NoteNextra-origin/content/Math4201/Math4201_L3.md

Math4201 Topology I (Lecture 3)

Recall form last lecture

Topological Spaces

Basis for a topology

Let X be a set. A basis for a topology on X is a collection \mathcal{B} (elements of \mathcal{B} are called basis elements) of subsets of X such that:

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

Example of standard basis in real numbers

Let X=\mathbb{R} and \mathcal{B}=\{(a,b)|a,b\in \mathbb{R},a<b\} (collection of all open intervals).

Check properties 1:

for any x\in \mathbb{R}, \exists (x-1,x+1)\in \mathcal{B} such that x\in (x-1,x+1)

Check properties 2:

let B_1=(a,b) and B_2=(c,d) be two basis elements, and x\in B_1\cap B_2=(\max(a,c),\min(b,d))\in \mathcal{B}.

Example of lower limit basis in real numbers

Let X=\mathbb{R} and \mathcal{B}_{LL}=\{[a,b)|a,b\in \mathbb{R},a<b\} (collection of all open intervals).

Check properties 1:

for any x\in \mathbb{R}, \exists [x,x+1)\in \mathcal{B}_{LL} such that x\in [x,x+1)

Check properties 2:

let B_1=[a,b) and B_2=[c,d) be two basis elements, and x\in B_1\cap B_2=[max(a,c),min(b,d))\in \mathcal{B}_{LL}.

Extend this to \mathbb{R}^2.

Definition for cartesian product

Let X and Y be sets. The cartesian product of X and Y is the set X\times Y=\{(x,y)|x\in X,y\in Y\}.

Example of open rectangles basis for real plane

Let X=\mathbb{R}^2 and \mathcal{B} be the collection of rectangle of the form (a,b)\times (c,d) where a,b,c,d\in \mathbb{R} and a<b,c<d. (boundary is not included)

Check properties 1:

for any (x,y)\in \mathbb{R}^2, \exists (x,y)\in \mathcal{B} such that (x,y)\in (x,y)

Check properties 2:

let B_1=(a,b)\times (c,d) and B_2=(e,f)\times (g,h) be two basis elements, and (x,y)\in B_1\cap B_2=(max(a,e),min(b,f))\times (max(c,g),min(d,h))\in \mathcal{B}.

Example of open disks basis for real plane

Let X=\mathbb{R}^2 and \mathcal{B} be the collection of open disks.

Check properties 1:

for any x\in \mathbb{R}^2, \exists B_1(x)\in \mathcal{B} such that x\in B_1(x).

Check properties 2:

let B_{r_1}(x) and B_{r_2}(y) be two basis elements, for every z\in B_{r_1}(x)\cap B_{r_2}(y), \exists B_{r_3}(z)\in \mathcal{B} such that z\in B_{r_3}(z)\subseteq B_{r_1}(x)\cap B_{r_2}(y).

(even B_{r_1}(x)\cap B_{r_2}(y)\notin \mathcal{B})

Topology generated by a basis

Let \mathcal{B} be a basis for a topology on X. The topology generated by \mathcal{B}, denoted by \mathcal{T}_{\mathcal{B}}.

U\in \mathcal{T}_{\mathcal{B}}\iff \forall x\in U, \exists B\in \mathcal{B} such that x\in B\subseteq U

Proof

\mathcal{T}_{\mathcal{B}} is a topology on X because:

1. \emptyset \in \mathcal{T}_{\mathcal{B}} because \emptyset \in \mathcal{B}. X\in \mathcal{T}_{\mathcal{B}} because \forall x\in X, \exists B\in \mathcal{B} such that x\in B\subseteq X (by definition of basis (property 1)))

2. \mathcal{T}_{\mathcal{B}} is closed under arbitrary unions.

   Want to show \{U_\alpha | U_\alpha\in \mathcal{T}_{\mathcal{B}}\}_{\alpha \in I}\implies \bigcup_{\alpha \in I} U_\alpha\in \mathcal{T}_{\mathcal{B}}.

   Because \forall x\in \bigcup_{\alpha \in I} U_\alpha, \exists \alpha_0 such that x\in U_{\alpha_0}. Since U_{\alpha_0}\in \mathcal{T}_{\mathcal{B}}, \exists B\in \mathcal{B} such that x\in B\subseteq U_{\alpha_0}\subseteq \bigcup_{\alpha \in I} U_\alpha.

3. \mathcal{T}_{\mathcal{B}} is closed under finite intersections.

   Want to show U_1,U_2,\ldots,U_n\in \mathcal{T}_{\mathcal{B}}\implies \bigcap_{i=1}^n U_i\in \mathcal{T}_{\mathcal{B}}.

   If n=2, since \forall x\in U_1\cap U_2, x\in U_1 and x\in U_2, \exists B_1\in \mathcal{B} such that x\in B_1\subseteq U_1 and \exists B_2\in \mathcal{B}.

   Applying the second property of basis, \exists B_3\in \mathcal{B} such that x\in B_3\subseteq B_1\cap B_2\subseteq U_1\cap U_2.

   By induction, we can show that \bigcap_{i=1}^n U_i\in \mathcal{T}_{\mathcal{B}}.

Example of topology generated by a basis

Let X be arbitrary.

Let \mathcal{B}=\{x|x\in X\} (collection of all singleton subsets of X).

Then \mathcal{T} is the discrete topology.

Properties of basis in generated topology

Observation 1: Any B\in \mathcal{B} is an open set in \mathcal{T}_{\mathcal{B}}.

By the defining property of basis, \forall x\in B, \exists x\in B\subseteq B.

Observation 2: For any collection \{B_\alpha\}_{\alpha \in I}, \bigcup_{\alpha \in I} B_\alpha\in \mathcal{T}_{\mathcal{B}}.

By observation 1, each B_\alpha\in \mathcal{T}_{\mathcal{B}}. Since \mathcal{T}_{\mathcal{B}} is a topology, \bigcup_{\alpha \in I} B_\alpha\in \mathcal{T}_{\mathcal{B}}.

Lemma

Let \mathcal{B} and \mathcal{T}_{\mathcal{B}} be a basis and the topology generated by \mathcal{B} on X. Then,

U\in \mathcal{T}_{\mathcal{B}}\iff there are basis elements \{B_\alpha\}_{\alpha \in I} such that U=\bigcup_{\alpha \in I} B_\alpha.

Proof

(\Rightarrow)

If U\in \mathcal{T}_{\mathcal{B}}, we want to show that U is a union of basis elements.

For any x\in U, by the definition of \mathcal{T}_{\mathcal{B}}, there is a basis element B_x such that x\in B_x\subseteq U.

So, U\subseteq \bigcup_{x\in U} B_x.

Since \forall B_x\in \{B_x\}_{\alpha \in I}, B_x\subseteq U, we have U\supseteq\bigcup_{x\in U} B_x.

So, U=\bigcup_{x\in U} B_x.

(\Leftarrow)

Applies observation 2.

Note

A basis for a topology is like a basis for a vector space in the sense that any open set/vector can be represented in terms of basis elements.

But unlike linear algebra, it's not true that any open set can be written as a union of basis element in a unique way.