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Are closed cells of a $CW$-complex always compact?

In my definition, a $CW$-complex is a Hausdorff space $X$ with a partition (the cells) with for each cell $C$ a map $\phi : D^n \to C$ ($n$ is the "dimension" of $C$) which is a homeomorphism from $\mathring{D^n}$ to $C$ (hence $n$ being unique) so that $\phi(S^{n-1})$ is contained in a finite union of cells of dimension $\leqslant n-1$, and equipped with the following topology on $X$ : $A$ is closed iff $A \cap \overline{C}$ is closed in $\overline{C}$ for each cell $C$.

Edit: my definition is flawed. I need to define $\phi : D^n \to X$ and not $C$.

Loulou
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  • What is your definition of a closed cell? The closure of a cell? – Kritiker der Elche Nov 08 '24 at 17:22
  • No, a cell that is closed in $X$. – Loulou Nov 08 '24 at 17:28
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    By that definition, no cell is closed in $X$. If $C$ is an $n$-cell then, letting $A = C$, the set $A \cap \overline C = C \cap \overline C = C$ is not closed in $\overline C$. – Lee Mosher Nov 08 '24 at 17:38
  • I do not see why. – Loulou Nov 08 '24 at 17:46
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    A subset $C$ of a topological space is closed if and only if it is equal to its closure $\overline C$. But $C \ne \overline C$. Therefore $C$ is not closed. – Lee Mosher Nov 08 '24 at 18:52
  • Where does your definition of a CW complex come from? A textbook? – Paul Frost Nov 08 '24 at 19:28
  • @LeeMosher But why $C \ne \overline{C}$ ? (You are going in circles.) – Loulou Nov 09 '24 at 08:03
  • Remark that $B$ is closed in $\overline{C}$ iff it is included in $\overline{C}$ and closed in $X$, since $\overline{C}$ is closed in $X$. – Loulou Nov 09 '24 at 08:09
  • You write that for each cell $C$ there is a map $\phi : D^n \to C$ which is homeomorphism from $\mathring{D^n}$ to $C$. Since $\mathring{D^n}$ is not compact, also $C$ is not compact. On the other hand $C = \phi(\mathring{D^n}) \subset \phi(D^n) \subset C$, thus $C =\phi(D^n)$ so that $C$ is compact. – Paul Frost Nov 09 '24 at 08:59
  • You should ask your teacher for clarification. – Paul Frost Nov 09 '24 at 09:05

2 Answers2

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The correct definition is this:

For each cell $C$ of $X$ there exists a map $\phi_C : D^n \to X$ such that $\phi_C$ maps $\mathring{D^n}$ homeomorphically onto $C$. The map $\phi_C$ is called the characteristic map of $C$.

Some authors call the cells of a CW-complex "open cells". However, this may lead to confusion because the cells are in general no open subspaces of $X$. The reason for this notation is that they homeomorpic to open disks.

A closed cell is usually defined as the closure $\overline C$ of a cell $C$.

  1. Each closed cell $\overline C$ is compact.

Proof. $\phi_C(D^n)$ is compact, it is closed in $X$. Since $C \subset \phi_C(D^n)$, we conclude that $\overline C \subset \phi_C(D^n)$. Hence $\overline C$ is a closed subset of a compact space, thus compact.

  1. $\overline C = \phi_C(D^n)$. Thus the closed cells are the images of the characteristic maps of the cells. [This result can be taken as an alternative definition of closed cells.]

Proof. We already know that $\overline C \subset \phi_C(D^n)$. Since $\phi_C$ is continuous, we conclude (see eg. here) $$\phi_C(D^n) = \phi_C(\overline{\mathring{D^n}})\subset \overline{\phi_C(\mathring{D^n})} = \overline C .$$

  1. For $n > 0$, the $n$-cells $C$ of $X$ are not closed in $X$. Thus $C \subsetneqq \overline C$.

Proof. $C \approx \mathring{D^n}$ is not compact. If $C$ were closed then it would be compact (as a closed subset of $\phi_C(D^n)$ which is compact), a contradiction.

Remark. The $0$-cells are closed one-point sets. This comes from the fact that $\mathring{D^0} = D^0$ which is a one-point space.

Paul Frost
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  • Did I make an error then and characteristic maps are not defined from $D¨n \to C$ ? – Loulou Nov 08 '24 at 17:50
  • @Loulou There is something wrong in your understanding of a cell. See the comments to your question. – Paul Frost Nov 08 '24 at 19:31
  • OK, so there is clearly a problem in my definition. Let replace $\phi : D^n \to C$ by $\phi : D^n \to X$. I have another question to see if my teacher's definition is definitely flawed: can we deduce that $\phi$ is a surjection from $D^n \to \overline{C}$? (in order to match Wikipedia's characterisation in : https://en.wikipedia.org/wiki/CW_complex). – Loulou Nov 09 '24 at 09:40
  • Or do I have to include it in the definition? – Loulou Nov 09 '24 at 13:06
  • Thank you, very interesting! – Loulou Nov 09 '24 at 22:22
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Contrary to what you said in the comments, the inequality $C \ne \overline C$ follows directly from the definition of a CW complex. Here's why.

Consider a CW complex $X$ and let $X^{(n)}$ denote its $n$-skeleton. By definition, the $n$-cells of $X$ are the path components of $X^{(n)} - X^{(n-1)}$. Also by definition, for each $n$-cell $C$ there exist a "characteristic function" $\chi_C : D^n \to X^{(n)}$ such that $\chi_C$ is continuous, $\chi_C(\text{interior}(D^n)) = C$, and $\chi_C(\partial D^n) \subset X^{(n-1)}$, where \begin{align*} D^n &= \{p \in \mathbb R^n \mid |p| \le 1\} \\ \text{interior}(D^n) &= \{p \in D^n \mid |p| < 1\} \\ \partial D^n &= \{p \in D^n \mid |p| = 1\} \end{align*} Notice that each $p \in \partial D^n$ is in the closure of $\text{interior}(D^n)$, thereore $\chi_C(p) \in X^{(n-1)}$ is in the closure of $C=\chi_C(\text{interior}(D^n))$, in other words $\chi_C(p) \in \overline C$. But $X^{(n-1)}$ and $C$ are disjoint so $\chi_C(p) \not\in \overline C$.

So, since $\chi_C(p) \in \overline C - C$ for each $p \in \partial D^n$, it follows that $\overline C \ne C$.

To address your question directly, one must do a bit more work to prove that $\overline C$ is actually equal to $\chi_C(D^n)$, by applying the characterization of closed subsets of $X$ which you have mentioned as part of the definition of a CW complex.

Since $D^n$ is compact, its continuous image $\overline C$ is therefore also compact, proving that closed cells are compact.

Lee Mosher
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