Math4201 Topology I (Lecture 2)

Topology

Distance in $\mathbb{R}$ , or more generally in $\mathbb{R}^n$ . (metric space )

Intervals in $\mathbb{R}$ , or more generally open balls in $\mathbb{R}^n$ . (topological space)

Topological Spaces

Definition of Topological Space

A topological space is a pair of set $X$ and a collection of subsets of $X$ , denoted by $\mathcal{T}$ (imitates the set of “open sets” in $X$ ), satisfying the following axioms:

$\emptyset \in \mathcal{T}$ and $X \in \mathcal{T}$
$\mathcal{T}$ is closed with respect to arbitrary unions. This means, for any collection of open sets $\{U_\alpha\}_{\alpha \in I}$ , we have $\bigcup_{\alpha \in I} U_\alpha \in \mathcal{T}$
$\mathcal{T}$ is closed with respect to finite intersections. This means, for any finite collection of open sets $\{U_1, U_2, \ldots, U_n\}$ , we have $\bigcap_{i=1}^n U_i \in \mathcal{T}$

The elements of $\mathcal{T}$ are called open sets.

The topological space is denoted by $(X, \mathcal{T})$ .

Examples of topological spaces

Trivial topology: Let $X$ be arbitrary. The trivial topology is $\mathcal{T}_0 = \{\emptyset, X\}$

Discrete topology: Let $X$ be arbitrary. The discrete topology is $\mathcal{T}_1 = \mathcal{P}(X)=\{U \subseteq X\}$

Understanding all possible topologies on a set.

Let’s say $X=\{a,b\}$

The trivial topology is $\mathcal{T}_0 = \{\emptyset, \{a,b\}\}$

The discrete topology is $\mathcal{T}_1 = \{\emptyset, \{a\}, \{b\}, \{a,b\}\}$

Other topologies:

$\mathcal{T}_2 = \{\emptyset, \{a\}, \{a,b\}\}$

$\mathcal{T}_3 = \{\emptyset, \{b\}, \{a,b\}\}$

Non-example of topological space

Let $X=\{a,b,c\}$

The set $\mathcal{T}_1=\{\emptyset, \{a\}, \{b\}, \{a,b,c\}\}$ is not a topology because it is not closed under union $\{a\} \cup \{b\} = \{a,b\} \notin \mathcal{T}_1$

Definition of Complement finite topology

Let $X$ be arbitrary. The complement finite topology is $\mathcal{T}\coloneqq \{U\subseteq X|X\setminus U \text{ is finite}\}\cup \{\emptyset\}$

The topology is valid because:

Proof

$\emptyset \in \mathcal{T}$ because $X\setminus \emptyset = X$ is finite.
Let $\{U_\alpha\}_{\alpha \in I}$ be an arbitrary collection such that $X\setminus U_\alpha$ is finite for each $\alpha \in I$ .

Without loss of generality, we can assume that $U_\alpha \neq \emptyset$ for each $\alpha \in I$ , since the union of arbitrary set with $\emptyset$ is the set itself.

If all of them are empty, then the union is empty, which complement is $X\subset \mathcal{T}$ .

Otherwise,
$X\setminus \bigcup_{\alpha \in I} U_\alpha = \bigcap_{\alpha \in I} (X\setminus U_\alpha)$
is finite because each $X\setminus U_\alpha$ is finite. Therefore, $\bigcup_{\alpha \in I} U_\alpha \in \mathcal{T}$ .
Let $\{U_1, U_2, \ldots, U_n\}$ be a finite collection such that $X\setminus U_i$ is finite for each $i=1,2,\ldots,n$ .

Without loss of generality, we can assume that $U_i \neq \emptyset$ for each $i=1,2,\ldots,n$ , since the intersection of arbitrary set with $\emptyset$ is $\emptyset$ .

If all of them are empty, then the intersection is $X\subset \mathcal{T}$ .

Otherwise,
$X\setminus \bigcap_{i=1}^n U_i = \bigcup_{i=1}^n (X\setminus U_i)$
is finite because each $X\setminus U_i$ is finite. Therefore, $\bigcap_{i=1}^n U_i \in \mathcal{T}$ .

Another non-example is $\mathcal{T} = \{U\subseteq X|U \text{ is finite}\}\cup \{X\}$

The topology is invalid because an arbitrary union of points is not a finite set.

Consider $X=\mathbb{Z}$ but take $U_1=\{1\}, U_2=\{2\}, U_3=\{3\}, \ldots$

Then $\bigcup_{i=1}^\infty U_i = \mathbb{Z}^+$ is not a finite set and is not $\{X\}$ .

Note

If $X$ is finite, then finite complement topology is the same as the discrete topology.

Definition of Topology basis

For a set $X$ , a topology basis, denoted by $\mathcal{B}$ , is a collection of subsets of $X$ , such that the following properties are satisfied:

For any $x \in X$ , there exists a $B \in \mathcal{B}$ such that $x \in B$
If $B_1, B_2 \in \mathcal{B}$ and $x \in B_1 \cap B_2$ , then there exists a $B_3 \in \mathcal{B}$ such that $x \in B_3 \subseteq B_1 \cap B_2$