# The Fundamentals of Topological Spaces

We will soon discuss a special type of topological space. Before we do, we will define what a topological space is and describe some fundamental concepts regarding them.

Definition: Let $X$ be a nonempty set. A Topology on $X$ is a collection $\tau$ of subsets of $X$ which satisfies the following properties:1) $X \in \tau$ and $\emptyset \in \tau$.2) A union of any arbitrary collection of sets in $\tau$ is also in $\tau$.3) An intersection of any finite collection of sets in $\tau$ is also in $\tau$.The pair $(X, \tau)$ is said to be a Topological Space. |

In general, $X$ can have many topologies defined on it, each resulting in a different topological space. Given a topological space $(X, \tau)$, we can define many important notions on it. The most fundamental are open and closed sets.

## Open Sets

Definition: Let $(X, \tau)$ be a topological space. A set $U$ is said to be Open in $X$ if $U \in \tau$. |

So by definition, if $(X, \tau)$ is a topological space then $\tau$ is the collection of all open sets in $X$.

The definition of a topology can easily translate to open sets as follows.

Proposition 1: Let $(X, \tau)$ be a topological space. Then:a) $X$ and $\emptyset$ are open.b) Any arbitrary union of open sets is an open set.c) Any finite intersection of open sets is an open set. |

Definition: Let $(X, \tau)$ be a topological space and let $x \in X$. An Open Neighbourhood of $x$ (or simply Neighbourhood of $x$) is an open set containing $x$. |

The following proposition tells us exactly when a set is open.

Proposition 2: Let $(X, \tau)$ be a topological space and let $O \subseteq X$. Then $O$ is open if and only if for every $x \in O$ there exists a $U_x \in \tau$ such that $x \in U_x \subseteq O$. |

**Proof:**$\Rightarrow$ Suppose that $O$ is open. Let $x \in O$. Then $O$ itself is an open neighbourhood of $x$. So set $U = O$. Then $U \in \tau$ and $x \in U \subseteq O$.

- $\Leftarrow$ Suppose that for every $x \in O$ there exists a $U \in \tau$ such that $x \in U_x \subseteq O$. Then $\displaystyle{O = \bigcup_{x \in X} U_x}$. So $O$ is an arbitrary union of open sets and is thus open. $\blacksquare$

Definition: Let $(X, \tau)$ be a topological space and let $O \subseteq X$. A point $x \in O$ is said to be an Interior Point of $O$ if there exists an open set $U$ such that $x \in U \subseteq O$. The set of all interior points of $O$ is called the Interior of $O$ and is denoted by $\mathrm{int}(O)$. |

Observe that $\mathrm{int}(O) \subseteq O$. In fact, proposition 1 tells us that $\mathrm{int}(O)$ is always open and that $O$ is open if and only if $O = \mathrm{int}(O)$. The interior of a set can be characterized in a different way as noted in proposition 3 below.

Proposition 3: Let $(X, \tau)$ be a topological space and let $O \subseteq X$. Then $\mathrm{int}(O)$ is the largest open set contained in $O$. |

**Proof:**Clearly $\mathrm{int}(O) \subseteq O$. Suppose that there exists an open set $U$ such that $\mathrm{int}(O) \subset U$ and $U \subseteq O$. Then $U \setminus \mathrm{int}(O) \neq \emptyset$. Let $x \in U \setminus \mathrm{int}(O)$. Then $x \in U$ and $x \not \in \mathrm{int}(O)$. But since $x \in U$ by definition we have that $x \in \mathrm{int}(O)$ - a contradiction. Hence $\mathrm{int}(O)$ must be the largest open set contained in $O$. $\blacksquare$

## Closed Sets

Definition: Let $(X, \tau)$ be a topological space. A set $C$ is said to be Closed in $X$ if $C^C$ is open. |

So the set of all closed sets can be obtained by taking the complement of each set in $\tau$.

Like from above, the definition of a topology can be translated to tell us which sets are closed.

Proposition 4: Let $(X, \tau)$ be a topological space. Then:a) $X$ and $\emptyset$ are closed.b) Any finite union of closed sets is a closed set.c) Any arbitrary intersection of closed sets is a closed set. |

Definition: Let $(X, \tau)$ be a topological space and let $C \subseteq X$. A point $x \in X$ is said to be a Closure Point of $C$ if every open neighbourhood of $x$ contains some point of $C$. The set of all closure points of $C$ is called the Closure of $C$ and is denoted by $\bar{C}$ or $\mathrm{cl}(C)$. |

Observe that if $x \in C$ then $x \in \bar{C}$ trivially. So $C \subseteq \bar{C}$. Furthermore, $\bar{C}$ is closed. To see this, consider $\bar{C}^C$. Let $x \in \bar{C}^C$. Observe by definition that there exists an open neighbourhood of $x$ which does not contain any point of $C$. That is, there exists a $U \in \tau$ for which $x \in U \subseteq \bar{C}^C$. Hence $\bar{C}^C$ is open and so $\bar{C}$ is closed. The following proposition will give us an analogue to proposition $2$ regarding the closure of a set.

Proposition 5: Let $(X, \tau)$ be a topological space and let $C \subseteq X$. Then $\bar{C}$ is the smallest closed set containing $C$. |

**Proof:**Clearly $C \subseteq \bar{C}$. Suppose that there exists a closed set $D$ such that $C \subseteq D \subset \bar{C}$. Then $\bar{C} \setminus D \neq \emptyset$. Let $x \in \bar{C} \setminus D$. Then $x \in \bar{C}$ and $x \not \in D$. Consider $D^C$. Since $x \not \in D$ we have that $x \in D^C$. Since $D$ is closed, $D^C$ is open and so there is an open neighbourhood $U$ of $x$ for which $x \in U \subseteq D^C$. But then $U$ is an open neighbourhood of $x$ containing no point of $C$, which is a contradiction since $x \in \bar{C}$. Hence $\bar{C}$ is the smallest closed set containing $C$. $\blacksquare$