NoteNextra-origin/content/Math4201/Math4201_L4.md

Math4201 Topology I (Lecture 4)

Recall from last lecture

Assignment due next Thursday. 10PM

Let \mathcal{B} be a basis for a topology. Then the topology (\mathcal{T}_{\mathcal{B}}) generated by \mathcal{B} is \{U\in \mathcal{T}_{\mathcal{B}} \mid \forall x\in U, \exists B\in \mathcal{B} \text{ such that } x\in B\subseteq U\}.

New materials

Topology basis

Given a topology on a set X, When is a given collection of subsets of X a basis for a topology?

Suppose U\in\mathcal{T} is an open set in X. If an arbitrary set \mathcal{C} is a basis for \mathcal{T}, then by the definition of a topology generated by a basis, we should have the following:

\exists C\in \mathcal{C} \text{ such that } x\in C\subseteq U

Theorem of basis of topology

Caution

In this course, we use lowercase letters to denote element of a set, and uppercase letters to denote sets. We use \mathcal{X} to denote set of subsets of X.

Let (X,\mathcal{T}) be a topological space. Let \mathcal{C}\subseteq \mathcal{T} be a collection of subsets of X satisfying the following property:

\forall U\in \mathcal{T}, \exists C\in \mathcal{C} \text{ such that } U\subseteq C

Then \mathcal{C} is a basis and the topology generated by \mathcal{C} is \mathcal{T}.

Proof

We want to show that \mathcal{C} is a basis.

Recall the definition of a basis:

1. \forall x\in X, there is B\in \mathcal{B} such that x\in B
2. \forall B_1,B_2\in \mathcal{B}, \forall x\in B_1\cap B_2, there is B_3\in \mathcal{B} such that x\in B_3\subseteq B_1\cap B_2

First, we want to show that \mathcal{C} satisfies the first property.

Take x\in X. Since X\in \mathcal{T}, we can apply the given condition ($ \forall U\in \mathcal{T}, \exists C\in \mathcal{C} \text{ such that } U\subseteq C $) to get C\in \mathcal{C} such that x\in C\subseteq X.

Next, we want to show that \mathcal{C} satisfies the second property.

Let C_1,C_2\in \mathcal{C} and x\in C_1\cap C_2. Since C_1,C_2\in \mathcal{T}, by the definition of \mathcal{T}, we have U=C_1\cap C_2\in \mathcal{T}.

We can apply the given condition to get C_3\in \mathcal{C} such that x\in C_3\subseteq U=C_1\cap C_2.

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Then we want to show that the topology generated by \mathcal{C} is \mathcal{T}.

Recall the definition of the topology generated by a basis:

To prove this, we need to show that \forall U\in \mathcal{T}\implies U\in \mathcal{T}_{\mathcal{C}} and \forall U\in \mathcal{T}_{\mathcal{C}}\implies U\in \mathcal{T}.

Moreover, from last lecture, we have U\in \mathcal{T}_{\mathcal{B}}\iff U=\bigcup_{\alpha \in I} B_\alpha for some \{B_\alpha\}_{\alpha \in I}\subseteq \mathcal{B}.

First, we want to show that \forall U\in \mathcal{T}_{\mathcal{C}}\implies U\in \mathcal{T}.

Let U=\bigcup_{\alpha \in I} C_\alpha for some \{C_\alpha\}_{\alpha \in I}\subseteq \mathcal{C}. Then since C_\alpha\in \mathcal{T}, by the definition of \mathcal{T}, we have U\in \mathcal{T}.

Next, we want to show that \forall U\in \mathcal{T}\implies U\in \mathcal{T}_{\mathcal{C}}.

Let U\in \mathcal{T}. Then \forall x\in U by the given condition, we have C\in \mathcal{C} such that x\in C\subseteq U.

So, U=\bigcup_{\alpha \in I} C_\alpha\in \mathcal{T}_{\mathcal{C}}. (using the same trick last time)

Let \mathcal{T} be the topology on X. Then \mathcal{T} itself satisfies the basis condition.

Definition of subbasis of topology

A subbasis of a topology on a set X is a collection \mathcal{S}\subseteq \mathcal{T} of subsets of X such that their union is X.

\mathcal{S}=\{S_{\alpha}\mid S_\alpha\subseteq X\}_{\alpha \in I}\text{  and  }\bigcup_{\alpha \in I} S_\alpha=X

Definition of topology generated by a subbasis

If we consider the basis generated by the subbasis \mathcal{S} by the following:

\mathcal{B}=\{B\mid B\text{ is the intersection of a finite number of elements of }\mathcal{S}\}

Then \mathcal{B} is a basis.

Proof

First, \forall x\in X, there is S_\alpha\in \mathcal{S} such that x\in S_\alpha. In particular, x\in \mathcal{B}.

Second, let B_1,B_2\in \mathcal{B}. Since B_1 is the intersection of a finite number of elements of \mathcal{S}, we have B_1=\bigcap_{i=1}^n S_{i_1}, B_2=\bigcap_{i=1}^n S_{i_2} for some S_{i_1},S_{i_2}\in \mathcal{S}.

So B_1\cap B_2 is the intersection of finitely many elements of \mathcal{S}.

So B_1\cap B_2\in \mathcal{B}.

We call the topology generated by \mathcal{B} the topology generated by the subbasis \mathcal{S}. Denote it by \mathcal{T}_{\mathcal{S}}.

An open set with respect to \mathcal{T}_{\mathcal{S}} is a subset of X such that it can be written as a union of finitely intersections of elements of \mathcal{S}.

Example (standard topology on real numbers)

Let X=\mathbb{R}. Take \mathcal{S}=\{(-\infty, a)|a\in \mathbb{R}\}\cup \{(a,+\infty)|a\in \mathbb{R}\}.

We claim this is a subbasis of the standard topology on \mathbb{R}.

The basis \mathcal{B} associated with \mathcal{S} is the collection of all open intervals.

\mathcal{B}=\{(a,b)=(-\infty, b)\cap (a,+\infty)\}

So, \mathcal{B}=\mathcal{B}_{st} (the standard basis).

This topology on \mathbb{R} is the same as the standard topology on \mathbb{R}.

Example (finite complement topology)

Let X be an arbitrary set. Let \mathcal{S} defined as follows:

\mathcal{S}=\{S\subseteq X\mid S=X\setminus \{x\} \text{ for some } x\in X\}

Let x,y\in X and x\neq y. Then S_x=X\setminus \{x\} and S_y=X\setminus \{y\} are two elements of \mathcal{S}. Since x\neq y, we have S_x\cup S_y=X\setminus \{x\}\cup X\setminus \{y\}=X. So \mathcal{S} is a subbasis of X.

So, the basis associated with \mathcal{S}, \mathcal{B}, is the collection of subsets of X with finite complement.

This is in fact a topology, which is the finite complement topology on X.