A})^\circ=X\\ &\iff\overline{X\setminus \overline{A}}=X\\ &\iff\overline{(X\setminus A)^\circ}=X\\ &\iff (X\setminus A)^\circ\text{ is dense in }X\\ &\iff(X\setminus A)\text{ contains a dense open subset}. \end{align*}

The last equivalence may not be so obvious if you're not very used to metric spaces. See below, if necessary:

If $(X\setminus A)^\circ$ is dense in $X$, then $(X\setminus A)^\circ$ is a dense open subset of $X\setminus A$.

Conversely, if $(X\setminus A)$ contains a dense open subset $D$, then $D\subseteq (X\setminus A)^\circ$, so $(X\setminus A)^\circ$ is dense as well.

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