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I'm having a bit of difficulty understanding the following proof from Lee's An Introduction to Topological Manifolds:

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I don't understand the deduction that since $e_0\cap e \subseteq \overline{e}\setminus e$ and $\overline{e}\setminus e$ is contained in the union of finitely many cells of dimension less than $n$, that $e_0\cap e$ is empty since $e_0$ is an $n$-cell.

Why is this exactly?

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By defintion (the definition I learned) two cells of a CW complex are either disjoint or equal. Hence if you have cells of different dimensions they are disjoint. Since the author proved that $e_0\cap\bar{e}$ is contained in the finite union of cells of dimension less than the one of $e_0$, $e_0$ has empty intersection with each of them, hence with the union, hence the conclusion.

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Thomas has already explained the answer to your question. For my answer, it does not answer your question directly but it provides another reason why each $n$ -cell $e_\alpha^n$ is open in $X$.

You already know that your $n$ -cells are attached via a characteristic map $\varphi_\alpha : S^{n-1} \to Y$. $Y$ now is the space that when you attach $n$ -cells $e_\alpha^n$ you get $X$. Now each $e_\alpha^n$ has its characteristic map $\Phi_\alpha$ which is defined to be the composition

$$D_\alpha^n \hookrightarrow Y \sqcup D_\alpha^n \to Y \sqcup D_\alpha^n/\sim $$

where the quotient space is just $X$. Recall that $\sim$ is the smallest equivalence relation generated by the subset of all pairs $(x,\varphi(x))$ for $x \in S^{n-1}$ in

$$(Y \sqcup D_\alpha^n) \times (Y \sqcup D_\alpha^n).$$

Now it is not hard to see that a subset $A$ of $X$ is open iff $\Phi_\alpha^{-1}(A)$ is open in $D_\alpha^n$ for each characteristic map $\Phi_\alpha$. Now suppose you have an $n$ -cell $e_\alpha^n$. Then if $\alpha \neq \beta$, then it is clear that $\Phi^{-1}_\beta(e_\alpha^n ) = \emptyset$ that is open in $D_\beta^n$. Now if $\beta = \alpha$, this is trivial to see because we would have that $\Phi$ is a homeomorphism of the interior of $D_\alpha^n$ and $e_\alpha^n$.

Is it clear to you why an $n$ - cell is open now?

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  • $\begingroup$ Yes, thank you. $\endgroup$ – Holdsworth88 Aug 17 '12 at 14:17

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