product topology in subspaces Let $A_i$ be an arbitrary subspace of $X_i$. Show that the product topology on $\prod_i A_i$ is equal to the relative topology on $\prod_i A_i$ as a subset of the product space $\prod_i X_i$.
The product topology $\prod_i X_i$ would be the set $\int=\bigcup_i\{\prod_i^{-1}(U):U\in \tau_i\}$, then
the subspace topology $\prod_i A_i$ is the set $\int_{A_i}=\{\prod_i A_i \cap H: H\in \int\}$
                            =$\{\prod_i A_i\cap (\bigcup_i \prod_i^{-1}(U):U\in \tau_i\} = \{\bigcup_i (\prod_i A_i\cap \prod_i^{-1}(U):U\in \tau_i\}$
But, the topology of $A_i$ $\tau_{A_i}=\{A_i \cap U: U\in \tau_i\}$ from this part I find it a little hard to see equality between these relative topologies 
 A: To begin with, your notation $\prod_i^{-1}(U)$ makes no sense: $\prod$ here is the symbol for the Cartesian product, not a function with an inverse. Thus, it’s not really clear what you think the product topology is.
For $i\in I$ let $\tau_i$ be the topology on $X_i$. Let 
$$\mathscr{U}=\left\{\prod_{i\in I}U_i:U_i\in\tau_i\text{ for each }i\in I\right\}\;.$$
For each $U=\prod_{i\in I}U_i\in\mathscr{U}$ let $\operatorname{supp}(U)=\{i\in I:U_i\ne X_i\}$, and let 
$$\mathscr{B}=\{U\in\mathscr{U}:\operatorname{supp}(U)\text{ is finite}\}\;.$$
That is, $\mathscr{B}$ is the set of all products of the form $\prod_{i\in I}U_i$ such that $U_i\in\tau_i$ for each $i\in I$, and $U_i=X_i$ for all but finitely many $i\in I$. By definition $\mathscr{B}$ is a base for the product topology $\tau$ on $X=\prod_{i\in I}X_i$, and that topology therefore consists of all possible unions of members of $\mathscr{B}$:
$$\tau=\left\{\bigcup\mathscr{V}:\mathscr{V}\subseteq\mathscr{B}\right\}\;.$$
Now let $A=\prod_{i\in I}A_i$, where $A_i\subseteq X_i$ for each $i\in I$. For $i\in I$ let
$$\tau_i'=\{A_i\cap U:U\in\tau_i\}\;,$$
the relative topology that $A_i$ inherits from $X_i$. Let $\tau'$ be the product topology on $A=\prod_{i\in I}A_i$, where each $A_i$ has the topology $\tau_i'$. Finally, let $\tau_A$ be the relative topology on $A$ as a subspace of $X$; the problem is to show that $\tau_A=\tau'$.
We know that $\tau_A=\{A\cap U:U\in\tau\}$. We can describe $\tau'$ the same way that we described $\tau$. Let
$$\mathscr{U}'=\left\{\prod_{i\in I}U_i:U_i\in\tau_i'\text{ for each }i\in I\right\}$$
and
$$\mathscr{B}'=\{U\in\mathscr{U}':\operatorname{supp}(U)\text{ is finite}\}\;,$$
where this time $\operatorname{supp}(U)=\{i\in I:U_i\ne A_i\}$; then
$$\tau'=\left\{\bigcup\mathscr{V}:\mathscr{V}\subseteq\mathscr{B}'\right\}\;.$$
That gets all of the relavant definitions out of the way. Once you understand them, the proof itself is quite straightforward.


*

*Prove that $\{A\cap U:U\in\mathscr{B}\}$ is a base for $\tau_A$.  

*Then prove that $\{A\cap U:U\in\mathscr{B}\}=\mathscr{B}'$. Conclude that $\tau_A=\tau'$.



This exercise is really about two things: understanding the product topology, and understanding that it’s often sufficient (and easier!) to work with bases for topologies than with the topologies themselves.
A: Recall that the product topology on $\prod_{i}X_i$ has as a basis the collection of all sets of the form  $\prod_{i}U_i$, with $U_i$ open in $X_i$ and $U_i\neq X_i$ for only finitely many $i$. It follows that the collection of all sets of the form $\big(\prod_{i}U_i\big)\cap\big(\prod_{i}A_i\big)=\prod_{i}(U_i\cap A_i)$, with $U_i$ open in $X_i$ and $U_i\neq X_i$ for only finitely many $i$,  is a basis for the subspace topology on $\prod_{i}A_i$. But this collection is precisely the collection all sets of the form $\prod_{i}U'_i$, with $U'_i$ relatively open in $A_i$ and $U'_i\neq A_i$ for only finitely many $i$, which a basis for the product topology on $\prod_{i}A_i$.
