# 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 ${x}^{\prime}\ne {x}_{0}$ and ${y}^{\prime}\ne {y}_{0}$ (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 $({x}_{0},{y}_{0})\notin \sigma (\mathrm{I})$.

Let $({x}_{1},{y}_{1})\in X\times Y\setminus \{({x}_{0},{y}_{0})\}$. Therefore either ${x}_{1}\ne {x}_{0}$ or ${y}_{1}\ne {y}_{0}$. Assume that ${y}_{1}\ne {y}_{0}$ (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:

$${\sigma}^{\prime}:\mathrm{I}\to X\times Y\mathit{\hspace{1em}}\mathrm{such}\mathrm{that}{\sigma}^{\prime}(t)=(\sigma (t),{y}_{1});$$ |

$${\tau}^{\prime}:\mathrm{I}\to X\times Y\mathit{\hspace{1em}}\mathrm{such}\mathrm{that}{\tau}^{\prime}(t)=({x}^{\prime},\tau (t)).$$ |

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

$$({\sigma}^{\prime}*{\tau}^{\prime})(t)=\{\begin{array}{cc}{\sigma}^{\prime}(2t)\hfill & \mathrm{when}\mathrm{\hspace{0.25em}\hspace{0.25em}0}\le t\le \frac{1}{2}\hfill \\ {\tau}^{\prime}(2t-1)\hfill & \mathrm{when}\hspace{1em}\frac{1}{2}\le t\le 1\hfill \end{array}$$ |

is a desired path. $\mathrm{\square}$

Proposition^{}. 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. $\mathrm{\square}$

Title | the real numbers are indecomposable as a topological space |
---|---|

Canonical name | TheRealNumbersAreIndecomposableAsATopologicalSpace |

Date of creation | 2013-03-22 18:30:59 |

Last modified on | 2013-03-22 18:30:59 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 12 |

Author | joking (16130) |

Entry type | Theorem |

Classification | msc 54F99 |