A set $X \subset \mathbb{R}^{n}$ is said topologically homogeneous when given $a,b \in X$ there is a homoemorphism such that $h(a) = b$. Prove that, for all $n \in \mathbb{N}$, the space $\mathbb{R}^{n}$ and the sphere $S^{n}$ are topologically homogeneous. Show that $[0,1]$ is not topologically homogeneous.
I know that if $M, N$ are homeomorphic metric spaces, then $M$ is topologically homogeneous iff N is too. Maybe it help.
For $\mathbb{R}^{n}$, I think in any function of the form $x \to x + \alpha$ with $\alpha$ fixed, since are Lipschitz functions.
For $S^{n}$, maybe an aplication that change the base, but I don't know to explain this function.
For $[0,1]$ I don't have any idea.
Hint on $X=[0,1]$: Both $X\backslash\{0\}$ and $X\backslash\{1\}$ are connected. Is this true for any other point of $X$?
Consider $S^n$ to be the subset of $\mathbb{R}^{n+1}$ consisting of all points a unit distance from the origin of $\mathbb{R}^{n+1}$. Let $P,Q\in S^n$. In $R^{n+1}$ consider the intersection of the plane $OPQ$ ($O$ being the origin of $\mathbb{R}^{n+1}$) and $S^n$. The intersection is a circle. Rotate $\mathbb{R}^{n+1}$ about the line through $O$ with direction vector $\vec{OP}\times\vec{OQ}$ by a sufficient angle to move $P$ to $Q$. Restricted to $S^n$ this will be a homeomorphism mapping $S^n$ onto $S^n$ taking $P$ onto $Q$.
For $\mathbb{R}^n$ a simple translation by $\vec{PQ}$ will do.