# Can we show the product of two connected topological space is connected by a proof by contradiction?

Most proofs that shows the product of two connected spaces are connected uses the theorem that a space is connected iff every continuous function to {0,1} is constant or it uses the horizontal or vertical slice trick and take the union of all slices.

I was wondering if there is a "direct" proof by contradiction possible. If $$X \times Y$$ is the product of two connected space, and we assume $$X \times Y$$, then there is a disconnection possible: we can write it as a union of two disjoint non-empty open sets. Can we then make use of this to prove that either $$X$$ or $$Y$$ has a disconnection? Basically we can express $$U$$ and $$V$$ as unions as basic open sets in the product topology.

• @FSrike that is not true, there are plenty of easy counterexamples of disconected sets in $\mathbb{R}^2$ whose projections are connected. Commented Apr 2, 2022 at 14:09
• @FShrike for some reason, i thought we needed to express U and V as basic open sets first... but I just remembered the projection is itself a open map. Commented Apr 2, 2022 at 14:11
• @Marcos Apologies. I am right in thinking however that they cannot possibly have connected projections in all coordinates? Commented Apr 2, 2022 at 14:13
• @FShrike No, take two concentric (disjoint) circles in $\Bbb R^2$, their projection is always a segment Commented Apr 2, 2022 at 14:14
• @AlessandroCodenotti Wow my intuition is really off today.. apologies again and thank you for the example Commented Apr 2, 2022 at 14:15

I don't think there is a direct proof by contradiction as you want. Why? Because if we say $$X\times Y$$ is not connected, then there exists open sets $$U,V$$ such that $$U\cap V=\emptyset$$ and $$X\times Y=U\cup V$$. If we can say that $$U=U_X\times U_Y$$ and $$V=V_X\times V_Y$$ then we will be done. But sadlly, this is not true, since every subset of a cartesian product need not be equal to a cartesian product. Of course it is if we consider basis elements, but we don't have any reason tho think that $$U$$ and $$V$$ belong to the basis.
• I thought it might have been possible to write $U=\bigcup (A_i \times B_i)$ and $V=\bigcup (A_j \times B_j)$ and then we have $\bigcup (A_i \times B_i) \bigcap \bigcup(A_j \times B_j)=\emptyset$ Is there any set manipulation we can do to show that $X=(\bigcup A_i) \bigcup (\bigcup A_j)$ With this, we can express X as the union of two disconnected open set. Commented Apr 2, 2022 at 14:33
• The problem here is that $\bigcup A_i$ and $\bigcup A_j$ need not be disjoint but $\bigcup (A_i \times B_i) \bigcap \bigcup(A_j \times B_j)=\emptyset$. Same happens with $Y$. Commented Apr 2, 2022 at 14:48