# Is the union of an increasing sequence of topological copies of $\mathbb{R}^n$ homeomorphic to $\mathbb{R}^n$?

Let $M$ be an $n$-dimensional topological manifold, and let $(U_k)_{k \in \mathbb{N}}$ be an increasing sequence of open sets $U_k \subset M$ such that for each $k \in \mathbb{N}$, $U_k$ is homeomorphic to $\mathbb{R}^n$. Is it necessarily the case that $\,U\!:=\bigcup_{k=1}^\infty U_k\,$ is homeomorphic to $\mathbb{R}^n$?

If so, does anyone know any nice reference for this fact (either as a theorem/lemma/etc. or as an exercise)?

• Cannot you write $\mathbb{R}^2 \setminus \{(0,0)\}$ as such a union? Nov 18, 2015 at 4:30
• @HennoBrandsma: No. In such a limit, $\pi_k(M) = \lim \pi_k(U_n)$, so $M$ is automatically contractible.
– user98602
Nov 18, 2015 at 4:34
• Because every curve lies entirely within one $U_k$ by compactness, I see. Nov 18, 2015 at 4:36
• Right (and every homotopy between such too). The claim in the OP should be true by the annulus theorem but I haven't checked the details.
– user98602
Nov 18, 2015 at 4:38

Yes, this is correct. You could prove it as a corollary of the annulus theorem (sketch of a sketch: at each stage the closure $\overline U_n$ is an open $n$-ball, and you're attaching an annulus, and the limit of this procedure is $\Bbb R^n$), but this wasn't known in all dimensions until the 80s. An early proof of this theorem was given in 1961 by Morton Brown here; it appears to use a modification of the above idea and fact, though I have not at all read it in detail. It appears that this was been known slightly earlier as a corollary of Mazur's proof of the topological Schoenflies theorem.
• As stated this proof is false since there's no reason the closure should be a ball (and obvious not an open one). But if the closure of each is contained in the next, and you have a homeomorphism $U_n \to \Bbb R^n$ you can pull back a sufficiently large closed ball in the latter and consider an increasing sequence of those and apply the annulus theorem.