The Open and Closed Sets of a Topological Space

# The Open and Closed Sets of a Topological Space

Consider a topological space $(X, \tau)$. We will now define exactly what the open and closet sets of this topological space are.

Definition: Let $(X, \tau)$ be a topological space. If $A \subseteq X$ is such that $A \in \tau$ then $A$ is said to be Open. A subset $A \subseteq X$ is said to be Closed if $A^c = X \setminus A$ is open. If $A \subseteq X$ are both open and closed, then $A$ is said to be Clopen. |

For example, consider the set $X = \{ a, b, c \}$ and the nested topology $\tau = \{ \emptyset, \{ a \}, \{ a, b \}, \{ a, b, c \} \}$. Then all elements in $\tau$ are open and then the sets $\{ b, c \}$ and $\{ c \}$ are closed sets since:

(1)\begin{align} \quad \{ b, c \}^c = \{ a \} \: \mathrm{is \: open} \quad \mathrm{and} \quad \{ c \}^c = \{ a, b \} \: \mathrm{is \: open} \end{align}

Further we note that $\emptyset^c = X$ and $X^c = \emptyset$ so by definition, $\emptyset$ and $X$ are both open AND closed, i.e., clopen! In general, it is possible that other subsets of $X$ are both open and closed, or neither.

We will now look at an important result regarding closed sets.

Theorem 1: Let $(X, \tau)$ be a topological space. Then:a) $\emptyset$ and $X$ are closed sets.b) If $\{ U_i \}_{i \in I}$ is an arbitrary collection of closed subsets of $X$ for some index set $I$ then $\displaystyle{\bigcap_{i \in I} U_i}$ is closed.c) If $\{ U_1, U_2, ..., U_n \}$ is a finite collection of closed subsets of $X$ then $\displaystyle{\bigcup_{i=1}^{n} U_i}$ is closed. |

**Proof of a)**The complement of $\emptyset$ is $\emptyset^c = X$ which is open, so $\emptyset$ is closed. Similarly, the complement of $X$ is $X^c = \emptyset$ which is open. Therefore $\emptyset$ and $X$ are closed. $\blacksquare$

**Proof of b)**Let $\{ U_i \}_{i \in I}$ be an arbitrary collection of closed subsets of $X$ for some index set $I$. Consider the intersection $\displaystyle{\bigcap_{i \in I} U_i}$. The complement of this set (using De Morgan's Laws) is:

\begin{align} \quad \left ( \bigcap_{i \in I} U_i \right )^c = \bigcup_{i \in I} U_i^c \end{align}

- Since $U_i$ is closed for each $i \in I$ we have that $U_i^c$ is open for each $i \in I$. Therefore the complement above is the union of an arbitrary collection of open sets which is open. Therefore $\displaystyle{\bigcap_{i \in I} U_i}$ is closed. $\blacksquare$

**Proof of c)**Let $\{ U_1, U_2, ..., U_n \}$ be a finite collection of closed subsets of $X$ for some index set $I$. Consider the union $\displaystyle{\bigcup_{i=1}^{n} U_i}$. The complement of this set (using De Morgan's Laws) is:

\begin{align} \quad \left ( \bigcup_{i=1}^{n} U_i \right )^c = \bigcap_{i=1}^{n} U_i^c \end{align}

- Since $U_i$ is closed for each $i \in \{ 1, 2, ..., n \}$ we must have that $U_i^c$ is open for each $i \in \{ 1, 2, ..., n \}$. Therefore the complement above is the intersection of a finite collection of open sets which is open. Therefore $\displaystyle{\bigcup_{i=1}^{n} U_i}$ is closed. $\blacksquare$