TITLE: The real line is not homeomorphic to any non-trivial product space
QUESTION [5 upvotes]: I came across a question that interested me recently. It asked the following:

Prove that if $\mathbb R$ is homeomorphic to $X \times Y$, then $X$ or $Y$ is a singleton set.

I have an easy proof using path-connectedness. I was interested if there is an even more elementary argument. The notion of connectedness had not yet been introduced in the text.

REPLY [2 votes]: If $X$ and $Y$ are connected topological spaces, each containing at least two points, then the product space $X\times Y$ has no cut point.
Proof. Consider any point $(a,b)\in X\times Y;$ I have to show that $X\times Y\setminus\{(a,b)\}$ is connected.
Choose $x_0\in X\setminus\{a\}$ and $y_0\in Y\setminus\{b\}.$ Now consider any point $(x,y)\ne(a,b);$ I will show that $(x,y)$ and $(x_0,y_0)$ are in the same component of $X\times Y\setminus\{(a,b)\}.$
Case I. If $x\ne a$ then $(\{x\}\times Y)\cup(X\times\{y_0\})$ is a connected subset of $X\times Y\setminus\{(a,b)\}$ containing $(x,y)$ and $(x_0,y_0).$
Case II. If $y\ne b$ then $(X\times\{y\})\cup(\{x_0\}\times Y)$ is a connected subset of $X\times Y\setminus\{(a,b)\}$ containing $(x,y)$ and $(x_0,y_0).$