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Let $\{X_i \ | \ i \in I\}$ be a topological spaces family, and: $$B_C=\{\prod_{i \in I}U_i \ | \ U_i \ is \ open\ of\ X_i \}$$ It's not difficult to see that $B_C$ is base of a topology, and it will be denoted by $\tau_c$. We have to show that if $I$ is infinite, the product topology is strictly contained in the box topology. This is $\tau_p \subsetneq \tau_c$.

We know that the product topology is generated by the base: $$B_P=\{\bigcap_{j\in J}\pi_j^{-1}U_j \ | \ U_j \ is \ open \ of \ X_j \ y \ J\subseteq I \ is \ finite \}$$ The interesing of this excercise is to found the open set that is in $\tau_c$ and is not in $\tau_p$.

Nicolay Avendaño
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  • See https://math.stackexchange.com/questions/871610/why-are-box-topology-and-product-topology-different-on-infinite-products-of-topo – Reiner Martin Jan 24 '19 at 21:54
  • Just choose a collection where $J$ is not finite and each of the $U_j$ are nontrivial, right? – MPW Jan 24 '19 at 22:08

1 Answers1

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Take an infinite subset $I_0$ of $I$ such that there are open sets $\emptyset \neq U_i \subsetneq X_i$ for all $i \in I_0$ (this need not be possible, e.g. in the trivial case where all $X_i$ are indiscrete, but it's a very minor condition to impose; also all $X_i$ must be non-empty to avoid trivialities).

Then define $O=\prod_{i \in I} O_i$ where $O_i = U_i$ for $i \in I_0$ and $O_i = X_i$ for all other $i$ is open in the box topology (as all $U_I$ and $X_i$ are open) but not open in the product topology as it cannot even contain a single basic open subset (so cannot be a union of them, so cannot be open): suppose $B=\prod_i B_i$ is a basic open set, which means by definition that there is a finite subset $F$ of $I$ such that $B_i = V_i$, some non-empty open subset of $X_i$ for all $i \in F$ and $B_i = X_i$ for all $i \notin F$.

Then there is some index $j \in I_0 \setminus F$ (here we use $I_0$ is infinite and $F$ is finite) and some $p_j \in X_j \setminus U_j$ (as $U_j$ is a proper subset) and also picking $p_i \in B_i$ for all $i \neq j$ we have a point $(p_i)_i$ that lies in $B$ (as $j \notin F$, we are free to choose $p_j$ and still be in $B$) but not in $O$ as $p_j \notin O_j=U_j$.

This shows that under this mild assumption we indeed have that the product topology is strictly weaker than the box topology.

Henno Brandsma
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