What are some of the various ways of proving that $(0,1)^3$ is not homeomorphic to $[0,3)^3$ using the fundamental group and homology groups? I feel like I have various ways of understanding why this is true in terms of point-set topology but I would like some help with developing an algebraic approach.
Both spaces are contractible, so the fundamental group won't be able to tell them apart. Maybe if i remove a certain point from both of them it will? For that matter, is it wrong for me to think of $[0,1)^3$ as a closed octant of $\mathbb{R}^3$? That's what it looks like to me, as I believe $[0,1) \cong [0, \infty)$.
Thanks in advance!
Best Answer
Hint: Note that $[0,3)^3 \setminus \{ (0,0,0) \}$ is contractible, but $(0,1)^3 \setminus \{ x_0 \}$ is not contractible for any $x_0 \in (0,1)^3$.