I have the following doubt. Consider a manifold $(M,\tau)$ and its product topology $\tau^2$.

How is then an open set $U\in\tau^2$ defined?

Is it $U=\bigcup_{i\in I} U_i\times V_i$, where $U_i,V_i\in\tau$ or $U=\bigcup_{i,j\in I}U_i\times V_j$?

It seems to me that it is the first one because in the second one

\begin{align} U=\bigcup_{i,j\in I}U_i\times V_j=\bigcup_{i\in I}(\bigcup_{j\in I}U_i\times V_j)=\bigcup_{i\in I}U_i\times(\bigcup_{j\in I}V_j)=(\bigcup_{i\in I}U_i)\times\text{Im }U=\text{Dom }U\times\text{Im }U \end{align}

And, so, for example, $U=A\times B\cup C\times D$ could not be an open set if $A$ and $C$, and $B$ and $D$ are disjoint open sets in $\tau$.

Am I right? Thanks.

  • 1
    $\begingroup$ I didn't follow your reasoning with the equalities, but it should be the second one. Think for example of what open sets look like in the plane $\mathbb R \times \mathbb R$. They are unions of little open boxes $(a,b) \times (c,d)$. $\endgroup$ – D_S Aug 28 at 13:42
  • $\begingroup$ Note that both sets are open. Every open set in $\tau^2$ has the first form, but not nessecarily the second. An example is $(0,1)\times (0,1) \cup (2,3) \times (2,3) \subset \mathbb{R}^2$. $\endgroup$ – G. Chiusole Aug 28 at 14:11

It should be the first. An open set is a union of products of open sets, so it would take the form $\bigcup_{i \in I} U_i \times V_i$. It is a union of single terms (the terms just happen to have two factors) so only one index is needed. The second option you present is ``too big."

You should think of an open set as $\bigcup_i \mathcal{O_i}$, where each $\mathcal{O}_i$ has the form $\mathcal{O}_i = U_i \times V_i$. This makes it clear that one index suffices.


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