NoteNextra-origin/content/Math4201/Math4201_L6.md

Math4201 Topology I (Lecture 6)

Product topology

Define topological spaces on cartesian product of two topological spaces

Let (X,\mathcal{T}_X) and (Y,\mathcal{T}_Y) be two topological spaces.

X\times Y=\{(x,y)|x\in X,y\in Y\}

Goal: Define a topology on X\times Y.

One way is to take \mathcal{B}_{X\times Y}=\{(U\times V)|U\in \mathcal{T}_X,V\in \mathcal{T}_Y\} is a basis for the topology on X\times Y.

There are two ways to define the topology on \mathbb{R}^2: By rectangles (\mathcal{B}_{rect}=\{(a,b)\times (c,d)|a,b,c,d\in \mathbb{R},a<b,c<d\}) or by open disks (\mathcal{B}_{disk}=\{(x,y)\in \mathbb{R}^2|d((x,y),(a,b))<r\}). (check bonus video)

Proof

Take U=X,V=Y, then (x,y)\in (U\times V)=X\times Y.

So the first property of basis is satisfied.

Check the second property of basis:

Let B_1=U_1\times V_1,B_2=U_2\times V_2 be two basis elements, and (x,y)\in B_1\cap B_2.

B_1\cap B_2=(U_1\times V_1)\cap (U_2\times V_2)=(U_1\cap U_2)\times (V_1\cap V_2)

Since U_1\cap U_2\in \mathcal{T}_X and V_1\cap V_2\in \mathcal{T}_Y, we have (U_1\cap U_2)\times (V_1\cap V_2)\in \mathcal{B}_{X\times Y}.

Take B_3=(U_1\cap U_2)\times (V_1\cap V_2), then (x,y)\in B_3=B_1\cap B_2.

Caution

\mathcal{B}_{X\times Y} is not a topology on X\times Y.

\mathcal{B}_{X\times Y} is not closed with respect to arbitrary unions.

Definition of product topology

Let (X,\mathcal{T}_X) and (Y,\mathcal{T}_Y) be two topological spaces.

\mathcal{B}_{X\times Y}=\{(U\times V)|U\in \mathcal{T}_X,V\in \mathcal{T}_Y\}

The product topology or X\times Y is the topology generated by \mathcal{B}_{X\times Y}.

Let \mathcal{B}_X be a basis for \mathcal{T}_X and \mathcal{B}_Y be a basis for \mathcal{T}_Y.

If we define

\mathcal{B}'_{X\times Y}=\{(U\times V)|U\in \mathcal{B}_X,V\in \mathcal{B}_Y\}

Proposition

\mathcal{B}'_{X\times Y} is a basis for the product topology on X\times Y.

Note

This basis is smaller than \mathcal{B}_{X\times Y}.

Consider X=Y=\mathbb{R} and U=(a,b)\cap (c,d) and V=(e,f)\cap (g,h). (Assume a<b,c<d,e<f,g<h) The union of U and V is four rectangles, which is not in \mathcal{B}'_{X\times Y}. but it is in \mathcal{B}_{X\times Y}.

Proof

Using the lemma from Friday. it suffices to show that:

Let W\in X\times Y and (x,y)\in W, we need to show that there are B\in \mathcal{B}_X, B'\in \mathcal{B}_Y such that (x,y)\in (B\times B')\subseteq W.

Since W is open with respect to the topology generated by \mathcal{B}_{X\times Y}, in particular, there is U\times V such that (x,y)\in U\times V\subseteq W. And x\in U and y\in V.

Since U\in \mathcal{B}_X and V\in \mathcal{B}_Y, by property of basis \mathcal{B}_X and \mathcal{B}_Y, \forall x\in U, \exists B\in \mathcal{B}_X such that x\in B\subseteq U and \forall y\in V, \exists B'\in \mathcal{B}_Y such that y\in B'\subseteq V.

So (x,y)\in (B\times B')\subseteq U\times V\subseteq W.

Definition of standard topology on \mathbb{R}^n

The standard topology on \mathbb{R}^n is defined as the product topology on \mathbb{R}^{n-1}\times \mathbb{R}.

Subspace topology

Let (X,\mathcal{T}) be a topological space and Y\subseteq X.

Want to define a topology on Y.

\mathcal{T}_Y=\{U\cap Y|U\in \mathcal{T}\}

We claim that \mathcal{T}_Y is a topology on Y, called as the subspace topology on Y.

Proof

First, \emptyset \cap Y=\emptyset \in \mathcal{T}_Y and Y=X\cap Y\in \mathcal{T}_Y.

Second, let \{U_\alpha\cap Y\}_{\alpha \in I} be collection of open sets in \mathcal{T}_Y. Note that U_\alpha\in \mathcal{T} for all \alpha \in I.

So, \bigcup_{\alpha \in I} U_\alpha\cap Y=\left(\bigcup_{\alpha \in I} U_\alpha\right)\cap Y\in \mathcal{T}_Y because \bigcup_{\alpha \in I} U_\alpha\in \mathcal{T}.

Third, let \{U_i\cap Y\}_{i=1}^n be a finite collection of open sets in \mathcal{T}_Y. Note that U_i\in \mathcal{T} for all i=1,2,\ldots,n. So, \bigcap_{i=1}^n U_i\cap Y=\left(\bigcap_{i=1}^n U_i\right)\cap Y\in \mathcal{T}_Y because \bigcap_{i=1}^n U_i\in \mathcal{T}.

Generate basis for subspace topology

Let \mathcal{B} is a basis for (X,\mathcal{T}). We'd like to use \mathcal{B} to define a basis for (Y,\mathcal{T}_Y).

\mathcal{B}_Y=\{B\cap Y|B\in \mathcal{B}\}

Proposition for basis of subspace topology

\mathcal{B}_Y is a basis for a topology on Y that generates the subspace topology on Y.

Proof as exercise. (same as the proof for the basis of product topology)

Example: not every open set in subspace topology is open in the original space

Let X=\mathbb{R} with standard topology and Y=[0,1]\cup [2,3]. equipped with subspace topology generated by the standard basis for \mathbb{R}.

so [0,1]=(-1,\frac{3}{2})\cap Y In particular, [0,1] is open set in Y, but not an open set in \mathbb{R}.

Lemma of open set in subspace topology

Let (X,\mathcal{T}) be a topological space and Y\subseteq X is open. Z\subseteq Y is open subset with respect to the subspace topology on Y. Then Z is open in X.