# Every n-cell of an n-dimensional CW complex is open.

I am confused about a step in the following proof:

Proposition: Every $n$-cell of an $n$-dimensional CW complex $X$ is open.

Proof: Suppose $e_0$ is an open $n$-cell of $X$, $e$ is any other (i.e. $e \neq e_0$) cell of $X$, then $e_0 \cap e = \emptyset$, so $e_0 \cap \bar e$ is contained in $\bar e\setminus e$, 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 $e_0 \cap \bar e = \emptyset$.

Why must $\bar e \setminus e$ contained in the union of finitely many cells of dimension less than $n$? It seems intuitive that the boundary of an n-cell must has dimension less than its dimension, however I cannot figure out a way to prove this.

(I am using the definition of CW complex from page 132 of Lee's Introduction to Topological Manifolds.)

This is part of the definition of a CW-complex. Specifically, following Lee's definition, the definition of a cell decomposition of a space requires that the characteristic map $\Phi:D\to X$ of an $m$-cell maps $\partial D$ to the union of all the cells of dimension strictly less than $m$. In your case, this means that since the dimension of $e$ is at most $n$, $\bar{e}\setminus e$ (which is equal to the image of $\partial D$ under the characteristic map of $e$) is contained in the union of all the cells of dimension strictly less than $n$. The fact that you need only finitely many cells of smaller dimension to cover $\bar{e}\setminus e$ follows from the closure-finite axiom for CW complexes.
• Eric, i have some confusion why the dimension of $e_0$ is necessary to conclude that $e_0 \cap \bar{e} = \emptyset$. My guess is that if $\bar{e}\smallsetminus e$ is contain in a union of many cells of dimension less than or equal to $n$ (say $e_i, i=1,\dots,k$ althogh i know this may not finite), then we cannot conclude $e_0 \cap \bar{e} = \emptyset$ because $e_0$ itself probably in the set $\{e_i\}$ ? – Sou Apr 15 '18 at 20:23
• The cells $e_i$ have dimension less than $n$ (not less than or equal to $n$). So $e_0$ cannot be one of them. – Eric Wofsey Apr 15 '18 at 20:30
• Yes. But why dimension $e_0$ is $n$ imply that $e_0\cap \bar{e} = \emptyset$ ? – Sou Apr 15 '18 at 20:32
• Because $e_0$ does not intersect $e$, and it also does not intersect $\bar{e}\setminus e$. – Eric Wofsey Apr 15 '18 at 20:36