Let $\{ X_s \}_{s \in S}$ be a family of topological spaces, then the product topology defined on the cartesian product $X := \prod_{s\in S} X_s$ is the coarsest (i.e. smallest) topology such that every projection map $\tau_s: X \to X_s$ is continuous (see PlanetMath).

Now I am interested how the open sets look like in this product topology. In my notes and textbook's I find, that the sets of the form $$ \prod_{s \in S} W_s $$ with $W_s$ open in $X_s$ and $W_s \ne X_s$ only for finitely many $s \in S$ form a base of this topology. Now I know how does the base sets look, but how does the open sets look? I know every open set could be written as an union of base sets, but because in general $$ (A \times B) \cup (C \times D) \ne (A \cup B) \times (C \cup D) $$ (just $\subset$ holds) I can not say for example that the open sets are the sets $\prod_{s \in S} W_s$ with $W_s$ open and $W_s \ne X_s$ only for finitely many $s \in S$. So, could something be said about the form of the open sets?

  • $\begingroup$ Little more can be said than what is in the definition. For example one such open neighboorhodd of a point $x\in\mathbb R^{\mathbb N}$ says that $y\approx x$ if either $y_1\approx x_1$ or both $x_2\approx y_2$ and $x_3\approx y_3$ ... $\endgroup$ Commented Jul 13, 2013 at 14:35
  • $\begingroup$ I think this question is a little too vague. $\endgroup$
    – tomasz
    Commented Jul 13, 2013 at 14:40
  • 3
    $\begingroup$ This is indeed a great question to keep in your mind. But it does not take you anywhere just think about abstract definitions and set-theoretic operations. You need to see product topology in work in some specific examples. $\endgroup$
    – Hui Yu
    Commented Jul 13, 2013 at 14:54
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    $\begingroup$ btw. if I just say a set in the product topology is open iff every component is an open set, in the infinite case I get the box topology, which differs from the product topology, so this straightforward characterisation is wrong. $\endgroup$
    – StefanH
    Commented Jul 13, 2013 at 15:02
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    $\begingroup$ That a Cartesian product is closed iff all factors are closed in no way implies that all closed sets are Cartesian products. For example, in $\mathbb{R}^2$ every line is closed, but only the horizontal and vertical ones are products. $\endgroup$ Commented Jul 13, 2013 at 16:18

2 Answers 2


As Hui Yu says, you are not going too far with just the definition and set-theoretic operations: think about specific examples.

For instance, before fighting against scary monsters like infinite arbitrary products, how about looking for examples in the humble $\mathbb{R}^2$? Are you sure you could find out a simple characterization of the open sets there (which happen to be the same for the product topology and for the Euclidian, usual one)?

E.g., what about a set like this one:

$$ U = \left\{ (x,y) \in \mathbb{R}^2 \ \vert \ xy > 1 \ , \ x > 0 \right\} \quad \text{?} $$

It's an open set. Do you think you could describe it easily (I mean, without just repeating the definition of open sets in $\mathbb{R}^2$) in terms of the open sets of the basis of the product topology?

  • $\begingroup$ Ok, got the point. Guess so it is simple not possible. $\endgroup$
    – StefanH
    Commented Jul 13, 2013 at 17:44

Note that the following are true (in all spaces): (fix a base $\mathcal{B}$ for a space $X$)

  1. $f: X \to Y$ is open iff $f[B]$ is open in $Y$ for every $B \in \mathcal{B}$.

  2. $f: X \to Y$ is continuous at $x$ iff for every open set $O$ that contains $f(x)$, there exists some $B \in \mathcal{B}$ such that $x \in B$ and $f[B] \subset O$.

  3. $X$ is compact iff every cover of $X$ with elements from $\mathcal{B}$ has a finite subcover.

  4. $D \subset X$ is dense iff every non-empty $B \in \mathcal{B}$ intersects $D$.

  5. $f : Y \to X$ is continuous iff $f^{-1}[B]$ is open in $Y$ for all $B \in \mathcal{B}$.

Note that we can reason about continuity, openness, compactness, knowing only a base for the topology. So in most cases, all we really need is a good description for a base.

This is analogous to the situation of metric spaces $(X,d)$, where a base is specified (all sets of the form $B(x,r) = \{ y \in X: d(x,y) < r \}$, where $r>0$) and continuity between metric spaces is often expressed using the $\epsilon$-$\delta$ definition, which is just like 2., except using this base in both spaces. For product spaces as well, all proofs involving them essentially uses this base (or the subbase of all $\pi_s^{-1}[O]$ for open sets $O$ in $X_s$). We really do not need a description beyond the fact that they are unions of basic sets.

  • $\begingroup$ Sorry for necropost. Can you think of an open set in the product AxB that is not of the form $ U_a \times U_b $; $U_a, U_b$ open in $A,B$ respectively? $\endgroup$
    – MSIS
    Commented Nov 19, 2019 at 2:19
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    $\begingroup$ @MSIS E.g. in $\Bbb R \times \Bbb R$ the open set $\{(x,y): x \neq y\}$ or $\{(x,y): x^2+y^2 > 1\}$ $\endgroup$ Commented Nov 19, 2019 at 5:03
  • $\begingroup$ @HennoBrandsma Could you give an example of an open set in product topology where infinitely many factors are proper subsets? $\endgroup$
    – ZSMJ
    Commented Apr 16, 2020 at 15:50
  • $\begingroup$ @AkashGaur a product set with infinitely many factors are proper subsets has empty interior, so doesn’t contain a non-empty open set. $\endgroup$ Commented Apr 16, 2020 at 15:57
  • $\begingroup$ +1 I liked the answer $\endgroup$ Commented Feb 24, 2022 at 3:27

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