Intereting Posts

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Suppose $E$ is a simply-connected open subset of $\mathbb R^n$. Must there be a sequence of compact subsets $K_n$ such that $E = \bigcup_{n=1}^\infty K_n$, $K_n \subseteq K_{n+1}$ for all $n$, and each $K_n$ is simply connected? This is trivially true if $n=1$, and is true if $n=2$ by the Riemann mapping theorem, but topology can become counter-intuitive in higher dimensions.

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This is false in dimension 3, Whitehead manifold is an example.

One way to see it is to observe that simply connected compact 3-manifolds with connected nonempty boundary are balls, while Whitehead manifolds cannot be exhausted by balls.

Here is a more general theorem:

**Theorem.** Suppose that $M$ is a simply-connected 1-ended connected 3-dimensional manifold which is not *simply-connected at infinity* (e.g., Whitehead manifold). Then $M$ does not admit an exhaustion by simply-connected compact submanifolds with boundary.

Proof. The assumption that $M$ is not simply-connected at infinity means that there exists a compact subset $K\subset M$, such that for every compact subset $K’\subset M$ containing $K$, there exists a loop $\gamma\subset M – K’$ which does not contract in $M-K$. Suppose that such $M$ admits an exhaustion by simply-connected compact submanifolds with boundary $M_i$. Then the boundary of each $M_i$ is a disjoint union of 2-spheres. It follows that for each component $C_i$ of $M-M_i$, the boundary of $C_i$ is a 2-sphere. Therefore, $\pi_1(C_i)\to \pi_1(M)$ is injective. But this contradicts the property that $M$ is not simply-connected at infinity. qed

**Note.** In the solution I assumed that you are only asking about exhaustions by compact submanifolds with boundary. (The proof also works if you exhaust by simplicial subcomplexes.) Exhaustions by arbitrary compact subsets would be much more difficult to analyze and I am not sure how to solve the problem in this case.

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