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A CW complex $X$ is the union of open cells $\{e^i_\alpha\}$ and these open cells are disjoint subsets of $X$, so these open cells form a partition of $X$.

Now if we take the closed cells $\{\bar e^i_\alpha\}$, then if the number of cells is finite we can also write $X$ as a union of closed cells $$X=\bigcup_{i,\alpha}\bar e^i_\alpha$$ where the $\bar e^i_\alpha$ are not necessarily disjoint sets. In sum, the open cells form a partition of $X$ whereas the closed cells form a covering of $X$.

When the number of cells is infinite we don't have a covering of $X$ by closed cells, we only have that the union of closed cells is contained in $X$.

Is this correct?

palio
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    Every point of the CW complex is in an open cell, which is contained itself in a closed cell. – Angina Seng Oct 23 '18 at 19:38
  • So $X$ is at the same time the union of the open cells and also the union of the closed cells ? why do authors define CW complex as a union of open cells then, although most of the conditions are put on the closed cells: for example the closed cells meet only a finite number of cells, and also the weak topology on $X$ is relative to the closed cells ? – palio Oct 23 '18 at 21:02

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As Lord Shark the Unknown pointed out, you always have $X=\bigcup_{i,\alpha}\bar e^i_\alpha$ (simply because $e^i_\alpha \subset \bar e^i_\alpha$).

The set of open cells on its own does not determine the topology of the whole space $X$. The open cells are topologically fairly trivial as they are all homeomorphic to standard open balls in some Euclidean space, and elementary examples that non-homeomorphic spaces may be partitioned into "the same" collection of open cells (consider $X= [0,1]$ and $X' = S^1 \cup \{ 0 \}$ which both can be decomposed in two $0$-cells and one $1$-cell).

Thus we need to know how the open cells are pieced together, and that is why the closed cells enter the arena: Only the intersections $\bar e^i_\alpha \cap e^j_\beta = (\bar e^i_\alpha \setminus e^i_\alpha) \cap e^j_\beta$ are able to provide information about the topology of $X$, and for this purpose the characteristic maps $\phi^i_\alpha : (D^i,\mathring{D}^i) \to (\bar e^i_\alpha, e^i_\alpha)$ are needed.

In fact, there are various equivalent approaches to define CW complexes. See Show that any cell complex is homotopy equivalent to a CW complex and consult a textbook on that material.

Paul Frost
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  • is the inclusion $e^i_\alpha \subset \bar e^i_\alpha$ an inclusion of a subset into its closure ? If yes then don't we have that the closure of an open cell is not necessarily a closed cell ? so closed cells are not simply closures of open cells, Am i correct ? Also your answer seems to confirm that closed cells are important (maybe more important than the open cells if I understand you well) so again why all authors define $X$ as a union of open cells instead of a union of closed cells ? – palio Oct 23 '18 at 23:27
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    The open cells and the closed cells play different roles in the definition of a CW complex. For example, the topology of open cells is easily described, as one sees from the definition of a CW complex; but the topology of the closed cells is hard to describe. On the other hand, the topology on the CW complex itself is easily described using the closed cells, again as one sees from the definition of the CW complex; but the topology on the CW complex itself is hard (impossible?) to describe from just the open cells. – Lee Mosher Oct 23 '18 at 23:35
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    @palio I think we can say that the closed cells are more important. I recommend to consult the appendix of Hatcher, Allen. "Algebraic topology. 2002." Cambridge UP, Cambridge 606.9 (2002). http://pi.math.cornell.edu/~hatcher/AT/AT.pdf. It seems that it depends on the taste of authors whether open or closed cells are the primary constituents. Anyway, you need both to define CW complexes. Closed cells are in fact the closures of the open cells, but depending on the approach this might require a proof. The closed cells are quotient spaces of closed balls in some $\mathbb{R}^n$ – Paul Frost Oct 24 '18 at 08:04
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    in which points of the boundary $S^{n-1} \subset D^n$ are identified in a nontrivial way Therefore the closed calls are in general not simply connected. As an exmaple consider $S^1$ with one $0$- and one $1$-cell. – Paul Frost Oct 24 '18 at 08:08
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    @LeeMosher I think it is impossible to describe to topology of a CW complex just from the open cells. We definitely need information where we arrive if we "approach the boundary" of an open cell, and this is more than the open cells alone can tell us. – Paul Frost Oct 24 '18 at 08:29