Why is the product topology generally coarser than the box topology?

Question

This maybe very obvious to some but I can't understand it for the life of me. I'll try explain the difficulty in me trying to understanding this. In Munkres (sect-19), the box topology is defined through a basis and the product topology by a subbasis.

Now, the thing is, for the basis of the box topology, we consider the subsets for which all the projections are simultaneously continuous, but, even for subbasis of the product topology , we have less of a restraint as we don't require this simultaneously all the projections continuous (we only need some of them at a time for a set to be open), yet we have that the box topology has more open sets than the product????

Answer 1

We only see the difference when we are dealing with an infinite product of sets. The box topology includes all products of open sets within each of the individual sets that make up the product.

The product topology includes just the sets that make the projections continuous. Consider, for example, $\mathbb R\times\mathbb R\times ...$ as a topological space.

Each of the individual copies of $\mathbb R$ has its own projection map, say $p_i$ that takes $(x_1,x_2,.....)$ to $x_i\in\mathbb R_i$ where $\mathbb R_i$ is the $i$th copy of $\mathbb R$ in the product. We require $p_i$ to be continuous so we require that, for each open $U\in\mathbb R_i$, that $p_i^{-1}(U)$ to be open in the product space.

$p_i^{-1}(U)=\mathbb R_1\times\mathbb R_2\times...\times ...p_i^{-1}(U)\times\mathbb R_{i+1}\times ...$

Sets of this form, make up a basis for the product topology. If you take unions and finite intersections of these sets you will see that an open set in the product topology will be all of $\mathbb R$ in all but a finite number of the components of the product.

Therefore, a set like $(0,1)\times (0,1)\times ...$ will be open in the box topology but not in the product topology.

Answer 2

I think you misunderstood the use of projections in the definition of the box / product topology.

The product topology $\tau_p$ is the coarsest topology on $P = \prod_{i \in I} X_i$ such that all projections $p_i : P \to X_i$ are continuous. Thus they are simultaneously continuous. For any topology $\tau$ on $P$ having this property we must therefore have $p_i^{-1}(V_i) \in \tau$ for all $i \in I$ and all open $V_i \subset X_i$. The coarsest topology with this property has therefore as a subbasis all these sets $p_i^{-1}(V_i)$. Thus a basis for $\tau_p$ is given by all finite intersections of such sets. These have the form $$\bigcap_{i \in K} p_i^{-1}(V_i)$$ with arbitrary finite $K \subset I$. But clearly $$\bigcap_{i \in K} p_i^{-1}(V_i) = \prod_{i \in I} V_i$$ where $V_i = X_i$ for $i \in I \setminus K$. In other words, a basis for $\tau_p$ is given by all products $$\prod_{i \in I} V_i$$ where $V_i \subset X_i$ is open and $V_i = X_i$ for all but finitely many $i$.

The box topology $\tau_b$ on $P$ has as a basis all products $$\prod_{i \in I} V_i$$ where $V_i \subset X_i$ is open. It is allowed that $V_i \ne X_i$ for infinitely many $i$.

Clearly the box topology is finer than the product topology, in particular all projections $p_i$ are continuous.

Obviously the box topology agrees with product topology if $I$ is finite because their bases are identical. They also agree if one $X_i = \emptyset$ because then $P = \emptyset$ and the only topology on $P$ is $\{ \emptyset \}$. So let us assume that $I$ is infinite and $X_i \ne \emptyset$ for all $i$. Under this assumption we do not need to consider the case that some $V_i = \emptyset$ because then $\prod_{i \in I} V_i = \emptyset$ which does not make a real contribution to a basis. Thus is the sequel we only consider basic open sets in which all $V_i \ne \emptyset$.

1. For all but finitely many $i$ the space $X_i$ has the trivial topology. Then $\tau_b = \tau_p$.
   This follows from the fact that if $\prod_{i \in I} V_i \in \tau_b$ has all $V_i \ne \emptyset$, thus $V_i = X_i$ for all but finitely many $i$, hence $\prod_{i \in I} V_i \in \tau_p$.

2. For infinitely many $i$ the space $X_i$ has an open subset $V_i \ne \emptyset, X_i$. Then $\tau_b$ is strictly finer than $\tau_p$.
   This can be seen as follows:
   We have $p_i(\prod_{i \in I} V_i) = V_i$. Thus if $\prod_{i \in I} V_i \in \tau_p$, then all but finitely many $p_i(\prod_{i \in I} V_i) = X_i$. Since each open $U \in \tau_p$ is a union of basic open sets, we have $p_i(U) = X_i$ for all $U \in \tau_p$ and all but finitely many $i$. Now we can pick $V_i \ne \emptyset, X_i$ open in $X_i$ for infinitely many $i$. Take all other $V_i = X_i$. Then $p_i(\prod_{i \in I} V_i) = V_i \ne X_i$ infinitely many $i$. This shows that $\prod_{i \in I} V_i \in \tau_b$, but $\prod_{i \in I} V_i \notin \tau_p$.