An open set in a CW complex

Question

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!

Answer 1

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

Answer 2

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 $\overline{e}^m_\beta \cap e^n_\alpha \ne \emptyset$.