Math4201 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

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-examples:

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

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