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 ?

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    $\begingroup$ Because box topology has more open sets. $\endgroup$
    – Asaf Karagila
    Jul 19, 2014 at 9:28

6 Answers 6


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.

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    $\begingroup$ Small nitpick: $U_n \neq \emptyset$ (which is why the box topology and the product topology do coincide if all but finitely many $X_n$ have at most one element). $\endgroup$ Jul 19, 2014 at 10:11
  • $\begingroup$ Yes, very good point. I've edited to assume each $X_n$ has an open $U_n$ such that $\emptyset \varsubsetneq U_n \varsubsetneq X_n$. $\endgroup$
    – cws
    Jul 19, 2014 at 22:23
  • $\begingroup$ @cws , This is very insightful , thank you :) $\endgroup$ Jul 22, 2014 at 23:59
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    $\begingroup$ @cws $U := \prod\limits_{n\in\mathbb{N}} U_n$ is open in the box topology on $\prod\limits_{n\in\mathbb{N}} X_n$ but not the product topology. How it is not open in product topology? $\endgroup$ Jul 5, 2020 at 12:19
  • $\begingroup$ Why finite intersection? It seems we first get the sub base then generate the topology. Why do we consider the finite intersection? Why not infinite intersection? And it seems you use some category theory terms, which will create some problems for beginner topology readers. @cws $\endgroup$
    – falamiw
    May 15, 2021 at 14:17

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.

  • $\begingroup$ I don't get the notion $X\in (2^{-n},2^n)$ , what is that ? do you mean that $x$ is a coordinate ? or do you use a set-theoretic definition for $(a,b)$ ? Also , Isn't $\prod_{n \ge 1} (2^{-n},2^n)$ a subset of $\mathbb{R^\mathbb{N}} \times \mathbb{R^\mathbb{N}}$ not $\mathbb{R^\mathbb{N}}$ ? $\endgroup$ Jul 22, 2014 at 12:47
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    $\begingroup$ Sorry, here I mean the open interval $(a,b)$, i.e. $\{x \in \mathbb R : a < x < b \}$. I see now that that's pretty confusing notation when I'm also using $(x,x,x,\dots)$ to mean a vector in $\mathbb R^\mathbb N$, but I can't immediately think of a better way to write it. $\endgroup$ Jul 22, 2014 at 15:43
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    $\begingroup$ I get it now , it's a great example! I don't know how you have got this example! thank you. $\endgroup$ Jul 22, 2014 at 23:58
  • 2
    $\begingroup$ You could write open intervals as $]a,b[$. $\endgroup$
    – jinawee
    Feb 11, 2015 at 16:51
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    $\begingroup$ It means that a function to a product, e.g. $f : A \to B \times C$, is continuous exactly when the separate components, in this example $\pi_1f : A \to B$ and $\pi_2 f : A \to C$, are continuous. If any component is not continuous, the whole is not continuous, while if all the components are continuous, so is the whole. $\endgroup$ Apr 7, 2019 at 12:50

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.

  • $\begingroup$ More generally, in any product of nontrivial Hausdorff spaces, the subspace which is a product of two fixed points in each coordinate is discrete in box topology and compact in Tychnoff topology. $\endgroup$
    – tomasz
    Jul 19, 2014 at 11:34
  • $\begingroup$ Well, obviously. :-) $\endgroup$
    – Asaf Karagila
    Jul 19, 2014 at 11:35
  • $\begingroup$ My point was, thusly your example shows that the two topologies are (bar obvious exceptions) distinct, as opposed to being just an example. $\endgroup$
    – tomasz
    Jul 19, 2014 at 11:36
  • $\begingroup$ Ah, well, yes, I tried to give an example where the topologies can't be equal because of cardinality arguments. Thanks for the added point! $\endgroup$
    – Asaf Karagila
    Jul 19, 2014 at 11:37

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.


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.

  • $\begingroup$ Could you tell me where I can read more about the relation between "closure of finite intersections" and "behaving of topological spaces close to our intuitive expecations"? and between "closure under unions" and " interior operator" ? Because I have wondered asked where topological spaces are defined the way they are. Thank you in advance. $\endgroup$ Jul 22, 2014 at 11:53
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    $\begingroup$ @MathsLover The set theoretical topological spaces where defined by Felix Hausdorff in his book "Grundzüge der Mengenlehre", which appeared in 1914. In 1912, Jan Brouwer had started intuitionism, but Felix Hausdorff's work is not really based on it. The linked publication page of Dirk van Dalen is a good source for the connections between formal intuitionistic logic and topological spaces. If the links in that page don't work in your browser, copy the desired "link address" and replace "papers.html" in the current addresss by "articles/..." from the copied "link address". $\endgroup$ Jul 22, 2014 at 14:42

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!


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