Proof of topological isomorphism I remember reading in a section in plato.stanford.edu that the interval $(-∞, t)$ is topologically isomorphic to the interval $(0, t)$. I am not that good with topology, so could someone show me the proof, if this is true? 
 A: A picture homeomorphism, updated with animation

A: If $t>0$, then you can try the function $\phi:(-\infty,t)\to(0,t)$, defined as $$\phi(x)\equiv t\exp(x-t)\quad\forall x\in(-\infty,t).$$
Show that


*

*$\phi$ maps to $(0,t)$, indeed;

*$\phi$ is bijective (hint: it's strictly increasing);

*$\phi$ is continuous;

*the inverse function $\phi^{-1}:(0,t)\to(-\infty,t)$ exists and is continuous.


You will then have established that $\phi$ is a homeomorphism (“topological isomorphism,” if you will) between $(-\infty,t)$ and $(0,t)$.
A: I'm don't know topology, but I'm guessing that continuous real open intervals are homeomorphic if they are transformed by continuous, invertible functions over the domain [and that the inverse is continuous - thanks triple sec]. Is that right?
Some ideas here:
Homeomorphism of the real line-Topology
Here is my guess:
(0,t) f(x) = x/t => (0,1)
(0,1) f(x) = 1/x => ($∞$, 1) (1/x is continuous over that domain)
($∞$, 1) f(x) = -x => ($-∞$, -1) 
($-∞$, 1) f(x) = x + t + 1 => ($-∞$, t) 
