# 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. |

By the definition above we see that a set $A$ is closed by definition if and only if $X \setminus A$ is open. From this, we get a criterion for whether or not a set is open.

Proposition 1: Let $(X, \tau)$ be a topological space and let $A \subseteq X$. Then $A$ is open if and only if $X \setminus A$ is closed. |

**Proof:**$\Rightarrow$ Suppose that $A$ is open. Since $(X \setminus A)^c = A$ and $A$ is open, we see that $X \setminus A$ is closed.

- $\Leftarrow$ Suppose that $X \setminus A$ is closed. Then by definition, $(X \setminus A)^c = A$ is open. $\blacksquare$

The following theorem follows directly from the definition of closed sets above and the definition of a topological space.

Theorem 2: 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:

- 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:

- 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. $\blacksquare$

## Example 1

As proven in the theorem above if $(X, \tau)$ is a topological space then the whole set $X$ and the emptyset $\emptyset$ are always closed in a topological space. By definition, they are also always open in a topological space. Therefore, $X$ and $\emptyset$ are always clopen sets.

Sometimes $X$ and $\emptyset$ are the only clopen sets for a particular topology $\tau$ on $X$, but in general, a topological space may have many clopen sets.

## Example 2

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:

(3)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.