option: $X$ can't be homeomorphic to an open subset of $\mathbb R$ to be false Let $X$ be an infinite set(topological space) homeomorphic to $X\times X$.The book gives the option 
option:  $X$ can't be homeomorphic to an open subset of $\mathbb R$ 
to be false
I cant figure out why?Please help
 A: If $O$ is an open subset of $\mathbb{R}$, then you can show that $O \times O$ cannot be homeomorphic to $O$:
If $O$ is connected, then $O \times O$ is connected, and a general fact is that removing a point from a product set cannot disconnect the remaining points. I.e. $O \times O \setminus \{(x,y)\}$ is always connected. But if $O$ is open and connected in the reals, removing any point from it disconnects the remaining points. So $O$ cannot be homeomorphic to $O \times O$.
If $O$ is disconnected, and it has $k > 1$ components, where $k$ is finite, then $O \times O$ has $k^2 > k$ components and cannot be homeomorphic to $O$.
If $O$ is disconnected and has (open) connected components $O_1,O_2,\ldots$, then the connected components of $O \times O$ are exactly all sets $O_i \times O_j$, $i,j = 1,2,\ldots$. So if $h$ were a homeomorphism between $O \times O$ and $O$, for every $(i,j) \in \mathbb{N}^2$, $h[O_i \times O_j] = O_{k(i,j)}$ for some $k(i,j) \in \mathbb{N}$. This is because components map exactly to components under a homeomorphism.
But then the argument for connected open $O$ shows that the latter cannot happen.
A: Because, e.g., $(0,1)$ and $(0,1)\times(0,1)$ are not homeomorphic. More precisely: any open set in $\mathbb R$ is a countable union of disjoint open intervals. Arguments, that it is not homeomorphic to its (Cartesian) square are similar as in the case of an interval.
HINT for the rest of argumentation: 
A point can disconnect an interval, but not its Cartesian product.
