1
$\begingroup$

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?

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.