If $\prod (X_i,T_i)$ is connected, then each $(X_i, T_i)$ is connected.

Question

Let $(X_i,T_i)$ be a topological space and $i=1, 2, ..., n.$

If $\prod (X_i,T_i)$ is connected, then each $(X_i, T_i)$ is connected.

I attempt a proof by contrapositive.

Let $U_i$ and $V_i$ be two open sets in $T_i$ such that $U_i \cap V_i = \emptyset$ and $U_i \cup V_i = X_i.$

Then I have to find sets $\prod O_i$ and $\prod B_i$ such that $\prod O_i \cap \prod B_i = \emptyset$ and $\prod O_i \cup \prod B_i = \prod X_i$.

Since $\prod O_i \cap \prod B_i = \prod (O_i \cap B_i)$ and $U_i \cap V_i = \emptyset \Rightarrow \prod (O_i \cap B_i) = \emptyset$.

Next is where I am confused. $\prod O_i \cup \prod B_i \subseteq \prod O_i \cup B_i.$ So if $\prod O_i \cup \prod B_i = \prod X_i$ then $\prod O_i \cup B_i$ must $= \prod X_i$. But the only way for this to be true is if every $X_i$ is not connected. But then doesn't that change the quantifiers of the problem?

Answer 1

A much easier method would be using the projection map.

The image of a connected space under a continuous map is connected, and the projection maps are continuous.

Moreover, the image of $\pi_i: \prod X_i \to X_i$ is $X_i$. Hence, $X_i$ is connected.

Answer 2

Proof by Contrapositive: Suppose there is an $i_0$ such that $X_{i_0}$ is disconnected with separating open sets $A$ and $B$. Now for all $i$ in the indexing set we define the following sets:

\begin{align*} U_i&:=\begin{cases}X_i&\text{if }i\neq i_0\\A&\text{if }i=i_0\end{cases} & V_i&:=\begin{cases}X_i&\text{if }i\neq i_0\\B&\text{if }i=i_0\end{cases} \end{align*}

Now define $U:=\prod_{i}U_i$ and $V:=\prod_{i}V_i$.

It is easy to check that $U$ and $V$ are separating open sets for $\prod_iX_i$ in the product topology.