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Math4201Topology I (Lecture 6)

Math4201 Topology I (Lecture 6)

Product topology

Define topological spaces on cartesian product of two topological spaces

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

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

Goal: Define a topology on X×YX\times Y.

One way is to take BX×Y={(U×V)∣U∈TX,V∈TY}\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×YX\times Y.

There are two ways to define the topology on R2\mathbb{R}^2: By rectangles (Brect={(a,b)×(c,d)∣a,b,c,d∈R,a<b,c<d}\mathcal{B}_{rect}=\{(a,b)\times (c,d)|a,b,c,d\in \mathbb{R},a<b,c<d\}) or by open disks (Bdisk={(x,y)∈R2∣d((x,y),(a,b))<r}\mathcal{B}_{disk}=\{(x,y)\in \mathbb{R}^2|d((x,y),(a,b))<r\}). (check bonus video)

Proof

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

So the first property of basis is satisfied.

Check the second property of basis:

Let B1=U1×V1,B2=U2×V2B_1=U_1\times V_1,B_2=U_2\times V_2 be two basis elements, and (x,y)∈B1∩B2(x,y)\in B_1\cap B_2.

B1∩B2=(U1×V1)∩(U2×V2)=(U1∩U2)×(V1∩V2)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 U1∩U2∈TXU_1\cap U_2\in \mathcal{T}_X and V1∩V2∈TYV_1\cap V_2\in \mathcal{T}_Y, we have (U1∩U2)×(V1∩V2)∈BX×Y(U_1\cap U_2)\times (V_1\cap V_2)\in \mathcal{B}_{X\times Y}.

Take B3=(U1∩U2)×(V1∩V2)B_3=(U_1\cap U_2)\times (V_1\cap V_2), then (x,y)∈B3=B1∩B2(x,y)\in B_3=B_1\cap B_2.

Caution

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

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

Definition of product topology

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

BX×Y={(U×V)∣U∈TX,V∈TY}\mathcal{B}_{X\times Y}=\{(U\times V)|U\in \mathcal{T}_X,V\in \mathcal{T}_Y\}

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

Let BX\mathcal{B}_X be a basis for TX\mathcal{T}_X and BY\mathcal{B}_Y be a basis for TY\mathcal{T}_Y.

If we define

BX×Y′={(U×V)∣U∈BX,V∈BY}\mathcal{B}'_{X\times Y}=\{(U\times V)|U\in \mathcal{B}_X,V\in \mathcal{B}_Y\}

Proposition

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

Note

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

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

Proof

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

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

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

Since U∈BXU\in \mathcal{B}_X and V∈BYV\in \mathcal{B}_Y, by property of basis BX\mathcal{B}_X and BY\mathcal{B}_Y, ∀x∈U\forall x\in U, ∃B∈BX\exists B\in \mathcal{B}_X such that x∈B⊆Ux\in B\subseteq U and ∀y∈V\forall y\in V, ∃B′∈BY\exists B'\in \mathcal{B}_Y such that y∈B′⊆Vy\in B'\subseteq V.

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

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

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

Subspace topology

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

Want to define a topology on YY.

TY={U∩Y∣U∈T}\mathcal{T}_Y=\{U\cap Y|U\in \mathcal{T}\}

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

Proof

First, ∅∩Y=∅∈TY\emptyset \cap Y=\emptyset \in \mathcal{T}_Y and Y=X∩Y∈TYY=X\cap Y\in \mathcal{T}_Y.

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

So, ⋃α∈IUα∩Y=(⋃α∈IUα)∩Y∈TY\bigcup_{\alpha \in I} U_\alpha\cap Y=\left(\bigcup_{\alpha \in I} U_\alpha\right)\cap Y\in \mathcal{T}_Y because ⋃α∈IUα∈T\bigcup_{\alpha \in I} U_\alpha\in \mathcal{T}.

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

Generate basis for subspace topology

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

BY={B∩Y∣B∈B}\mathcal{B}_Y=\{B\cap Y|B\in \mathcal{B}\}

Proposition for basis of subspace topology

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

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

so [0,1]=(−1,32)∩Y[0,1]=(-1,\frac{3}{2})\cap Y In particular, [0,1][0,1] is open set in YY, but not an open set in R\mathbb{R}.

Lemma of open set in subspace topology

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

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