This is not homework
I am dealing with infinite product topology
Source:Conover,A First Course in Topology
First a definition for reference
3.1 Definition Let $\Lambda$ be a any nonempty set such that for each $\lambda \in \Lambda$ there is a nonempty set $X_{\lambda}$ The Cartesian product of {$X_{\lambda}$: $\lambda \in \Lambda$ } is defined to be:
$\Pi_{\lambda\in\Lambda}$ $X_{\lambda}$= {x:$\Lambda$$\to \bigcup_{\lambda\in\Lambda} X_{\lambda}$: x($\lambda$)$\in X_{\lambda}$ for each $\lambda\in\Lambda$}
the set of all functions x from the the index set $\Lambda$ to the union of the factors($\bigcup_{\lambda\in\Lambda}$$X_{\lambda}$) such that for each $\lambda\in\Lambda$, x( $\lambda$)$\in$ $X_{\lambda}$
Exercise 3.7.1 page 100
Let {$X_n$:n$\in$ $Z^{+}$} be a a countable collection with $X_n$=R with the usual topology. Consider the sequence {($\frac{1}{n}$,$\frac{1}{n}$,…):n$\in $$Z^{+}$} $\subset $$\Pi X_i$Certainly the sequence should converge to the point $(0,…,0)$ $\in \Pi X_i$ But if we try to over control the product space by requiring that the product of open sets to be open be it will not
For each n$\in$ $ Z^{+}$,put $U_n$=(-$\frac{1}{n}$, $\frac{1}{n}$)
(1) Show that each $U_n$ is open in R but the sequence {($\frac{1}{n}$, $\frac{1}{n}$…):n$\in $$Z^{+}$} is never in $\Pi U_i$, so if the set is open in the product {($\frac{1}{n}$,$\frac{1}{n}$,…):n$\in $$Z^{+}$ } will not converge to $(0,…,0)$ which is not reasonable.
I think it is reasonable that it doesn’t converge ,since the sequence is not in the product space.
My attempt
Let {$X_n$:n$\in$ $Z^{+}$} be a countable collection with $X_n$=R with the usual topology.
Let $\Lambda$ be our indexing set and |$\Lambda$|$\geq$ $\aleph_{0}$.
For each n$\in$ $ Z^{+}$,Put $U_n$=(-$\frac{1}{n}$, $\frac{1}{n}$)
Let U=$\Pi U_i$ be basic open set containing $(0,..,0)$ If $U_i\ne\emptyset$ and $U_i =X_i$, $U_i$ is open. Since {($\frac{1}{n}$,$\frac{1}{n}$,…):n$\in $$Z^{+}$ }$\notin$ $\Pi U_i$ then for each $U_i$ ,$\frac{1}{n} \notin U_i$ Also $\bigcap U_i=${ ${0}$} which is not open l Thus the sequence diverges
As to openess,
R$\setminus$(-$\frac{1}{n}$,$\frac{1}{n} $)= (-$\infty$,-$\frac{1}{n}$)$\cup$($\frac{1}{n}$,$\infty$)
Interior point is the empty set.
Comment: I doubt if I got any of this correct. I proved it converges for a finitely many of them ,which was another question. I found enough info on how to do it. How do you define over control? If you reword Def 3.1 so you force all sets to be open, that could make the sequence diverge.l would like to know how to it properly.
Though this problem seems a contradiction.