202301161201

Type :Note Tags : Topology

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Product topology

Definition

Note

$(X,{\tau }_X);(Y,{\tau }_Y)$. Then the product topology on the set $X\times Y$ is the topology generated by the basis ${\mathcal{B}}_{X\times Y}=\{U\times V;U\in {\tau }_X,V\in {\tau }_Y\}$

exercise: Check that this is a basis.

Lemma:

${\mathcal{B}}_{X\times Y}=\{B_1\times B_2:B_1\in {\mathcal{B}}_X,B_2\in {\mathcal{B}}_Y\}$ is a basis.

Examples

1. Any finite product of discrete spaces is discrete.
2. $({\mathbb{R}}_{std}{)}^n={\mathbb{R}}_{std}^n$

Projections

Let $(X_i,{\tau }_i)$ be finitely many topo spaces with the kth projection map: ${\pi }_k:\prod X_i\to X_i$ ${\pi }_k((x_i))=x_k$

Fact:

$A\subset X$, $B\subset Y$ then ${\pi }_1^{-1}(A)=A\times Y$ and ${\pi }_2^{-1}(B)=X\times B$.

Theorem

Note

The collection $S_{X\times Y}=\{{\pi }_1^{-1}(U);U\in {\tau }_X\}\cup \{{\pi }_2^{-1}(V):V\in {\tau }_Y\}$ is a Sub-basis for ${\tau }_{X\times Y}$.

Proof

Let $\tau$ be the topo gen by $S:=S_{X\times Y}$. Every element of S is in ${\tau }_{X\times Y}$ implies that $\tau \subset {\tau }_{X\times Y}$. Every basis element $U\times V$ of ${\tau }_{X\times Y}$ is ${\pi }_1^{-1}(U)\cap {\pi }_2^{-1}(V)\subset \tau$ This gives that ${\tau }_{X\times Y}\subset \tau$.

This definition with the sub-basis generalises to more general settings.

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Theorem

Product of subspace is same as subspace of product.

If A is a subspace of X, B is a subspace of Y. Then ${\tau }_{A\times B}$ = subspace topo on $A\times B$ from $X\times Y$.

proof:

${\mathcal{B}}_{A\times B}=\{M\times N:M\in {\tau }_A,N\in {\tau }_B\}=\{(U\cap A)\times (V\cap B):U\in {\tau }_X,V\in {\tau }_Y\}=\{(U\times V)\cap (A\times B)\}=$ basis for subspace topo on $A\times B$.

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Products and continuity

1. Let $X,Y$ be topological spaces and give $X\times Y$ the product topology. We have the two maps ${\pi }_1:X\times Y\to X$ and ${\pi }_2:X\times Y\to Y$ . Then ${\pi }_1$ and ${\pi }_2$ are continuous.
2. Let Z be any topological space. A function $f:Z\to X\times Y$ is continuous if and only if both of its components $f_1={\pi }_1\circ f:Z\to X$ and $f_2={\pi }_2\circ f:Z\to Y$ are continuous.
3. The product topology is the coarsest topology on $X\times Y$ such that both the projection maps are continuous.

Idea

There is a natural bijection between $\mathrm{Func}(X\times Y,Z)$ and $\mathrm{Func}(X,\mathrm{Func}(Y,Z))$. Here Func is the set of all set theoretic funtions between the two sets in consideration. But such a thing does not happen if we consider continuous functions.

• Suppose $f:X\times Y\to Z$ is a continuous function. Now for each $x$, we have $i_x:Y\to X\times Y$ which is just an inclusion $i_x(y)=(x,y)$. Now the map $H_f(x):=f\circ i_x:Y\to Z$ is a continuous map.
• So we get a map $\mathrm{Cont}(X\times Y,Z)\to \mathrm{Func}(X,\mathrm{Cont}(Y,Z))$. But this is not surjective since we can have functions $f$ that are continuous in one direction for all fixed $x$, but is not continuous. So to fix this we need to talk about continuity of $H_f$ but then we want a topology on $\mathrm{Cont}(Y,Z)$.
• We can think of $Func(Y,Z)=\prod _{y\in Y}Z$, so we need to look at product topologies for arbitrary products.

Product topology on arbitrary products

Note

The product topology on $\prod _{\alpha \in J}{X}_{\alpha }$ is the topology generated by the sub-basis $S=\{{\pi }_{\beta }^{-1}(U_{\beta }):\beta \in J,U_{\beta }\text{ open in }{X}_{\beta }\}$

• Note that this is the coarsest topology which makes all the projection maps continuous.

Topology on function spaces

Given the space of functions $f:X\to Y$, for topo spaces X and Y, we can put a topology on Func(X,Y) which is the product space $\prod _{x\in X}Y$, with the product topology as defined above. Convergence of sequences in this topology is the same as pointwise convergence.

Lemma

Note

Let $(f_n{)}_{n=1}^{\infty }$ be a sequence of functions $f_n:Y\to Z$ and let $f:Y\to Z$ be another such function. Then $f_n\to f$ as $n\to \infty$ in Func(Y, Z) if and only if the functions $f_n$ converge pointwise to f , i.e. if and only if for each y $\in$ Y , we have $f_n(y)\to f(y)$ as $n\to \infty$ in Z.

Some properties of products:

1. Let $A_{\alpha }$ be a subspace of $X_{\alpha }$ for each $\alpha$. Then the product topology on $\prod _{\alpha }{A}_{\alpha }$ equals the subspace topology induced from ${\prod }_{\alpha }{X}_{\alpha }$.
2. If each $X_{\alpha }$ is hausdorff then the product is hausdorff.
3. If each $X_{\alpha }$ has at least two points and the product is hausdorff then each $X_{\alpha }$ is hausdorff.
4. Let $A_{\alpha }$ be a subspace of $X_{\alpha }$ for each $\alpha$. Then closure of the product of $A_{\alpha }$ is equal to the product of the closures.

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Related Problems

1. Show that ${\mathbb{R}}^{\mathbb{R}}$ is separable in the product topology. Proof: Look at each element in R^R as a graph in R^2, then partition the plane into a grid of resolution 1/n at the nth step and pick heights for the function in the first n^2 columns on each side of the origin. This gives a countable collection of functions which can be used to approximate any function in ${\mathbb{R}}^{\mathbb{R}}$.

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References

Sub-basis Bases for a topology Hausdorff Property Continuous Functions