# An open set in a CW complex

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 $$\mathbb{E}^{n}$$.

• 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 $S^1$ ? – palio Oct 23 '18 at 16:08
• 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. – drhab Oct 23 '18 at 17:00
• 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 ! – palio Oct 23 '18 at 17:34
• 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. – 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 $$\overline{e}^m_\beta \cap e^n_\alpha \ne \emptyset$$.