# Prove the equivalent conditions for nowhere dense subset.

Let $$(X,d)$$ be a metric space and $$A$$ be a subset of $$X$$. Then the following statements are equivalent.

1. $$A$$ is nowhere dense.

2. $$\overline{A}$$ doesn't contain any non-empty open set.

3. Each non-empty open set has a non-empty open subset which is disjoint from
$$\overline{A}$$

4. Each non-empty open set has a non-empty open subset which is disjoint from
$${A}.$$

5. Each non-empty open set has a non-empty open sphere which is disjoint from
$${A}.$$

It is the statements given in the Textbook 'Topology and modern analysis' by G.F Simmons.

My attempt:- $$(1)\implies (2)$$

Let $$A$$ be a nowhere dense subset of $$X$$. Suppose there is an open set $$U$$ lie inside $$\overline A$$. That is Every point inside the open set $$U$$ will be an interior point of $$\overline A.$$ Which contradict the fact that $$A$$ is no where dense.

$$(2)\implies (1)$$

Every open ball is an open set. So, any open ball centred at the point from $$\overline A$$ does not lie in $$\overline A$$. So, interior of $$\overline A$$ is empty. Hence, $$A$$ is a Nowhere dense subset of $$X$$.

$$(1)\implies (3)$$

Suppose $$A$$ is nowhere dense and Suppose there exists a nonempty open set $$U$$ of $$X$$ such that if $$V$$ is any nonempty subset of $$U$$ then $$V\cap \overline A \neq \emptyset.$$ Let $$x\in U$$, where $$U$$ is an open set. By our assumption, If $$W$$ is an open set contain $$x$$. So, $$x\in W\cap \overline A \neq \emptyset.$$ Then, $$x\in W\cap U\subseteq U$$, so $$x\in W\cap U\cap \overline A \neq \emptyset \implies$$ $$U\cap \overline A \neq \emptyset.$$ Hence, $$x\in \overline{\overline{A}}=\overline{A}.$$ Hence, interior of $$\overline{A}\neq \emptyset.$$ Which contradict to the fact that $$A$$ is nowhere dense.

$$(3)\implies (4)$$

Each non-empty open set has a non-empty open subset which is disjoint from $$\overline{A}$$. We know that $$A\subseteq \overline A.$$ Each non-empty open set has a non-empty open subset which is disjoint from $${A}.$$

$$(4)\implies (5)$$.

For every nonempty open set $$U$$ there is an open ball which lies in $$U.$$ Hence, Each non-empty open set has a non-empty open sphere which is disjoint from $${A}.$$

$$(5)\implies (1)$$

Suppose $$A$$ is not nowhere dense subset of $$X$$ and satisfies $$(5).$$ $$A$$ is not nowhere dense subset of $$X \implies$$ interior of $$\overline {A}$$ is non empty. Let $$x\in$$ interior of $$\overline {A}$$ So there exists a an open set containing $$x$$, $$U$$, lies inside $$\overline {A}.$$ Hence there is an open sphere centred at $$x$$ lies inside $$\overline {A}$$. Which contradict $$(5)$$.

Is my arguments correct?

• (2) to (1) is superfluous once you have a full circle of implications. – Henno Brandsma Feb 17 at 11:28
• The rest of the arguments seem in essence correct. The differences between the statements are only slight, so most steps are trivial. (4) is the explanation of the name: $A$ is not dense in any non-empty open set (so "nowhere" dense). – Henno Brandsma Feb 17 at 11:32
• okay. Thank you. what do you mean at (4). I don't understand. Can you explain? – Unknown x Feb 17 at 11:42
• In every open set $U$ there is some open subset $V$ (all non-empty) that misses $A$ and this shows that the closure of $A\cap U$ in $U$ is not $U$, so “$A$ is not dense in $U$”. Hence the name. – Henno Brandsma Feb 17 at 11:45
• I find the last sentence of (5) to (1) to quickly concluded... – Henno Brandsma Feb 17 at 15:04

(1) equivalent to (2) is trivial: $$\operatorname{int}(\overline{A}) \neq \emptyset$$ iff $$\overline{A}$$ contains a non-empty open set (by definition of the interior as the largest open subset of $$A$$).
(2) to (3): if $$U$$ is non-empty open, then by (2): $$U \nsubseteq \overline{A}$$, so $$V:= U \setminus \overline{A}$$ is non-empty and open, and by definition it's disjoint from $$\overline{A}$$.
(3) to (4): trivial, as $$A \subseteq \overline{A}$$. Any $$V$$ disjoint from $$\overline{A}$$ is a fortiori disjoint from $$A$$.
(5) to (1). If $$x\in \operatorname{int}(\overline{A})$$, then for some $$r>0$$ we have $$B(x, r) \subseteq \overline{A}$$. Then then any open ball inside $$\operatorname{int}(A)$$ intersects $$\overline{A}$$. Which contradicts (5) directly, so $$A$$ is nowhere dense.