[proofplan] We use the subbasis definition of the [product topology](/page/Product%20Topology). A subbasic [open set](/page/Open%20Set) in the product is the preimage of an open set in one factor under a coordinate projection. Since each factor topology is indiscrete, that open set is either empty or the whole factor, and its preimage is therefore either empty or the whole product. Hence the topology generated by these subbasic sets contains no open sets beyond $\varnothing$ and the whole product. [/proofplan] [step:Define the product and its coordinate subbasis] Let \begin{align*} P:=\prod_{i\in I}X_i \end{align*} denote the Cartesian product of the family $(X_i)_{i\in I}$. For each $i\in I$, let \begin{align*} \pi_i:P&\to X_i \end{align*} be the coordinate projection map. By definition, the product topology $\tau_{\mathrm{prod}}$ on $P$ is generated by the subbasis \begin{align*} \mathcal{S}:=\{\pi_i^{-1}(U): i\in I,\ U\in\tau_i\}. \end{align*} If $I=\varnothing$, then this is the empty subbasis, and the same generated-topology argument below applies. [/step] [step:Show every subbasic open set is either empty or the whole product] Fix an index $i\in I$ and an open set $U\in\tau_i$. Since $\tau_i=\{\varnothing,X_i\}$, either $U=\varnothing$ or $U=X_i$. If $U=\varnothing$, then \begin{align*} \pi_i^{-1}(U)=\pi_i^{-1}(\varnothing)=\varnothing. \end{align*} If $U=X_i$, then \begin{align*} \pi_i^{-1}(U)=\pi_i^{-1}(X_i)=P. \end{align*} Thus every member of $\mathcal{S}$ belongs to $\{\varnothing,P\}$. [guided] The product topology is built from coordinate information. A basic test for openness in the product topology begins by choosing one coordinate $i\in I$ and asking that the $i$-th coordinate lie in some open subset $U\subset X_i$. Formally, this test set is the preimage $\pi_i^{-1}(U)$ under the projection map \begin{align*} \pi_i:P&\to X_i. \end{align*} Now the key point is that the topology on $X_i$ is indiscrete. Therefore the only possible choices for the open set $U\in\tau_i$ are $U=\varnothing$ and $U=X_i$. If $U=\varnothing$, no point of $P$ can have its $i$-th coordinate in $U$, so \begin{align*} \pi_i^{-1}(\varnothing)=\varnothing. \end{align*} If $U=X_i$, every point of $P$ has its $i$-th coordinate in $X_i$, so \begin{align*} \pi_i^{-1}(X_i)=P. \end{align*} This also covers the case where $P=\varnothing$, because then both $\varnothing$ and $P$ are the same set. Therefore each subbasic open set used to generate the product topology is either $\varnothing$ or the whole product $P$. [/guided] [/step] [step:Conclude that the generated topology is indiscrete] We have shown that \begin{align*} \mathcal{S}\subset \{\varnothing,P\}. \end{align*} The collection $\{\varnothing,P\}$ is a topology on $P$, and it contains $\mathcal{S}$. Since $\tau_{\mathrm{prod}}$ is the smallest topology on $P$ containing $\mathcal{S}$, we have \begin{align*} \tau_{\mathrm{prod}}\subset \{\varnothing,P\}. \end{align*} Conversely, every topology on $P$ contains both $\varnothing$ and $P$, so \begin{align*} \{\varnothing,P\}\subset \tau_{\mathrm{prod}}. \end{align*} Therefore \begin{align*} \tau_{\mathrm{prod}}=\{\varnothing,P\}. \end{align*} Thus the product topology on $\prod_{i\in I}X_i$ is the [indiscrete topology](/page/Indiscrete%20Topology). [/step]