Why are box topology and product topology different on infinite products of topological spaces?

Question

Why are box topology and product topology different on infinite products of topological spaces ?

I'm reading Munkres's topology. He mentioned that fact but I can't see why it's true that they are different on infinite products.

So , Can any one please tell me why aren't they the same on infinite products of topological spaces ?

Answer 1

Let $X_n$ be topological spaces for each $n\in\mathbb{N}$. To avoid the issues pointed out by Najib, assume for each $n$ that $X_n$ is not a point, and the topology on $X_n$ is not the trivial topology (i.e. there is an open set besides $\emptyset$ and $X_n$). For each $n$, let $U_n \subset X_n$ be a proper, nonempty open subset. Then the set $U := \prod\limits_{n\in\mathbb{N}} U_n$ is open in the box topology on $\prod\limits_{n\in\mathbb{N}} X_n$ but not the product topology.

The product topology is generated by sets of the form $\prod\limits_{n\in\mathbb{N}} U_n$ where each $U_n$ is open in $X_n$ and, for all but finitely many $n$, we have $U_n = X_n$. In other words, almost all of the factors have to be the entire space. For the box topology, each factor $U_n$ just has to be open in $X_n$.

Here is one way of understanding why the product topology is more important (even though the box topology seems more intuitive at first). The product topology is the smallest topology such that for each $k\in\mathbb{N}$, the projection map $\pi_k:\prod\limits_{n\in\mathbb{N}} X_n\to X_k$ is continuous. The preimage of an open set $U_k\subseteq X_k$ via $\pi_k$ is one of the basic sets for the product topology described above: specifically, it is $U_k$ in the $k$th factor and the whole space $X_n$ in each other factor. To generate a topology, we need to include finite intersections of such sets (so not the entire space in finitely many positions), but not infinite intersections. So thinking about wanting the $\pi_k$ to be continuous, the product topology has "enough" open sets, and the box topology adds in open sets that aren't needed.

Answer 2

Their basic open sets are the same: direct products of open sets. Except in the product topology, all but finitely many of those open sets must be the whole space in that coordinate. Any finite intersection of these kinds of sets also has the same property, so for any open set in the product topology, the image under that open set is the entire codomain for all but finitely many of the coordinate projections.

For instance, $(0,1)^\Bbb N\subset\Bbb R^\Bbb N$ is open in the box topology, but it is "too tight" to be open in the product topology. However $(0,1)\times\Bbb R\times\Bbb R\times\cdots$ is open in the product topology, since only one of the factors is not the whole space. (Of course, these are just basic sets. In general, open sets can't be described as direct products of open sets from each coordinate space, but rather as arbitrary unions of these basic open sets.) The product topology on $\prod X_i$ is all about making all of the coordinate projections continuous and nothing more, so the open sets in the product topology are generated by the preimages of open sets under these coordinate projections - that's where we get the basic open sets from.

But anything open in the product topology is open in the box topology. For infinite products, then, the box topology is strictly finer than the product topology.

Answer 3

The simplest way, perhaps, to see a particular difference is the product of a finite discrete space.

Consider the product of $\Bbb N$ copies of $\{0,1\}$ (with the discrete topology). The result is the Cantor space. This is a compact metric space, and therefore it has a countable basis, and only $2^{\aleph_0}$ open sets.

Consider the box topology on the same product, then you get a discrete space of size $2^{\aleph_0}$, which therefore has $2^{2^{\aleph_0}}$ open sets, and is most certainly not compact.

Answer 4

The other two answers [edit: at the time of writing] are essentially correct, but here's a concrete example as to why the box topology has "too many" open sets.

Consider the function $f : \mathbb R \to \mathbb R^\mathbb N$ given by $f(x) = (x,x,x,\dots)$. Consider the subset of $\mathbb R^\mathbb N$ given by $\prod_{n\ge 1}(-2^{-n},\,2^{-n})$. In the box topology this is open. Its pre-image under $f$ is $\{ x \in \mathbb R : \forall n. x \in (-2^{-n},\,2^{-n}) \}$, but it's easy to see that's just $\{ 0 \}$, which is not open. So $f$ is not continuous!

You may or may not decide that this is a surprising result, but the essential property of the product topology is that you can identify continuous functions purely by looking at each individual projection, and with $f$ that is simply not true: every projection is continuous, but $f$ itself is not.

Answer 5

The box topology is identical to the product topology on finite products of topological spaces, because the system of open sets is closed under finite intersections. Because it is not closed under arbitrary intersections, this is no longer true for infinite products.

In an "open system" (not really an established name), the system of open sets would only be closed under arbitrary unions, but nothing at all would be said about intersections. In that case, the box topology and product topology are different even for finite products of "open systems". (The box-topology nevertheless defines a symmetric monoidal product, but so what?) The closure under arbitrary unions allows to define an interior operator, which an important part of a topological space. The closure under finite intersections ensure that topological spaces behave close to our intuitive expectations, which also explains their close connections to formal intuitionistic logic.

Answer 6

Hypotheses: Take infinitely many topological spaces $(X_i, \mathcal T_i)_{i\in I}$ such that none of $X_i$ is empty, and infinitely many $\mathcal T_i$ are strictly finer than than the indiscrete topology on $X_i$.

Claim: There exists an open set it the box topology $\mathcal T_\square$ that is not open in the product topology $\mathcal T_\Pi$.

Arguments: We first observe the kind of open sets that $\mathcal T_\Pi$ admits. It contains arbitrary unions of finite (nonepmty) intersections, each of which is of the form $$ \bigcap_{i\in K} \biggl(\ \bigcap_{k = 1}^{N_i} \pi_i^{-1}(U_{i_k}) \biggr)\text, $$ where $K$ is a finite nonempty subset of $I$ and $U_{k_i}\in\mathcal T_i$. (Also, $N_i\ge 1$ for each $i$.) Now, for each $i$, since $U_{i_k}$'s belong to the same $\mathcal T_i$, we can pull $\pi_i^{-1}$'s, getting $$ \bigcap_{i\in K}\Biggl(\pi_i^{-1}\biggl( \underbrace{\bigcap_{k = 1}^{N_i} U_{i_k}}_{V_i} \biggr)\Biggr)\text. $$

Since each $V_i$ is a finite nonempty intersections of open sets in $\mathcal T_i$, we have $V_i\in\mathcal T_i$. Hence we have gotten a simpler form $\bigcap_{i\in K}\Bigl(\pi_i^{-1}(V_i)\Bigr)$. It is not difficult to see that for any $W\subseteq X_i$, we have that $\pi_{i_0}^{-1}(W) = \prod_{i\in I} S_i$ where $S_i = X_i$ for $i\ne i_0$ and $S_{i_0} = W$. Using this and that the intersection and Cartesian products commute, we get the above as $$ \prod_{i\in I} Y_i\text, $$ where $Y_i = V_i$ if $i\in K$, otherwise $Y_i = X_i$. Notice that this Cartesian product has $Y_i\ne X_i$ in only finitely many $i$'s.

We are now to analyze an arbitrary union $\mathcal U$ of the above sets. Note that we will have only finitely many $i$'s such that the $i$-th coordinate of $\mathcal U$ will not range through all of $X_i$. (If we get an $X_i$ is the $i$-th coordinate in one of the intersections (of which $\mathcal U$ is a union of ), it'll "swamp" everything.) That is, $\pi_i(\mathcal U)\ne X_i$ for only finitely many $i$'s.

Now we just need to find a $\mathcal V\in\mathcal T_\square$ such that $\pi_i(\mathcal V)\ne X_i$ for infinitely many $i$'s, and we will be done. Just choose nenempty $U_i$'s from $\mathcal T_i$'s such that $U_i\ne X_i$ for infinitely many $i$'s and this can be done because of our hypotheses!