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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.

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  • $\begingroup$ 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. $\endgroup$
    – Jakobian
    Commented Oct 18, 2021 at 7:59

1 Answer 1

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Paul Frost
    Commented Oct 18, 2021 at 14:36
  • $\begingroup$ @PaulFrost yes, you get a smaller but equivalent base then. Later in the chapter Munkres does explain it better. $\endgroup$ Commented Oct 18, 2021 at 14:54
  • $\begingroup$ 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 $V_j$. $\endgroup$
    – Paul Frost
    Commented Oct 18, 2021 at 15:32
  • $\begingroup$ Thank you everybody! $\endgroup$
    – Mingg
    Commented Oct 19, 2021 at 4:55

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