This is pretty basic, but I just learned about the product topology in class, and it seems to me that it doesn't agree with how it is defined - obviously, I'm wrong, but I was hoping someone could explain to me exactly where my thoughts are going wrong.

Working in the simple case of two sets for illustrative purposes: Let $(X,T_x)$ and $(Y,T_y)$ be topological spaces. Then we define the product topology on $X \times Y$ to be $T_{X \times Y} = \{U \times V: U \in T_x, V \in T_y\}$. This is supposed to be the weakest topology on $X \times Y$ such that the projection operator $p_i$ is continuous for all $i$. However, let us take the case $P \times Q \subset X \times Y$ such that $P \in T_x$ but $Q \notin T_y$. By the definition of product topology this isn't an open set. Then, $p_1[P \times Q] = P \in T_x$. Since $P$ is an open set, but $p_1^{-1}[P]$ isn't an open set, wouldn't this mean that $p_1$ is not continous, in contradition to the definition of the product topology?


2 Answers 2


No, we do not define the product topology to be $\{U\times V:U\in T_X,V\in T_Y\}$ we define it to the the topology with basis $\{U\times V:U\in T_X,V\in T_Y\}$.

In all cases, if $U\in T_X$ then $p^{-1}(U)=U\times Y$ which is open in $X\times Y$.


The set $\mathcal{B}:= \{ U \times V: U \in \mathcal{T}_X, V \in \mathcal{T}_Y\}$ is not the product topology (it's not closed under unions) but it is a base for that topology: all product-open sets are unions of sets from this family.

This makes all projections continuous: $(p_X)^{-1}[U] = U \times Y$ and $(p_Y)^{-1}[V] = X \times V$, for all open $U$ in $X$ and open $V$ in $Y$, and these products are all in $\mathcal{B}$, of course, hence open.

If $\mathcal{T}$ is any topology on $X \times Y$ that makes both projections continuous, then for any open $U$ in $X$ and any open $V$ in $Y$, $(p_X)^{-1}[U] \cap (p_Y)^{-1}[V]$ must be open in $\mathcal{T}$. And clearly $$(p_X)^{-1}[U] \cap (p_Y)^{-1}[V] = U \times V$$

which shows that then we know that $\mathcal{B} \subseteq \mathcal{T}$. So any topology that makes the projections continuous at least has all of $\mathcal{B}$ as open sets. So taking that collection as a base (which we can do, as it's closed under intersections, e.g.) gives us the smallest such topology.

Your set $P \times Q$ is not of the form $p^{-1}[O]$ for a projection $p$ and some open set so not a contradiction to continuity of such projections.

  • $\begingroup$ Ah! I see now. Thank you so much! I didn't realize that the pre-image of an open set would map to the pre-image cross the entirety of the other sets. $\endgroup$ Sep 19, 2018 at 2:50

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