Looking for a part of a proof involving connected spaces

Question

I am looking for the proof of the following theorem:

If $X_1 \times ... \times X_n$ is a connected space $\iff$ $X_i$ is a connected space $\forall i$

I got one of the implications ($\implies$) But I am looking for a proof to the other implication

Answer 1

Suppse $X_1$ and $X_2$ are connected spaces; I have to show that $X_1\times X_2$ is connected. It suffices to show that any two points of $X_1\times X_2$ lie in the same connected component. Consider two points $(x_1,x_2),\ (y_1,y_2)\in X_1\times X_2.$ Now $\{x_1\}\times X_2$ is connected (since it's homeomorphic to $X_2$), and $X_1\times\{y_2\}$ is connected (since it's homeomorphic to $X_1$), and $(\{x_1\}\times X_2)\cap(X_1\times\{y_2\})$ is nonempty (since it contains the point $(x_1,y_2)$), so $(\{x_1\}\times X_2)\cup(X_1\times\{y_2\})$ is a connected set containing $(x_1,x_2)$ and $(y_1,y_2).$

This proves that the product of two connected spaces is connected. It follows by induction that the product of any finite number of connected spaces is connected. To show that an arbitrary product of connected spaces is connected, choose a base point $b$ and show that the set of points differing from $b$ in only finitely many coordinates is connected and dense.

Answer 2

Assume $\prod_{i\in I} X_i=U\cup V$ where $U,V$ are disjoint open and assume $U\ne\emptyset$. We want to show $V=\emptyset$ (or $\prod_{i\in I} X_i=U$). Recall that a basis for the product consists of open sets of the form $\prod_{i\in I}U_i$ where $U_i=X_i$ for all but finitely many $i$. Among all non-empty open subsets of this form contained in $U$, pick one $\prod_{i\in I}U_i$ that minimizes the number of $i$ with $U_i\ne X_i$. As it is not empty, pick $(x_i)_{i\in I}\in \prod_{i\in I}U_i$. Consider any $j\in I$. Then the map $f\colon X_j\to \prod_{i\in I} X_i$, where $\pi_j(f(x))=x$ and $\pi_i(f(x))=x_i$ otherwise, is continuous. We can write $X_j=f^{-1}(U)\cup f^{-1}(V)$ as disjoint union of open sets. As $x_j\in f^{-1}(U)$, we conclude $f^{-1}(U)=X_j$. But tat means that we can replace $U_j$ with $X_j$ in $\prod_{i\in I}U_i$ and still have a subset of $U$. After finitely many such steps, we arrive at $\prod_{i\in I}X_i\subseteq U$, as desired.