I'm interested in the definition that says that a CW complex is defined as a partition of open cells $\{e^i_{\alpha}\}$ verifying "some conditions"(see Munkres page 214).

We endow $X$ with a topology called the weak topology relative to the collection of closed cells $\{\bar e^i_\alpha\}$ defined as follows: a set $A$ of $X$ is closed in $X$ if and only if $A\cap \bar e^i_\alpha$ is closed in $\bar e^i_\alpha$ for each closed cell $\bar e^i_\alpha$.

In this topology what would be an open set? can we say $A$ is an open set in $X$ if and only if $A\cap \bar e^i_\alpha$ is open in $\bar e^i_\alpha$? I know that a set $A$ is open if and only if its complement is closed, but I wanted a characterisation that does not contain complement operator, is this possible?

Also in this weak topology is the open cell $ e^i_\alpha$ open in $X$? I think it is not because otherwise any CW complex would be disconnected which is not true. Then why we call it an open cell, isn't this confusing ? thank you for your help!


A CW-complex $X$ goes along with a cell-decomposition $\mathcal E$ and is coherent with the collection of closed cells $\{\overline e\mid e\in\mathcal E\}$.

That means that a set $A\subseteq X$ is open if and only if $A\cap\overline{e}$ is open in $\overline e$ for every $e\in\mathcal E$. So "yes" to your first question.

Actually this statement is equivalent with the one that states that $A\subseteq X$ is closed if and only if $A\cap\overline{e}$ is closed in $\overline e$ for every $e\in\mathcal E$.

You are correct in thinking that an open cell $e$ is not necessarily open in $X$. For instance we can think of $S^1$ as a union $e^0\cup e^1$ where $e^0=\{0\}$ and $e^1$ are disjoint open cells. It is evident that $e^0=\{0\}$ is not an open subset of $S^1$.

Defining $\mathbb{E}^{n}=\left\{ x\in\mathbb{R}^{n}\mid\left\Vert x\right\Vert <1\right\} $ and $\mathbb{D}^{n}=\left\{ x\in\mathbb{R}^{n}\mid\left\Vert x\right\Vert \leq1\right\} $ as open and closed $n$-disk an $n$-cell is a space homeomorphic to $\mathbb{E}^{n}$ and inherits the label "open" from $\mathbb{E}^{n}$.

  • $\begingroup$ So if $T_O$ is the original topology of $X$ and $T'_O$ is the subspace topology of $\bar e$ and $T_C$ is the new coherent topology on $X$ we can say that $A$ is open in $(X,T_c)$ iff $A\cap \bar e$ is open in $(\bar e,T'_O)$ iff $A=U\cap \bar e $ where $U$ is an open subset of $(X,T_O)$. In particular, if $A$ is open in $(X,T_O)$ then it is open in $(X,T_C)$. is that correct ? Also, Isn't $S^1=e^0\sqcup e^1$ here ? I mean the union is disjoint because the open cells are disjoint and form a partition on $S^1$ ? $\endgroup$ – palio Oct 23 '18 at 16:08
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    $\begingroup$ That is correct. Topology $T_c$ is in principle a finer topology than $T_O$ in the sense that $T_O\subseteq T_c$. We speak of coherence if both topologies coincide, as is the case here. Indeed formally we have $S_{1}=e^{0}\sqcup e^{1}$. I have edited and added the word "disjoint" now. $\endgroup$ – drhab Oct 23 '18 at 17:00
  • $\begingroup$ what do you mean by both topologies coincide ? you mean $(X, T_c)$ is homeomorphic to $(X,T_O)$? and which case you mean when you say "as is the case here", do you mean the cw decomposition in general or the specific case of the circle ? Thanks a lot ! $\endgroup$ – palio Oct 23 '18 at 17:34
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    $\begingroup$ What I mean by that is that $T_c=T_O$ so that the spaces $(X,T_c)$ and $(X,T_O)$ are the same (which is even stronger than homeomorphic). By "as is the case here" I mean the general situation prescribed in my answer. So concerning CW-spaces and not just the circle. You are welcome. $\endgroup$ – drhab Oct 23 '18 at 17:40

A space $X$ is said to have the weak topology with respect to a collection of subspaces $S_\alpha$ if $U \subset X$ is open in $X$ iff all $U \cap S_\alpha$ are open in $S_\alpha$. This is equivalent to the requirement that $A \subset X$ is closed in $X$ iff all $A \cap S_\alpha$ are closed in $S_\alpha$ (simply observe that $(X \setminus M) \cap S_\alpha = S_\alpha \setminus (M \cap S_\alpha)$).

For each $n$-cell $e^n_\alpha$ there exists a surjective map $\varphi_\alpha : (D^n,S^{n-1}) \to (\overline{e}^n_\alpha, e^n_\alpha)$ which maps the interior $\mathring{D}^n$ homeomorphically onto $e_\alpha$. Since $\mathring{D}^n$ is an open ball, $e^n_\alpha$ is called an open cell.

As you said, open cells are in general not open in $X$. In fact, $e^n_\alpha$ is open in $X$ iff there exists no $\beta$ such that $\overline{e}^m_\beta \cap e^n_\alpha \ne \emptyset$.


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