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Is there an obvious/natural/simple CW complex structure for punctured Euclidean space, $\mathbb{R}^n \setminus \{0\}$?

EDIT: Here's what I've thought of. I can build $\mathbb{R}^n$ out of countably many cubes $e^n = [0,1]^n$ and attaching them along faces. Maybe I can build the interval $(0,1]$ as a CW complex by attaching infinitely many 1-cells $\{[1/(k+1),1/k]\}$ for $k \in \mathbb{N} \cup \{0\}$? I'm not sure if this is a CW complex. If it is, perhaps it can be generalized to higher dimensions.

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    What are your own thoughts about this? – 2017-02-11
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    Since Euclidean space is homeomorphic to the ball, you can not stress about countably many cubes, and just put a CW structure on the punctures ball. If you manage to do this, everything you could want out of a CW complex would be preserved by the homeomorphism between the two. – 2017-02-11
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    Are you saying that my guess can be generalized to higher dimensions? Would either of you mind posting a fleshed out answer? – 2017-02-11
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    OK, good job on $R^N$. Do you know that $R^n -\{0\}$ is homeomorphic to $R\times S^{n-1}$? Can you give $S^k$ a structure of a CW complex? – 2017-02-11
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    Yes I know how to put a CW complex structure on the $n$-sphere. Is there a CW complex structure for the punctured plane that has finitely many cells? – 2017-02-12
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    Any CW complex with finitely many chains would have to be compact (the image of a compact space under a continuous map is compact). – 2017-02-13
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    @MoisheCohen As a consequence of your statement, punctured Euclidean space cannot be written as a CW complex with finitely many cells? I'm not familiar with the the meaning of a "chain." – 2017-02-13
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    Sorry, I meant "cells", not "chains". Other than that, yes, you need infinitely many cells once your space is noncompact. – 2017-02-13

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