Basis for Product Topology ..

Question

Problem- Prove that ${P}$ is a basis for $\prod X_\alpha$, where $P$ is the product basis:

$$P=\Big\{\prod\limits_{\alpha \in I} U_\alpha\Big\}$$

Where each $U_\alpha$ is open set in the space $(X_\alpha)_{\tau_\alpha}$ and $U_\alpha=X_\alpha$ for all but finitely many $\alpha \in I$.

Attempt-

1.Let $x=(x_i)_{i=1}^\infty$ be any point of $\prod X_\alpha$. Then

$x\in U_{\alpha_1} ×U_{\alpha_2}×...×U_{\alpha_k} ×...×X×X×X...$.

2. Suppose $$x\in (\prod U_{\alpha_i})\cap(\prod V_{\alpha_i})=\prod( U_{\alpha_i} \cap V_{\alpha_i})$$

$U_{\alpha_i}=V_{\alpha_i}=X_\alpha$ for $i=1,\dots\ ,k$ otherwise $U_\alpha\neq X_\alpha$ and $ V_\alpha\neq X_\alpha$.

therefore we can find a basis element $W_{\alpha_i}$ such that

$x\in \prod W_{\alpha_i}\subset \prod( U_{\alpha_i} \cap V_{\alpha_i})$

$W_{\alpha_i}=X_\alpha=U_\alpha=V_\alpha$, for $i=1,\dots\ ,k$. otherwise $U_\alpha,V_\alpha,W_\alpha$ not equal to $X_\alpha$

Is it correct?

Thanks.

Answer 1

There are some problems with it.

In the first part you have no reason to think that all of the factor spaces are the same, so you can’t replace $X_\alpha$ by $X$ in the product defining a member of $P$. To describe a typical member of $P$, you must specify a finite $F\subseteq I$ and an open set $U_\alpha$ in $X_\alpha$ for each $\alpha\in F$; the corresponding member of $P$ is then

$$\prod_{\alpha\in F}U_\alpha\times\prod_{\alpha\in I\setminus F}X_\alpha\;.\tag{1}$$

Next, given $x=\langle x_\alpha:\alpha\in I\rangle\in X=\prod_{\alpha\in I}X_\alpha$ and a generic member of $P$ as in $(1)$, you have no reason to think that $x$ is in that member of $P$: what if $\alpha_1\in F$, and $x_{\alpha_1}\notin U_{\alpha_1}$, for instance? Then $x$ definitely is not in that product. You need to specify a member of $P$ that provably contains $x$. (This is actually trivial, since the set $X=\prod_{\alpha\in I}X_\alpha$ is in $P$, but you should think about exactly what has to be true for $x_\alpha$ in order for $x$ to belong to the set in $(1)$.)

In the second part you have the right idea for showing that the intersection of two members of $P$ that contain $x$ is a member of $P$ that contains $x$, but you have the definition of members of $P$ backwards: there are only finitely many coordinates on which $U_\alpha$ is not all of $X_\alpha$. You’ve also made the unwarranted assumption that the two members of $P$ restrict $U_\alpha$ and $V_\alpha$ (to be something other than the whole space) on the same finite set of coordinates. In fact one of the sets might be the one in $(1)$, and the other might be

$$\prod_{\alpha\in G}V_\alpha\times\prod_{\alpha\in I\setminus G}X_\alpha\;,$$

where $G$ is a finite subset of $I$ completely disjoint from $F$.