Basic Theorems Regarding Nowhere Dense Sets in a Topological Space

# Basic Theorems Regarding Nowhere Dense Sets in a Topological Space

Recall from the Dense and Nowhere Dense Sets in a Topological Space page that if $(X, \tau)$ is a topological space then a set $A \subseteq X$ is said to be nowhere dense if:

(1)\begin{align} \quad \mathrm{int}(\bar{A}) = \emptyset \end{align}

We will now look at some basic theorems regarding nowhere dense sets in topological spaces.

Theorem 1: Let $(X, \tau)$ be a topological space and let $A$ be a nowhere dense subset of $X$. If $B \subseteq A$ then $B$ is nowhere dense in $X$. |

**Proof:**Suppose that $A$ is nowhere dense in $X$. Then $\mathrm{int} (\bar{A}) = \emptyset$.

- Let $B \subseteq A$ and suppose that $\mathrm{int} (\bar{B}) \neq \emptyset$. Then let $x \in \mathrm{int}(\bar{B})$.

- Since $B \subseteq A$ we have that $\bar{B} \subseteq \bar{A}$ by the one of the theorems on the Basic Theorems Regarding the Closure of Sets in a Topological Space.

- But since $\bar{B} \subseteq \bar{A}$, we have that $\mathrm{int}(\bar{B}) \subseteq \mathrm{int} (\bar{A})$ by one of the theorems on the Basic Theorems Regarding the Interior Points of Sets in a Topological Space page.

- So $x \in \mathrm{int}(\bar{A})$ which contradicts $A$ being nowhere dense in $X$.

- Thus $\mathrm{int} (\bar{B}) = \emptyset$, i.e., $B \subseteq A$ is also a nowhere dense set in $X$. $\blacksquare$

Theorem 2: Let $(X, \tau)$ be a topological space and let $A$ be a nowhere dense subset of $X$. Then $\bar{A}$ is nowhere dense in $X$. |

- Let $A$ be a nowhere dense subset of $X$. Then $\mathrm{int} (\bar{A}) = \emptyset$. But $\bar{\bar{A}} = \bar{A}$, so $\mathrm{int} (\bar{\bar{A}}) = \emptyset$ which shows that $\bar{A}$ is nowhere dense in $X$. $\blacksquare$

Theorem 3: Let $(X, \tau)$ be a topological space and let $\{ A_1, A_2, ..., A_n \}$ be a finite collection of nowhere dense subsets of $X$. Then $\displaystyle{A = \bigcup_{i=1}^{n} A_i}$ is nowhere dense in $X$. |

**Proof:**Suppose that $\{ A_1, A_2, ..., A_n \}$ is a finite collection on nowhere dense subsets of $X$. Then $\mathrm{int}(\bar{A_i}) = \emptyset$ for all $i \in \{1, 2, ..., n \}$.

- Suppose that $\displaystyle{A = \bigcup_{i=1}^{n} A_i}$ is not nowhere dense in $X$. Then:

\begin{align} \quad \mathrm{int} \left( \overline{\bigcup_{i=1}^{n} A_i} \right ) \neq \emptyset \\ \end{align}

- But:

\begin{align} \quad \emptyset \neq \mathrm{int} \left ( \overline{\bigcup_{i=1}^{n} A_i} \right) = \mathrm{int} \left ( \bigcup_{i=1}^{n} \bar{A_i} \right ) \subseteq \bigcup_{i=1}^{n} \mathrm{int} (\bar{A_i}) \end{align}

- But $\mathrm{int} (\bar{A_i}) = \emptyset$ for all $i \in \{1, 2, ..., n \}$, so the inclusion above is a contradiction.

- Hence if $\{A_1, A_2, ..., A_n \}$ is a finite collection of nowhere dense subsets of $X$, then $\displaystyle{\mathrm{int}(\bar{A_i}) = \emptyset}$.