Box Topology

The box topology is an alternative topology on a product of spaces that allows all coordinates to be constrained simultaneously. While it coincides with the product topology for finite products, it is strictly finer for infinite products and has substantially different (often pathological) properties.

Definition

Definition3.6Box Topology

Let $\{(X_\alpha, \tau_\alpha)\}_{\alpha \in A}$ be a family of topological spaces. The box topology on $\prod_\alpha X_\alpha$ is the topology generated by the basis: $\mathcal{B}_{\text{box}} = \left\{\prod_{\alpha \in A} U_\alpha : U_\alpha \in \tau_\alpha \text{ for all } \alpha \in A\right\}.$

Unlike the product topology, there is no requirement that $U_\alpha = X_\alpha$ for all but finitely many $\alpha$ .

RemarkComparison with Product Topology

For a finite index set $A$ , the box topology and product topology coincide. For an infinite index set, the box topology is strictly finer than the product topology:

$\tau_{\text{product}} \subsetneq \tau_{\text{box}} \quad \text{when } |A| \geq \aleph_0 \text{ and each } X_\alpha \text{ has more than one point.}$

The set $\prod_{\alpha} U_\alpha$ with each $U_\alpha$ a proper open subset is open in the box topology but not in the product topology (it is not a union of basic open sets of the product topology, which require $U_\alpha = X_\alpha$ for all but finitely many $\alpha$ ).

Pathological Properties

The box topology lacks many desirable properties of the product topology. The following examples illustrate the key failures.

ExampleTychonoff's Theorem Fails

Tychonoff's theorem states that any product of compact spaces is compact in the product topology. This fails spectacularly in the box topology.

Consider $\prod_{n=1}^{\infty} [0, 1]$ with the box topology. The open cover $\mathcal{U} = \left\{\prod_{n=1}^\infty U_n^{(k)} : k \in \mathbb{N}\right\}$ for suitable choices of $U_n^{(k)}$ has no finite subcover, since each basic open set constrains all coordinates simultaneously but cannot control the "tail" coordinates uniformly.

More concretely, the box topology on $[0,1]^{\mathbb{N}}$ is not even Lindel"of.

ExampleProjections are Continuous but Products of Maps May Not Be

While each projection $\pi_\alpha: \prod_\alpha X_\alpha \to X_\alpha$ is continuous in the box topology (since the box topology is finer than the product topology), the characterization of continuous maps into products fails.

Consider $f: \mathbb{R} \to \mathbb{R}^{\mathbb{N}}$ defined by $f(t) = (t, t, t, \ldots)$ (the diagonal map). Each component $\pi_n \circ f = \operatorname{id}_{\mathbb{R}}$ is continuous. In the product topology, $f$ is continuous. In the box topology, $f$ is not continuous:

The set $U = \prod_{n=1}^{\infty} (-1/n, 1/n)$ is open in the box topology, but $f^{-1}(U) = \bigcap_{n=1}^{\infty} (-1/n, 1/n) = \{0\}$ , which is not open in $\mathbb{R}$ .

Connectedness in the Box Topology

Theorem3.5Box Product of Connected Spaces May Not Be Connected

In the box topology, $\mathbb{R}^{\mathbb{N}}$ is not connected.

Proof

Define the subsets:

$A = \{(x_n) \in \mathbb{R}^{\mathbb{N}} : (x_n) \text{ is bounded}\}$
$B = \{(x_n) \in \mathbb{R}^{\mathbb{N}} : (x_n) \text{ is unbounded}\}$

We claim both $A$ and $B$ are open in the box topology.

For $A$ : Let $\mathbf{x} = (x_n) \in A$ be bounded, say $|x_n| \leq M$ for all $n$ . Then $\prod_{n} (x_n - 1, x_n + 1)$ is an open neighborhood of $\mathbf{x}$ , and every element in it is bounded by $M + 1$ . So $A$ is open.

For $B$ : Let $\mathbf{x} = (x_n) \in B$ . Choose any box neighborhood $\prod_{n} (x_n - \epsilon_n, x_n + \epsilon_n)$ . Any element $\mathbf{y}$ in this neighborhood has $|y_n - x_n| < \epsilon_n$ . If $|x_n| \to \infty$ along a subsequence, then for small enough $\epsilon_n$ (we can always choose these since we're in the box topology), $|y_n|$ is also large along that subsequence. But more carefully: for any box neighborhood of $\mathbf{x}$ , we can find unbounded sequences in it. So $B$ is open.

Since $A$ and $B$ are nonempty, disjoint, open, and $A \cup B = \mathbb{R}^{\mathbb{N}}$ , the space is disconnected.

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When the Box Topology is Useful

RemarkUniform Topology

Despite its pathologies, the box topology arises naturally in some contexts. For instance, the uniform topology on $\mathbb{R}^{\mathbb{N}}$ (induced by the sup metric $\bar{d}((x_n), (y_n)) = \sup_n \min(|x_n - y_n|, 1)$ ) lies between the product and box topologies: $\tau_{\text{product}} \subsetneq \tau_{\text{uniform}} \subsetneq \tau_{\text{box}}.$

The uniform topology retains many good properties (metrizability, for instance) while being finer than the product topology.

ExampleSummary: Box vs. Product Topology

| Property | Product Topology | Box Topology | |---|---|---| | Projections continuous | Yes | Yes | | Universal property for maps into product | Yes | No | | Tychonoff theorem | Yes | No | | Product of connected spaces connected | Yes | No | | Product of Hausdorff spaces Hausdorff | Yes | Yes | | Coincides with product topology for finite products | Yes | Yes |

The product topology is almost always the "correct" topology to use. The box topology serves mainly as a counterexample factory.