# Why is closure finiteness necessary in this proof that every $n$-cell of an $n$-dimensional CW complex is open?

I am reading John Lee's Introduction to Topological Manifolds and I have a question about the proof of Proposition 5.5, which states that every $n$-cell of an $n$-dimensional CW complex $X$ is an open subset of $X$. The proof is as follows:

Suppose $e_0$ is an $n$-cell of $X$. If $\Phi\colon D\to X$ is a characteristic map for $e_0$, then $\Phi$, considered as a map onto $\overline{e_0}$, is a quotient map by the closed map lemma. Since $\Phi^{-1}(e_0)=\operatorname{Int}D$ is open in $D$, it follows that $e_0$ is open in $\overline{e_0}$. On the other hand, if $e$ is any other cell of $X$, then $e_0\cap e=\varnothing$, so $e_0\cap\overline{e}$ is contained in $\overline{e}\setminus e$, which in turn is contained in a union of finitely many cells of dimension less than $n$. Since $e_0$ has dimension $n$, it follows that $e_0\cap\overline{e}=\varnothing$. Thus the intersection of $e_0$ with the closure of every cell is open, so $e_0$ is open in $X$ by condition (W).

I don't understand why closure finiteness is necessary here. Is it not enough to just say that $\overline{e}\setminus e$ is contained in the union of all cells of dimension less than $n$?

## 1 Answer

Yeah, you seem to be right. I'm not sure why I mentioned "finitely many" in the proof; maybe I was just thinking about relating this fact directly back to the definition of a CW complex. But in fact, as you pointed out, the proof works without the closure-finiteness condition. (It does, however, require condition (W), the coherent topology condition.)

• I see, thanks. By the way, your book is great! – Zilliput Dec 19 '16 at 3:24
• @Zilliput: Thank you very much! – Jack Lee Dec 19 '16 at 21:07