the real numbers are indecomposable as a topological space

Let $ℝ$ be the set of real numbers with standard topology. We wish to show that if $ℝ$ 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

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

Proposition. If there exist topological spaces $X$ and $Y$ such that $ℝ$ 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 $ℝ$, 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 $r\in ℝ$ such that $X\times Y\setminus \{(x,y)\}$ is homeomorphic to $ℝ\setminus \{r\}$. Contradiction, since $ℝ\setminus \{r\}$ 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