Topology/Product Spaces

Let $\{(X_i,{\tau }_i){\}}_{i\in I}$ be an indexed collection of topological spaces and let $X=\prod _{i\in I}{X}_i$. For each $i\in I$ let ${\pi }_i:X\to X_i$ be the $i$th coordinate projection. The product topology is the topology $\tau$ on $X$ generated by sets of the form ${\pi }_i^{-1}[U_i]$ where $i\in I$ and $U_i\in {\tau }_i$.

Theorem: If two topological spaces are compact, then their product space is also compact.
Proof: Let X₁ and X₂ be two compact spaces. Let S be a cover of X₁×X₂. Let x be an element of X₁. Consider the sets A_x,y within S that contain (x,y) for each y in X₂. ${\pi }_2:(A_{(}x,y))$ forms a cover for X₂, with a finite subcover {A_x,yi}. Let B_x be the intersection of ${\pi }_1:(A_y)$ within {A_yi}, which is open. Thus, {B_x} forms an open cover, which has a finite subcover, {B_xi}. The corresponding sets {A_xi,yi} is finite, and forms an open subcover of the set.

Theorem: If two topological spaces are connected, then their product space is also connected.
Proof: Let X₁ and X₂ be two connected spaces. Suppose that there are two nonempty open disjoint sets A and B whose union is X₁×X₂. If for every x∈X, {x}×X₂ is either completely within A or within B, then π₁(A) and π₁(B) are also open, and are thus disjoint and nonempty, whose union is X₁, contradicting the fact that X₁ is connected. Thus, there is an x∈X such that {x}×X₂ contains elements of both A and B. Then π₂(A∩{(x,y)}) and π₂(B∩{(x,y)}), where y is any element of X₂, are nonempty disjoint sets whose union is X₂, and which are are a union of open sets in {(x,y)} (by the definition of product topology), and are thus open. This implies that X₂ is disconnected, a contradiction. Thus, X₁×X₂ is connected.