# Product topology in Munkres' books

In Topology, the second edition by Munkres, in section 19, on page 113 he says the following:

"So let us consider the cartesian product \begin{align*} X_1\times ...\times X_n \quad and \quad X_1\times X_2 \times ..., \end{align*} where each $$X_i$$ is a topology space. There are two possible ways to proceed. One way is to take as basis all sets of the form $$U_1\times ...\times U_n$$ in the first case, and of the form $$U_1\times U_2 ...$$ in the second case, where $$U_i$$ is an open set of $$X_i$$ for each $$i$$. This procedure does indeed define a topology on the cartesian product; we shall call it the box topology.

Another way to proceed is to generalize the subbasis formulation of the definition, given in §15. In this case, we take as a subbasis all sets of the form $$\pi_i^{-1}(U_i)$$, where $$i$$ is any index and $$U_i$$, is an open set of $$X_i$$,. We shall call this topology the product topology.

How do these topologies differ? Consider the typical basis element $$B$$ for the second topology. It is a finite intersection of subbasis elements say for $$i = i_1,..., i_k$$. Then a point $$\mathbf{x}$$ belongs to $$B$$ if and only if $$\pi_i(\mathbf{x})$$ belongs to $$U_i$$, for $$i = i_1,...,i_k$$; there is no restriction on $$\pi_i(\mathbf{x})$$ for other values of $$i$$."

My question is:

1. Why $$i = i_1,..., i_k$$? I think it should $$i=1,...,k$$.

2. I don't understand "there is no restriction on $$\pi_i(\mathbf{x})$$ for other values of $$i$$".

Can someone help me? Thanks.

• They differ in that if you take proper open subsets $U_i\subseteq X_i$ then $U_1\times U_2\times ...$ will be open in box topology but not in the product topology. Commented Oct 18, 2021 at 7:59

The subbase is all sets of the form $$\pi_i^{-1}[U_i]$$ where $$i \in I$$ (this can be any index set not just $$\Bbb N$$ BTW) and $$U_i \subseteq X_i$$ is open.
The generated subbase thus takes finitely many indices and open sets and not necessarily the first $$k$$ (if there even are “first indices”, $$I$$ could be $$\Bbb Z$$ or $$\Bbb R$$ or even larger sets) or “consecutive” ones (same remarks on arbitrariness of index sets apply). Hence the notation $$i_1,\ldots, i_k$$, it’s just a way of showing that the finite set is from the index set $$I$$ (hence small $$i$$) with subscripts to distinguish and count them.
And if $$x$$ in the product $$\prod_{i \in I} X_i$$ (a more neutral way of denoting it then the only “$$\Bbb N$$-suggesting” $$X_1 \times X_2 \times \ldots X_n \times \ldots$$) is in the set $$B=\pi_{i_1}^{-1}[U_{i_1}] \cap \ldots \pi_{i_k}^{-1}[U_{i_k}]$$ iff we have $$x_{i_j} = \pi_{i_j}(x) \in U_{i_j}$$ for all $$j=1,\ldots k$$ so we only have a condition on the finitely many coordinates $$\{i_1,\ldots,i_k\}$$ and no idea what $$x_i$$ could be for all other (infinitely many, usually) $$i \in I$$, only that it’s in $$X_i$$ by virtue of being in the product $$\prod_{i \in I} X_i$$.
Hope that clarifies and elucidates Munkres’ discussion. In the beginning Munkres only treats finite products and countable ones indexed by $$\Bbb N$$ but later on he will discuss the most general case, as I just did.
• For $I = \mathbb N$ you could equally well take all "initial segments" $\{1,\ldots,k\}$ instead of all general finite sets $\{i_1,..\ldots,i_k\}$. Perhaps this confused the OP. Commented Oct 18, 2021 at 14:36
• The bases are identical because $U_j = X_j$ is allowed. Each $B=\pi_{i_1}^{-1}[U_{i_1}] \cap \ldots \pi_{i_k}^{-1}[U_{i_k}]$ has the form $\pi_1^{-1}[V_1] \cap \ldots \pi_r^{-1}[V_r]$ with suitable $r$ and $V_j$. Commented Oct 18, 2021 at 15:32