\*\*Disclaimer:\*\*I posted the following question on [MSE](https://math.stackexchange.com/posts/3900271/edit), but since there were no answers. I'm migrating it here.


Let $Homeo\_0(\mathbb{R}^n)$ ($Homeo\_c(\mathbb{R}^n)$) be the space of all (compactly-supported) orientation-preserving homeomorphisms on $\mathbb{R}^n$ to itself equipped with the topology of compact convergence. Is $Homeo\_c(\mathbb{R}^n)$ dense in $Homeo\_0(\mathbb{R}^n)$?


I tried to use Kirby's local contractibility result, but I couldn't figure if this is correct.


**Note:** To avoid confusion, as in the comments, recall that a homeomorphism is said to be *compactly supported* if it is equal to the *identity map* outside some compact subset of its domain.  

See Definition 1.5 [of these nice notes.](https://math.berkeley.edu/%7Ekpmann/algdiff.pdf#page=1&zoom=auto,-101,792)



Density of compactly-supported homeomorphisms
I think this is true. It suffices to prove the


**Lemma**. Given an orientation-preserving homeomorphism $h$ of $\mathbb{R}^n$ there is a compactly-supported homeomorphsim $h\_1$ which agrees with $h$ on the unit ball.


By conjugating by dilations one can find $h\_r$'s that agree with $h$ on the ball of radius $r$, and this gives a sequence of compactly-supported homeomorphisms converging pointwise to $h$.


The lemma can be proved using the ideas in the proof of the Annulus Conjecture, but it is simpler here to deduce it from the Annulus Conjecture and the closely-related fact that $Homeo^+(S^{n-1})$ is path-connected.


*Proof*: Given $h$ consider the sphere $h(S^{n-1}) \subset \mathbb{R}^n$. This is compact so is contained inside a sphere $r S^{n-1}$ for some $r \gg 0$. Both these spheres are locally flat, so by the Annulus Theorem the region between them is homeomorphic to $[0,1] \times S^{n-1}$. Combined with $h\vert\_{D^n}$ this gives a homeomorphism
$$h': r D^n = D^n \cup [1,r] \times S^{n-1} \to r D^n = h(D^n) \cup \{annulus\}$$
which agrees with $h$ on the unit ball. On the boundary it induces an orientation-preserving homeomorphism of $rS^{n-1}$. Thisis topologically isotopic to the identity, and combined with $h'$ this gives a homeomorphism
$$h'' : (r+1)D^n = rD^n \cup [0,1] \times S^{n-1} \to (r+1)D^n = rD^n \cup [0,1] \times S^{n-1}$$
which agrees with $h$ on the unit ball and which is the identity on the boundary: extending by the identity gives the required $h\_1$.


