# Prove that if a set is nowhere dense iff the complement of the closure of the set is dense.

I am able to prove the iff in the forward direction. But, I am having trouble proving the statement in other direction. I am trying to use the definition of dense, but I am not getting anywhere with it.

## 4 Answers

Let $U \subseteq X$ be an open set and let $S \subseteq X$ be the set in question. Denseness is equivalent to saying that it intersects every open set non-trivially, hence $U \cap \bar{S}^c \neq \emptyset$. Considering properties of the complement, this is the same as saying $U$ is not a subset of $\bar{S}$. Since the closure of $S$ does not contain an open set, it has empty interior and so $S$ is nowhere dense.

Let $S^{c}$ denote the complement of $S$ in $X$ and $cl(S)$ the closure of $S$ in $X$.

To prove that $S$ is nowhere dense, any open set $U \subseteq cl(S)$ must be empty. Let $U$ be such a set.

The complements are included in the opposite order, that is $cl(S)^{c} \subseteq U^{c}$. Taking the closures of both sides and noting that $U^{c}$ is closed, this gives $cl( cl(S)^{c})) \subseteq U^{c}$.

But $cl(S)^{c}$ is by assumption dense, which (by definition) means that $cl( cl(S)^{c})) = X$. We thus got ourselves a relation $X \subseteq U^{c}$. This is possible only if $U = \emptyset$.

Hence any open subset of $cl(S)$ must be empty and thus $S$ is nowhere dense.

Let $A^c$ be complement of $A$, $\overline{A}$ be closure of $A$ and $A^o$ be interior of $A$.

We define:

1). $A$ is dense iff $\overline{A}=X$.

2). $A$ is nowhere dense iff $(\overline{A})^o=\varnothing$.

Edit:

We will use the facts that $(A^c)^c=A$ (Involution) in the following proof. First we prove following lemma $$\overline{A}=((A^c)^o)^c\tag1$$ By definition, $A^o$ (interior set of $A$) is the largest open contained in $A$. So $(A^c)^o$ is the largest open contained in $A^c$, i.e $(A^c)^o\subset A^c$. Thus $A=(A^c)^c\subset ((A^c)^o)^c$. Since $(A^c)^o$ is open, $((A^c)^o)^c$ is closed. So this means that $((A^c)^o)^c$ is the smallest closed set containing $A$. By definition of closure, $(1)$ follows.

Now if $A$ be nowhere dense, then $(\overline{A})^o=\varnothing$. Since by $(1)$ and Involution $$\overline{(\overline{A})^c}=((((\overline{A})^{c})^c)^o)^c=((\overline{A})^o)^c=\varnothing^c=X$$ i.e. $(\overline{A})^c$ is dense.

Second if $(\overline{A})^c$ is dense, then $\overline{(\overline{A})^c}=X$ and $$((\overline{A})^o)^c=((((\overline{A})^{c})^c)^o)^c=\overline{(\overline{A})^c}=X$$ So $(\overline{A})^o=\varnothing$, i.e $A$ is nowhere dense.

• I am not sure how you simplified the initial expressions in the two steps. – frank Jan 28 '16 at 6:28
• Use the involution, i.e. $(A^c)^c=A$. Also use the fact: $\overline{A}=(((A)^c)^o)^c$. – Math Wizard Jan 28 '16 at 6:29
• It is unclear to me how you went(for each line) went from the first step to the second. I understand the following steps. – frank Jan 28 '16 at 6:34
• By $\overline{A}=(((A)^c)^o)^c$, we have $\overline{(\overline{A})^c}=((((\overline{A})^{c})^c)^o)^c$. By involution, $((\overline{A})^{c})^c=\overline{A}$. So $((((\overline{A})^{c})^c)^o)^c=((\overline{A})^o)^c$ – Math Wizard Jan 28 '16 at 6:37
• Can you explain why the first fact is true? – frank Jan 28 '16 at 6:39

Assume that complement of (clA) is dense. Then cl[complement of (clA)] = X. So complement of (int clA) = X. Thus int clA is empty. Hence A is nowhere dense.