Product spaces

Question

• Permuting the indices?

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Overview §

There are two natural ways that topologies may be generated by a basis for a general, possibly infinite Cartesian product $\prod X_{\alpha }$:

• The box topology has as a basis all sets of the form $\prod U_{\alpha }$, where $U_{\alpha }\subseteq X_{\alpha }$ is an open subset for each index $\alpha$.
• The product topology has as subbasis all sets of the form ${\pi }_{\alpha }^{-1}(U_{\alpha })$ for some open subset $U_{\alpha }\subseteq X_{\alpha }$. This generates a basis containing all sets of the form $\prod U_{\alpha }$, where $U_{\alpha }=X_{\alpha }$ exactly except for finitely many values of $\alpha$.

In general, the box topology is finer than the product topology; they are exactly the same for finite products ${\prod }_{\alpha =1}^n{X}_{\alpha }$. When considering a general product space, assume the product topology unless otherwise specified.

Related: Projection and inclusion maps, Disjoint unions

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The product topology on $X\times Y$ §

Definition: Product topology on the Cartesian product X \times Y

Given topological spaces $(X,{\mathcal{T}}_X)$ and $(Y,{\mathcal{T}}_Y)$, the product topology on their Cartesian product $X\times Y$ is generated by the basis

$$\mathcal{B}=\{U\times V|U\in {\mathcal{T}}_X,V\in {\mathcal{T}}_Y\}.$$

Note that general open sets in $X\times Y$ are not of the form $U\times V$, because $\mathcal{B}$ is not closed under unions—that is, open sets are all unions of sets of the form $U\times V$.

The following result defines the product topology when $X,Y$ are specified by their bases.

Theorem (Munkres 15.1): Product topology specified by bases

If $\mathcal{B}$ is a basis for the topology of $X$ and $\mathcal{C}$ is a basis for the topology of $Y$, then the collection

$$\mathcal{D}=\{B\times C|B\in \mathcal{B},C\in \mathcal{C}\}$$

is a basis for the product topology on $X\times Y$.

Product topologies and subspaces §

Theorem: Two ways of defining subspace products on X \times Y

If $A$ is a subspace of $X$ and $B$ is a subspace of $Y$, then the product topology on $A\times B$ is the same as the topology $A\times B$ inherits as a subspace of $X\times Y$.

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Generalized products §

Terminology: Tuple notation for arbitrary indexing sets J be an indexing set. Given any set X, we define a J-tuple of elements of X as a function \mathbf x : J \to X. The function \mathbf x is often denoted by the quasi-tuple notation

Let

$$(x_{\alpha }{)}_{\alpha \in J}.$$

For each index $\alpha \in J$, the $\alpha$-coordinate of $\mathbf{x}$ is the value of the tuple at $\alpha$, and often denoted $x_{\alpha }$ rather than its technical definition $\mathbf{x}(\alpha )$.

Definition: Generalized Cartesian product

Given an indexed family of sets $\{X_{\alpha }{\}}_{\alpha \in J}$, the (generalized) Cartesian product

$$\prod _{\alpha \in J}{X}_{\alpha }$$

is the family of all functions $\mathbf{x}:J\to {\bigcup }_{\alpha \in J}{X}_{\alpha }$ such that, for all $\alpha \in J$, we get an element $\mathbf{x}(\alpha )\in X_{\alpha }$.

In other words, the Cartesian product is the set of all $J$-tuples $(x_{\alpha }{)}_{\alpha \in J}$ of elements of the union of $\{X_{\alpha }{\}}_{\alpha \in J}$, where for all $\alpha \in J$, the $\alpha$-coordinate $x_{\alpha }$ is an element of the set $X_{\alpha }$.

We can also phrase the notion of coordinates in terms of projection maps. Let $\mathbf{x}\in {\prod }_{\alpha \in J}{X}_{\alpha }$ be some function in the product. Then the $\beta$-coordinate of $\mathbf{x}$ for some $\beta \in J$ corresponds to an element $\mathbf{x}(\beta )$ in the set $X_{\beta }$, which is precisely the projection ${\pi }_{\beta }:{\prod }_{\alpha \in J}{X}_{\alpha }\to X_{\beta }$ that maps a function to an element by $\mathbf{x}\mapsto \mathbf{x}(\beta )$.

#concept-question Check connection between projection, functions to elements?

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Box and product topologies §

Definition: Box and product topologies

Suppose each set in the indexed family $\{X_{\alpha }{\}}_{\alpha \in J}$ is a topological space.

• The box topology on ${\prod }_{\alpha \in J}{X}_{\alpha }$ is the topology generated by the basis ${\prod }_{\alpha \in J}{U}_{\alpha }$, where each $U_{\alpha }$ is open in $X_{\alpha }$.
• The product topology on ${\prod }_{\alpha \in J}{X}_{\alpha }$ is the topology generated by the subbasis

$$\bigcup _{\alpha \in J}\{{\pi }^{-1}(U)|U\text{ open in }{X}_{\alpha }\},$$

which gives the basis

$$\left\{\prod _{\alpha \in J}{U}_{\alpha }|U_{\alpha }=X_{\alpha }\text{ for all but finitely many }\alpha \in J\right\}.$$

#wip

• Same for finite products
• Projection functions are continuous in both topologies

Transclude of (Theorem)-Continuity-of-maps-relative-to-the-product-topology#^d9fbb1

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The uniform topology §

Definition: Uniform metric, uniform topology

Consider the product ${\prod }_J\mathbb{R}$, where $J$ is any indexing set. As a set, ${\prod }_J\mathbb{R}$ is the set of all functions $J\to \mathbb{R}$, so we may also denote it ${\mathbb{R}}^J$. The uniform metric on ${\mathbb{R}}^J$ is defined by

$$\overline{\rho }(x,y)=\text{sup}\{\overline{d}(x_{\alpha },y_{\alpha })|\alpha \in J\},$$

where $\overline{d}(x,y)=\text{min}\{d(x,y),1\}$ is the standard bounded metric on $\mathbb{R}$ and $d(x,y)=|x-y|$ is the standard metric. It induces the uniform topology on ${\mathbb{R}}^J$.

Link to original

Theorem (Munkres 20.4): The uniform topology is “between” the box and product topologies

On the generalized product ${\mathbb{R}}^J$ with arbitrary indexing set $J$, the box topology is finer than the uniform topology, and the uniform topology is finer than the product topology. Further, $J$ is infinite, then all three are distinct.

Link to original

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Examples §

Continuity in the box topology vs. in the metric topology §

Let ${\mathbb{R}}^{\omega }$ be The countably infinite product of the real line with itself. This is an example of when (Theorem) Continuity of maps relative to the product topology fails for the box topology.

Define a map to the product $f:\mathbb{R}\to {\mathbb{R}}^{\omega }$ by

$$f(x)=(x,x,x,...),$$

so each coordinate function $f_n(x)=x$ is continuous, and thus, $f$ is continuous relative to the product topology. On the other hand, take the basic open set

$$B=(-1,1)\times \left(-\frac{1}{2},\frac{1}{2}\right)\times \left(-\frac{1}{3},\frac{1}{3}\right)\times \cdots =\prod _{n\ge 1}\left(-\frac{1}{n},\frac{1}{n}\right)$$

for the box topology. $B$ is clearly open in ${\mathbb{R}}^{\omega }$, but $f^{-1}(B)$ is not open in $\mathbb{R}$: we have $0\in f^{-1}B$, but there exists no $(-ϵ,ϵ)$ that is smaller than $(-1/n,1/n)$ for all $n$.#concept-question

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Review §

• Check that the basis for the product topology is indeed a basis. (Hint: is the intersection of basis elements also a basis element?)⭐
• Is the basis defined in the product topology itself a topology? Why or why not?
• Prove the theorem for a product topology specified by bases. (Hint: use (Lemma) Obtaining a basis from a topology.) ⭐
• Prove that the two ways of defining subspace products are equivalent.
• Check that the basis generating the box topology is indeed a basis.
• What is the difference between the bases for the box and product topologies? Specifically, for each basis element ${\prod }_{\alpha \in J}{U}_{\alpha }$, what is $U_{\alpha }$ representing? (See Munkres Theorem 19.1). ⭐
• Show that if $J$ is infinite, then the box, uniform, and product topologies are distinct (i.e., strict inclusions). ⭐
• Suppose $X_i=\{0,1\}$ has the discrete topology for $i\in {\mathbb{Z}}_{>0}$. Convince yourself that ${\prod }_{i=1}^{\infty }{X}_i$ is discrete in the box topology, but not in the product topology (in fact, in the product topology, this is isomorphic to the Cantor set). ⭐
• Give an example of when Theorem 19.6 fails for the box topology.

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Notes §

There are two natural ways that a topology arises for a general, possibly infinite Cartesian products

$$X_1\times \cdots \times X_n{X}_1\times X_2\times \cdots$$

where each $X_i$ is a topological space. One strategy, is to take as a basis the products of open sets $U_i\subset X_i$, while another is to take as a subbasis all sets of the form ${\pi }_i^{-1}(U_i)$, which generates basis elements that are finite intersections. In particular, for the second strategy, each tuple $\mathbf{x}$ is in a basis element

$$B={\pi }_{\alpha }^{-1}(U_1)\cap {\pi }_{\beta }^{-1}(U_2)\cap \cdots$$

if and only if ${\pi }_i(\mathbf{x})\in U_i$ for whatever indices $i$ are present in the intersection. For a finite product, these strategies are equivalent; for an infinite

• Other reasons the product topology is more important:
  • Coarsest topology such that the projection maps are continuous—preimage is the basic open set
  • We want finite intersections because the idea of openness has something to do with a neighborhood?