Any two closed and bounded intervals are homeomorhpic in $\mathbb{R}$:
If we want to show that the two sets $[a,b]$ and $[a_1,b_1]$ are homeomorphic we can consider the following map $f(x) \mapsto \frac{(x-a)}{b-a}(b_1-a_1) + a_1$. This map is continuous and bijective and the inverse of the map is also continuous.
However here are the follow up questions I have asked myself and I am stuck with –
1)What if the intervals are not bounded – Is $[a,b]$ hoemeomorphic to $[a_1,\infty)$?
I don't think it would hold because $[0,1]$ is probably not homeomorphic to $[1,\infty)$.
2)Is there a way to generalize this in $\mathbb{R}^n$– Any set in $\mathbb{R}^n$ (under the product metric ) is closed iff it is of the form $A_1 \times \cdots \times A_n$ where each $A_i$ is closed. Assume that each $A_i,B_i$ is closed and bounded then is $A_1 \times \cdots \times A_n$ homeomorphic to $B_1 \times \cdots \times B_n$?
I think this would hold true because $A_i$ and $B_i$ are closed and bounded in $\mathbb{R}$ and let us consider $f_i:A_i \mapsto B_i$ where $f_i$ is a homeomorphism. Then $f(x) = (f_1,\cdots f_n(x))$ might be the homeomorphism we are looking for.
Best Answer
$[0,1]$ is not homeomorphic to $[0,\infty)$ because the former has two non-cutpoints and the latter just one. Or the former is compact and the latter is not. So you cannot just blithely extend...
And in $\Bbb R^n$ it's very definitely not true that closed sets must be of the form $A_1 \times A_2 \ldots \times A_n$ as standard closed sets of the plane (lines, the $xy$-axes, graphs and circles...) already show. Such sets are closed (when all $A_i$ are) but the variation in more dimensions is immense. So get that notion out of your head ASAP!