Section 19: The Product Topology

Let $\{X_{\alpha}\}$ be an indexed family of topological spaces and $X=\sqcap_{\alpha}X_{\alpha}$ be their product.

The product topology on $X$ is the topology generated by the basis consisting of $\sqcap_{\alpha}U_{\alpha}$ where each $U_{\alpha}$ is an open subset (or, equivalently, a basis element) of $X_{\alpha}$ , and all but finite number of $U_{\alpha}$ equal $X_{\alpha}$ . $X$ with the product topology is called the product space.

Alternatively, the product topology is generated by the subbasis consisting of $\pi_{\alpha}^{-1}(U_{\alpha})$ where $U_{\alpha}$ is an open subset of $X_{\alpha}$ .

The box topology on $X$ is the topology generated by the basis consisting of $\sqcap_{\alpha}U_{\alpha}$ where each $U_{\alpha}$ is an open subset (or, equivalently, a basis element) of $X_{\alpha}$ .

These topologies agree on finite products.

...why we prefer the product topology to the box topology... We shall find that a number of important theorems about finite products will also hold for arbitrary products if we use the product topology, but not if we use the box topology.

Whenever we consider the product... we shall assume it is given the product topology unless we specifically state otherwise.

Common properties

If $X_{\alpha}$ ’s are all Hausdorff then $X$ is Hausdorff in either topology.
The product of subspaces $A=\sqcap_{\alpha}A_{\alpha}$ is a subspace of $X=\prod_{\alpha}X_{\alpha}$ if both products are given the same topology (box or product).
The closure of the product of subsets equals the product of the closures in either topology: $\overline{\sqcap_{\alpha}A_{\alpha}}=\sqcap_{\alpha}\overline{A_{\alpha}}$ .
If $f:A\rightarrow X=\sqcap_{\alpha}X_{\alpha}$ is continuous then $\pi_{\alpha}\circ f:A\rightarrow X_{\alpha}$ is continuous for every $\alpha$ (the other way works for the product topology only).
If a sequence of points $\{\mathbf{x}_{n}\}$ in $X$ converges to $\mathbf{x}$ then for each $\alpha$ the sequence $\{\pi_{\alpha}(\mathbf{x}_{n})\}$ converges to $\pi_{\alpha}(\mathbf{x})$ (the other way works for the product topology only).

$h((x_{1},x_{2},\ldots))=(a_{1}x_{1}+b_{1},a_{2}x_{2}+b_{2},\ldots)$ , $a_{i}>0$ for all $i\in\mathbb{Z}_{+}$ , is a homeomorphism of with itself regardless of whether it is given the product or box topology.

Properties of the product topology

$f:A\rightarrow X=\sqcap_{\alpha}X_{\alpha}$ is continuous iff $\pi_{\alpha}\circ f:A\rightarrow X_{\alpha}$ is continuous for every $\alpha$ .
- $f(x)=(x,x,\ldots):\mathbb{R}\rightarrow\mathbb{R}_{box}^{\omega}$ is not continuous in the box topology.
A sequence of points $\{\mathbf{x}_{n}\}$ in $X$ converges to $\mathbf{x}$ iff for each $\alpha$ the sequence $\{\pi_{\alpha}(\mathbf{x}_{n})\}$ converges to $\pi_{\alpha}(\mathbf{x})$ .
- The sequence $\mathbf{x}_{n}=(\underbrace{0,\ldots,0}_{n},1,1,\ldots)$ does not converge to any point in the box topology.

The topology induced by functions

Let $f_{\alpha}:B\rightarrow X_{\alpha}$ be an indexed family of functions.

There is a unique coarsest topology $\mathcal{T}$ on $B$ such that each $f_{\alpha}$ is continuous relative to it.
A map $g:A\rightarrow B$ is continuous relative to $\mathcal{T}$ iff each $f_{\alpha}\circ g:A\rightarrow X_{\alpha}$ is continuous.
If $f:A\rightarrow\sqcap_{\alpha}X_{\alpha}$ , $f(a)=(f_{\alpha}(a))$ , then the image of any set open in $\mathcal{T}$ is open in $f(A)$ (in the subspace topology).

Axiom of choice

The axiom of choice is equivalent to the statement that the product of every nonempty collection of nonempty sets is nonempty.

Subpages

Section 19: The Product Topology

Section 19: The Product Topology

Common properties

\mathbb{R}^{\omega}

Properties of the product topology

The topology induced by functions

Axiom of choice