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We denote the n-dimensional real euclidean space as $\mathbb{R}^n$ and the unit sphere in $\mathbb{R}^{n+1}$ as $S^n$.

I have read this example where we show that $\mathbb{R}^n$ is a CW complex by saying that the 0-cells are points with integer coordinates with n-cells being cubes with vertices as these points.

Also, $S^n$ is a CW complex by choosing a point to be 0-cell and the complement of it to be the n-cell.

My question is how do these 'decompostions' make these spaces into CW complexes - don't they need to be defined inductively? What happened to the attaching maps at the levels between 0 and n?

I have seen an inductive 'decomposition' of $S^n$ that clearly shows that it is a CW complex. However, it is the above 'decomposition' that I am not sure about. Also, I have no clue about inductive 'decomposition' of $\mathbb{R}^n$.

Any comments would be helpful.

Thanks.

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    They are constructing them inductively, they're just not writing it out in full detail. – 2017-01-18
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    What are the attaching maps? Are they trivial? – 2017-01-18
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    The $0$-skeleton is the set of points with integer coordinates. The attaching map for a $1$-cell attaches one end to a $0$-cell and another end to an adjacent $0$-cell. The attaching map for a $2$-cell is to attach its boundary to the square formed by 4 $1$-cells. – 2017-01-18
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    In case of $S^n$, we begin with two points as 0-cells, attach two intervals to obtain $S^1$, then attach two discs to obtain $S^2$ and continue this process. In this process, clearly, $S^n$ contains exactly one copy of $S^0, S^1,..., S^{n-1}$. But can we say such a thing for $\mathbb{R}^n$ : that is whether it contains exactly one copy of $\mathbb{R}, ..., \mathbb{R}^{n-1}$? – 2017-01-18
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    The $S^n$ example you read did it this way: we start with a single $0$-cell, and then we don't attach any $k$-cells for $0$n$-cell using the only boundary map we can. They both work just fine. – 2017-01-18

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