# the real numbers are indecomposable as a topological space

Let $\mathbb{R}$ be the set of real numbers with standard topology. We wish to show that if $\mathbb{R}$ is homeomorphic to $X\times Y$ for some topological spaces $X$ and $Y$, then either $X$ is one point space or $Y$ is one point space. First let us prove a lemma:

Lemma. Let $X$ and $Y$ be path connected topological spaces such that cardinality of both $X$ and $Y$ is at least $2$. Then for any point $(x_{0},y_{0})\in X\times Y$ the space $X\times Y\setminus\{(x_{0},y_{0})\}$ with subspace topology is path connected.

Proof. Let $x^{\prime}\in X$ and $y^{\prime}\in Y$ such that $x^{\prime}\neq x_{0}$ and $y^{\prime}\neq y_{0}$ (we assumed that such points exist). It is sufficient to show that for any point $(x_{1},y_{1})$ from $X\times Y\setminus\{(x_{0},y_{0})\}$ there exists a continous map $\sigma:\mathrm{I}\to X\times Y$ such that $\sigma(0)=(x_{1},y_{1})$, $\sigma(1)=(x^{\prime},y^{\prime})$ and $(x_{0},y_{0})\not\in\sigma(\mathrm{I})$.

Let $(x_{1},y_{1})\in X\times Y\setminus\{(x_{0},y_{0})\}$. Therefore either $x_{1}\neq x_{0}$ or $y_{1}\neq y_{0}$. Assume that $y_{1}\neq y_{0}$ (the other case is analogous). Choose paths $\sigma:\mathrm{I}\to X$ from $x_{1}$ to $x^{\prime}$ and $\tau:\mathrm{I}\to Y$ from $y_{1}$ to $y^{\prime}$. Then we have induced paths:

 $\sigma^{\prime}:\mathrm{I}\to X\times Y\ \ \mathrm{such}\ \mathrm{that}\ % \sigma^{\prime}(t)=(\sigma(t),y_{1});$
 $\tau^{\prime}:\mathrm{I}\to X\times Y\ \ \mathrm{such}\ \mathrm{that}\ \tau^{% \prime}(t)=(x^{\prime},\tau(t)).$

Then the path $\sigma^{\prime}*\tau^{\prime}:\mathrm{I}\to X\times Y$ defined by the formula

 $(\sigma^{\prime}*\tau^{\prime})(t)=\begin{cases}\sigma^{\prime}(2t)&\mathrm{% when}\ \ 0\leq t\leq\frac{1}{2}\\ \tau^{\prime}(2t-1)&\mathrm{when}\ \ \frac{1}{2}\leq t\leq 1\end{cases}$

is a desired path. $\square$

. If there exist topological spaces $X$ and $Y$ such that $\mathbb{R}$ is homeomorphic to $X\times Y$, then either $X$ has exactly one point or $Y$ has exactly one point.

Proof. Assume that neither $X$ nor $Y$ has exactly one point. Now $X\times Y$ is path connected since it is homeomorphic to $\mathbb{R}$, so it is well known that both $X$ and $Y$ have to be path connected (please see this entry (http://planetmath.org/ProductOfPathConnectedSpacesIsPathConnected) for more details). Therefore for any point $(x,y)\in X\times Y$ the space $X\times Y\setminus\{(x,y)\}$ is also path connected (due to lemma), but there exists a real number $r\in\mathbb{R}$ such that $X\times Y\setminus\{(x,y)\}$ is homeomorphic to $\mathbb{R}\setminus\{r\}$. Contradiction, since $\mathbb{R}\setminus\{r\}$ is not path connected. $\square$

Title the real numbers are indecomposable as a topological space TheRealNumbersAreIndecomposableAsATopologicalSpace 2013-03-22 18:30:59 2013-03-22 18:30:59 joking (16130) joking (16130) 12 joking (16130) Theorem msc 54F99