Prove that the product $ X = \Pi_{i \in I} X_{i} $ is discrete $\iff$ $I$ is finite Let  $(X_{i} \colon i \in I ) $ be a family of discrete topological spaces so that $card(X_{i}) \geq 2\  \forall i \in I$ . Prove that the product $ X = \Pi_{i \in I} X_{i}  $ is discrete $\iff$ $I$ is finite . 
I know that you will think this is a duplicate but I am looking for the proof for : if the product $ X = \Pi_{i \in I} X_{i}  $ is discrete then $I$ is finite. The other direction is easy.
 A: General fact about product spaces:
A base for $\prod_{i \in I} X_i$ in the product topology, is all sets of the form $\prod_{ i \in I} O_i$ where all $O_i$ are open in $X_i$ and such that there is a finite subset $F$ of $I$ such that $i \notin F \implies O_i=X_i$.
Now, if $X=\prod_{ i \in I} X_i$ is discrete, pick $x_i \neq y_i$ distinct points in $X_i$, which can be done by assumption. Then define $x=(x_i)_{i \in I} \in X$ and by assumption $\{x\}$ is open in $X$. So there is a basic open subset, as described above, of the form $O=\prod_{i \in I} O_i$ such $x \in O \subseteq \{x\}$, so that $\prod_{i \in I} O_i = \{x\}$. Now if $F \neq I$, then pick an index $i_0 \in I\setminus F$ and then $O_{i_0}=X_{i_0}$ by definition of the basis, and so $y_{i_0} \in \pi_{i_0}[O]$ but $y_{i_0} \notin \pi_{i_0}[\{x\}]=\{x_{i_0}\}$, contradiction. So $I=F$ and so $I$ is finite. 
A: Let $\pi_j:\prod_{i\in I}X_i\to X_j$ denote the projection for $j\in I$.
Let it be that $i_1,\dots,i_k\in I$ and we have sets $U_j$ open in $X_j$ such that $\pi_{i_1}^{-1}(U_1)\cap\cdots\cap\pi_{i_k}^{-1}(U_k)=\{x\}=\{(x_i)_{i\in I}\}$.
We claim that: $I=\{i_1,\dots,i_k\}$.
Proof: if $j\in I-\{i_1,\dots,i_k\}$  then we can choose $y_j\in X_j-\{x_j\}$ and also have $(y_i)_{i\in J}\in\pi_{i_1}^{-1}(U_1)\cap\cdots\cap\pi_{i_k}^{-1}(U_k)$ where $y_i=x_i$ if $i\neq j$ contradicting that $\pi_{i_1}^{-1}(U_1)\cap\cdots\cap\pi_{i_k}^{-1}(U_k)$ is a singleton.
If $\prod_{i\in I}X_i$ has discrete topology then every singleton is open, hence can be written as $\pi_{i_1}^{-1}(U_1)\cap\cdots\cap\pi_{i_k}^{-1}(U_k)$ because these sets form a basis of the product topology.
