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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?

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  • $\begingroup$ (2) to (1) is superfluous once you have a full circle of implications. $\endgroup$ Feb 17, 2019 at 11:28
  • $\begingroup$ 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). $\endgroup$ Feb 17, 2019 at 11:32
  • $\begingroup$ okay. Thank you. what do you mean at (4). I don't understand. Can you explain? $\endgroup$
    – user464147
    Feb 17, 2019 at 11:42
  • $\begingroup$ 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. $\endgroup$ Feb 17, 2019 at 11:45
  • $\begingroup$ I find the last sentence of (5) to (1) to quickly concluded... $\endgroup$ Feb 17, 2019 at 15:04

1 Answer 1

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In a more succint formulation:

(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 $A$.

(4) to (5): open spheres (I prefer the term balls) form a base so this is trivial: any non-empty open set contains an open ball.

(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.

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  • $\begingroup$ I don't understand where did you conclude open ball in int$(A)$ intersects $\overline{A}$ $\endgroup$
    – chesslad
    Nov 21, 2019 at 21:21
  • $\begingroup$ @Abhay $\operatorname{int}(A) \subseteq A \subseteq \overline{A}$ so that's trivial. $\endgroup$ Nov 21, 2019 at 23:08
  • $\begingroup$ But we have to show that it is disjoint from $A$ and not just $\bar{A}$ how do we know there is an open ball in $int(A)$ $\endgroup$
    – chesslad
    Nov 22, 2019 at 1:43
  • $\begingroup$ @Abhay it’s a proof by contrapositive, mind you. $\endgroup$ Nov 22, 2019 at 5:09

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