What are the typical basis elements of a product topology generated by $\mathcal{S}$?

78 Views Asked by At

In section 19 of James Munkres' Topology, the author defines $\mathcal{S}_{\beta}=\left\{ \pi_{\beta}^{-1}(U_{\beta}) \ | \ U_{\beta} \text{ open in} \ X_{\beta}\right\}$, and lets $\mathcal{S}$ denote the union of these collections: $\mathcal{S} = \bigcup_{\beta \in J}S_{\beta}$.

Then the author goes on to discuss the typical basis element in the basis $\mathscr{B}$ that $\mathcal{S}$ generates:

"the collection $\mathscr{B}$ consists of all finite intersections of elements of $\mathcal{S}$. If we intersect elements belonging to the same one of the sets $\mathcal{S}_{\beta}$, we don't get anything new, because $\pi_{\beta}^{-1}(U_{\beta})\cap\pi_{\beta}^{-1}(V_{\beta})=\pi_{\beta}^{-1}(U_{\beta}\cap V_{\beta})$. We get something new only when we intersect elements from different sets of $\mathcal{S}_{\beta}$. The typical element of the basis $\mathscr{B}$ can thus be described as follows: Let ${\beta}_1$, ..., ${\beta}_n$ be a finite set of distinct indices from the index set J, and let ${\beta}_i$ be an open set in $X_{{\beta}_i}$ for $i=1,..., n$. Then $B=\pi_{\beta_1}^{-1}(U_{\beta_1})\cap\pi_{\beta_2}^{-1}(U_{\beta_2})\cap\cdots\cap\pi_{\beta_n}^{-1}(U_{\beta_n})$ is the typical element of $\mathscr{B}$."

My confusion: $B=\pi_{\beta_1}^{-1}(U_{\beta_1})\cap\pi_{\beta_2}^{-1}(U_{\beta_2})\cap\cdots\cap\pi_{\beta_n}^{-1}(U_{\beta_n})$ is NOT an intersection of different $\mathcal{S}_{\beta_i}$'s. Instead, $\mathcal{S}_{\beta_1}\cap\mathcal{S}_{\beta_2}\cap\cdots\cap\mathcal{S}_{\beta_n}$ is. What am I missing here?

Please help. Many thanks in advance!

1

There are 1 best solutions below

5
On BEST ANSWER

The intersection $B$ is just the set of all $f$ in the product that obey the finitely many conditions imposed by being in $B=\pi_{\beta_1}^{-1}(U_{\beta_1})\cap\pi_{\beta_2}^{-1}(U_{\beta_2})\cap\cdots\cap\pi_{\beta_n}^{-1}(U_{\beta_n})$, i.e. $f \in \pi_{\beta_1}(U_{\beta_1})$ so $f(\beta_1) \in U_{\beta_1}$, and $f \in \pi_{\beta_2}(U_{\beta_2})$ so $f(\beta_2) \in U_{\beta_2}$ and so on till $f \in \pi_{\beta_n}(U_{\beta_n})$ so $f(\beta_n) \in U_{\beta_n}$

We take intersections of the different sets in those families, not between the families themselves!

And the point Munkres makes is that we can combine two "open conditions" in the same coordinate to one "open condition" in that same coordinate.