A somewhat general question about homeomorphisms. I have been asked to prove that $(0,1)$ is homeomorphic to $(0,1)$. Seems easy enough. If we assume the order topology an both, along with an identity mapping $f:x\to x$, we can show that both $f$ and $f^{-1}$ are continuous.  
Similarly, using the identity mapping $f$ and the order topology, we can show that $(0,1)$ is not homeomorphic to $[0,1]$. However, how can we prove that there exists no toplogy and suitable mapping such that $(0,1)$ can be proven to be homeomorphic to $[0,1]$?
Thanks in advance!

Well $(0,1)$ and $[0,1]$ have the same cardinality, so there exists a bijection $\Phi:(0,1)\rightarrow[0,1]$. Now, equip $(0,1)$ with any topology and define $U\subset [0,1]$ to be open if and only if $\Phi^{-1}(U)$ is open. Then it can easily be checked that $\Phi$ is a homeomorphism.
