Is $\mathbb{R}^n$ homeomorphic to a product $X \times Y$ with $X$ compact and not a point?


Bing's [Dogbone space](http://en.wikipedia.org/wiki/Dogbone_space) is a quotient of $\mathbb{R}^3$ with fibers points and arcs, and whose product with $\mathbb{R}$ is $\mathbb{R}^4$, so it doesn't seem to me to big a stretch to think that it may be possible. 


Or, is there a notion of dimension which takes care of it swiftly?



Does Euclidean space have a compact factor?
No it is not possible. Suppose that $X\times Y\cong\mathbb{R}^n$. Then, as the product is contractible, both $X$ and $Y$ must be contractible spaces. For any $x\in X$, I'll show that $\lbrace x\rbrace\times Y$ must be an open subset of $\mathbb{R}^n$, which will imply that $\lbrace x\rbrace$ is an open subset of $X$ and, hence, that $X$ is discrete. Discrete contractible spaces consist of a single point.


Choose any $p=(x,y)\in X\times Y$. We just need to show that this is contained in the interior of $\lbrace x\rbrace\times Y$.
As the spaces are contractible, there are deformation retractions $H\_X\colon X\times[0,1]\to X$ and $H\_Y\colon Y\times[0,1]\to Y$ respectively to the points $x,y$. So, $H\_X(u,0)=u$, $H\_X(u,1)=x$, $H\_Y(v,0)=v$, $H\_Y(v,1)=y$, for any $u\in X$ and $v\in Y$. Define the deformation retraction $J\colon(X\times Y)\times[0,1]\to X\times Y$ from $X\times Y$ to the point $p=(x,y)$ by
$$
J\left((u,v),t\right)=\begin{cases}
\left(H\_X(u,2t),v\right),&\textrm{if }t\le1/2,\cr
\left(x,H\_Y(v,2t-1)\right),&\textrm{if }t\ge1/2.
\end{cases}
$$


Identifying $X\times Y$ with $\mathbb{R}^n$, consider the (n-1)-sphere $S\_R=\lbrace a\in\mathbb{R}^n\colon\Vert a-p\Vert=R\rbrace$, for any fixed $R > 0$. As $K=X\times\lbrace y\rbrace$ is compact, it will have empty intersection with $S\_R$ so long as $R$ is chosen large enough. However, retricted to $S\_R\times[0,1]$, $J$ continuously deforms $S\_R$ down to the single point $\lbrace p\rbrace$. This implies that $J(S\_R\times[0,1])$ contains the open ball of radius $R$ centered at $p$. As $S\_R\cap K=\emptyset$, $J(S\_R\times[0,1/2])$ is a compact set not containing $p$. So, $J(S\_R\times[1/2,1])\subset\lbrace x\rbrace\times Y$ contains a neighborhood of $p$, showing that $\lbrace x\rbrace\times Y$ is open in $\mathbb{R}^n$.


