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Let $A=\{(x_1, \cdots, x_n)\in \mathbb{R}^n: x_1\ge 0,\|(x_1, \cdots, x_n)\|<1\}$. I want to show that this is not homeomorphic to any open set of $\mathbb{R}^n$. I can use the theorem of invariance of domain, which states that:

If $U$ is open in $\mathbb{R}^n$ and $f:U\rightarrow \mathbb{R}^n$ is continuous and injective then $f(U)$ is open in $\mathbb{R}^n$.

But is there a way to see this without invariance of domain?

For example, if I want to show that $A$ is not homeomorphic to an open ball in $\mathbb{R}^n$ then I can just show that $A\setminus\{0\}$ is contractible so has trivial homology groups $H_n(A\setminus\{0\})$ for $n\ge 1$ but an open ball minus a point is homotopy equivalent to $S^{n-1}$, whose $n-1$th homology group is $\mathbb{Z}$. Can we do something similar to show $A$ is not homeomorphic to a general open subset of $\mathbb{R}^n$?

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You can upgrade your realization that $A\setminus\{0\}$ is contractible into a proof that $A$ is not homeomorphic to any open set $W\subseteq \mathbb{R}^k$.

First, since $A$ is connected, we may as well assume wlog that $W$ itself is connected. Now we have the key lemma:

If $W$ is a connected open set, then for any two points in $W$, there is a homeomorphism of $W$ mapping the first point to the second.

Proof: First, note that $W$ is clearly locally path connected, hence path connected. Now, given $p$, $q \in W$, choose a smooth curve $\gamma$ connecting $p$ and $q$. We may assume wlog that this smooth curve has no self intersections. By compactness of $\gamma$, there is an $r >0$ so that if $U$ denotes the open tubular neighborhood of radius $r$ about $\gamma$, then $\overline{U}\subseteq W$. Now, take the tangent field of $\gamma$ and extend it to a vector field on $W$ which vanishes outside of $U$. Then the flow along this vector field exists for all time (because the flow is compactly supported) and flowing for a long enough time maps $p$ to $q$. $\square$

(In case you care, this actually proves the stronger statement that given any two points in $W$, there is an isotopy of $\mathbb{R}^k$ which preserves $W$ and maps $p$ to $q$.)

This is enough to show that $A$ is not homeomorphic to $W$. For in $W$, any point can be mapped homeomorphically onto any other. On the other hand, in $A$, $A\setminus \{0\}$ is contractible while if $p$ is an interior point of $A$, then $A\setminus \{p\}$ clearly deformation retracts onto a sphere, so is not contractible. This implies there is no homeomorphism of $A$ can $\{0\}$ to $p$.

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