# Characterizations of product topology and box topology

Given a collection of topological spaces $$X_i$$ indexed by the elements $$i$$ of a set $$I$$, we consider the set product $$P = \prod_{i \in I} X_i$$ with projections $$p_i : P \to X_i$$. There are two methods of topologizing $$P$$.

1. Take the product topology having as a subbase all sets of the form $$p_i^{-1}(U_i)$$ with $$i \in I$$ and $$U_i \subset X_i$$ open. Therefore a base for the product topology consists of all products of form $$\prod_{i \in I} U_i$$ with $$U_i \subset X_i$$ open and $$U_i \ne X_i$$ only for finitely many $$i$$.

2. Take the box topology having as a base all products of form $$\prod_{i \in I} U_i$$ with $$U_i \subset X_i$$ open.

By definition the product topology is the coarsest topology on $$P$$ such that all $$p_i^{-1}(U_i)$$, with open $$U_i \subset X_i$$, $$i \in I$$, are open. This is equivalent to defining the product topology as the coarsest topology on $$P$$ such that all projections $$p_i$$ become continuous.

It is standard to endow $$P$$ with the product topology. This makes $$(P,(p_i)_{i \in I})$$ the product of the objects $$X_i$$ in the category of topological spaces which is characterized by the following universal property:

A function $$f : Y \to P$$ defined on a topological space $$Y$$ is continuous if and only if all $$p_i \circ f : Y \to X_i$$ are continuous.

This seems to be the optimum what can be expected from a topology on $$P$$.

On infinite products the box topology is in general strictly finer than the product topology, thus it does in general not have this universal property. As far as I know, the box topology does not have any really nice universal property; see Does the box topology have a universal property?

On the other hand, the box topology is the coarsest topology such that all products of the form $$\prod_{i \in I} U_i$$ with open $$U_i \subset X_i$$, $$i \in I$$, are open. On the level of sets this seems to be a very natural requirement, even more natural than requiring that all special products of the form $$\prod_{i \in I} U_i$$ with open $$U_i \subset X_i$$ open and $$U_i \ne X_i$$ only for finitely many $$i$$, are open.

Question: Are there other characterizations of product or box topology than those described above? If so, do they occur somewhere in the literature?

• The box product can be characterised as the largest topology on $\prod_{i\in I}X_i$ for which each projection $pr_i:\prod_{i\in I}X_i\rightarrow X_i$ is a continuous open map (the product topology is the smallest such topology). Nov 9, 2022 at 21:42
• @Tyrone I suggest that you write an official answer. Nov 9, 2022 at 23:18
• @Tyrone This does not seem to be true. See my answer. Nov 24, 2022 at 23:14

Here is a fairly obvious alternative characterization of the product topology:

The product topology is the coarsest topology on $$P$$ such that all products of the form $$\prod_{i \in I} A_i$$ with closed $$A_i \subset X_i$$, $$i \in I$$, are closed.

This resembles the characterization of the box topology as the coarsest topology on $$P$$ such that all products of the form $$\prod_{i \in I} U_i$$ with open $$U_i \subset X_i$$, $$i \in I$$, are open; it is also a very natural requirement.

Note that the products of the form $$\prod_{i \in I} A_i$$ with closed $$A_i \subset X_i$$ are also closed in the box topology since it is finer than the product topology.

Let us prove the above characterization.

We know that the product topology is the coarsest topology on $$P$$ such that all $$p_i$$ become continuous. The continuity of all $$p_i$$ is equivalent to

• All $$p_i^{-1}(A_i)$$ with closed $$A_i \subset X_i$$, $$i \in I$$, are closed.

But this is equivelent to

• All products of the form $$\prod_{i \in I} A_i$$ with closed $$A_i \subset X_i$$, $$i \in I$$, are closed.

In fact, $$p_i^{-1}(A_i)$$ is a special case of such a product (take $$A_j = X_j$$ for $$j \ne i$$). Conversely, $$\prod_{i \in I} A_i = \bigcap_{i \in I} p_i^{-1}(A_i)$$ which is closed if all $$p_i^{-1}(A_i)$$ are closed.

It is obvious that the box topology on $$P$$ makes all projections $$p_i :P \to X_i$$ open maps.

In a comment it was conjectured that the box topology is the finest topology on $$P$$ such that all projections $$p_i :P \to X_i$$ are continuous open maps. This is not true.

Note that the requirement of continuity can also be omitted here because the box topology makes all projections continuous, therefore the same is true for each finer topology.

For finite products the box topology agrees with the product topology, and therefore the product topology on a finite product should be the finest topology on $$P$$ such that all projections $$p_i : P \to X_i$$ are open maps.

Consider $$P = [0,1] \times [0,1]$$ with projections $$p_1, p_2 : P \to [0,1]$$. As a subbase for a topology $$\tau$$ on $$P$$ take all $$p_i^{-1}(U)$$ with open $$U \subset [0,1]$$ and the set $$W_0 = \{(x,y) \in P \mid y < x \} \cup \{(0,0)\}$$. A base $$\mathcal B$$ for $$\tau$$ is obtained by taking all finite intersections of these sets. This yields all boxes $$U_1 \times U_2$$ with open $$U_i \subset [0,1]$$ plus all $$(U_1 \times U_2) \cap W_0$$ with open $$U_i \subset [0,1]$$. Clearly $$\tau$$ is strictly finer than the product topology $$\tau_p$$ because $$W_0 \notin \tau_p$$.

We claim that $$p_i(B)$$ is open in $$[0,1]$$ for all $$B \in \mathcal B$$ which shows that the $$p_i$$ are open maps with respect to $$\tau$$.

Since the $$p_i$$ are known to be open maps with respect to $$\tau_p$$, it suffices to consider $$B = (U_1 \times U_2) \cap W_0$$.

The set $$W = \{(x,y) \in P \mid y < x \}$$ belongs to $$\tau_p$$ and $$W_0 = W \cup \{(0,0\}$$. We get $$p_i(B) = p_i((U_1 \times U_2) \cap W \cup (U_1 \times U_2) \cap \{(0,0\}) \\= p_i((U_1 \times U_2) \cap W) \cup p_i((U_1 \times U_2) \cap \{(0,0\}).$$ The set $$(U_1 \times U_2) \cap W$$ belongs $$\tau_p$$ so that $$p_i((U_1 \times U_2) \cap W)$$ is open.

If $$(0,0) \notin U_1 \times U_2$$, then $$p_i(B) = p_i((U_1 \times U_2) \cap W)$$ is open.

If $$(0,0) \in U_1 \times U_2$$, then $$p_i((U_1 \times U_2) \cap \{(0,0\}) = \{0\}$$. Moreover there exists $$r \in (0,1]$$ such that $$[0,r) \times [0,r) \subset U_1 \times U_2$$. Thus $$(0,r) \subset p_i((U_1 \times U_2) \cap W)$$. This shows that $$p_i(B) = p_i((U_1 \times U_2) \cap W) \cup \{0\} = p_i((U_1 \times U_2) \cap W) \cup [0,r)$$ which is open in $$[0,1]$$.

• +1. I guess you need to assume additionally that for each $I\in\mathcal{I}$, each slice of $X_i$ in $\prod_\mathcal{I}X_i$ should be another copy of $X_i$. Maybe some other assumption makes it true? You probably have better intuition for it than I do. Do you think it's a lost cause? Nov 27, 2022 at 15:46