Define product topology for subspace I need to define a product topology for subspace $A_{i}$ of $X_{i}$. 
We have $X:=\prod_{i\in I} X_{i}$ and $A:=\prod_{i\in I} A_{i}$ is the subspace of $X$.
How I can define a product topology for this A? How about the basis of $A$? What are there?
My first thoughts were that I need to take open set $V$, from $A$. So I have $V_{i}\subseteq A_{i}$, and $V_{i}=A_{i}$ for only finite number of $i\in I$.
Then I am in trouble. I don't know what to do with the finite number of $i$'s.
 A: Well, the $A_i$ have a topology they inherit from $X_i$ as a subspace. Open sets in $A_i$ are exactly those of the form $O \cap A_i$ where $O$ is open in $X_i$.
The product topology on $A = \prod_{i \in I} A_i$ is defined as usual: the basic open sets are those of the form $\prod_{i \in I} U_i$, where there exists a finite subset $F$ of $I$ such that for all $i \in I \setminus F$ we have that $U_i = A_i$ and for $i \in F$ we have that $U_i$ is open in $A_i$ so there exist open sets $O_i \subseteq X_i$ such that $O_i \cap A_i = U_i$.
This topology coincides with the subspace topology that $A$ inherits as subspace of $X = \prod_{i \in I} X_i$. This is clear as basic product elements on $A$ (as above,s ame notation) are of the form $A \cap \prod_i O_i$ where $O_i$ is defined as above for $i \in F$, so $O_i \cap A_i = U_i$ and all other $O_i = X_i$. 
Proof: $(x_i)_i \in A \cap \prod_i O_i$ iff for all $i$, $x_i \in A_i$ and also $x_i \in O_i$. For $i \notin F$ this is equivalent to just $x_i \in A_i$ and for $i \in F$ this is equivalent to $x_i \in A_i \cap O_i = U_i$. So always this is equivalent to $(x_i)_i \in \prod_i U_i$ as required.  
So basic elements in the product topology on $A$ are subspace open (w.r.t $X$). And in general, if $\mathcal{B}$ is a base for the topology on any space $X$, then $\mathcal{B}_A = \{B \cap A: B \in \mathcal{B} \}$ is a base for the subspace topology on a subspace $A$. And this base is just equal to the product base, when starting with the standard product base on $X$. So the topologies coincide.
