Let it be that $X$ is a CW-complex. I practicize the following definition: $X$ is a Hausdorff space and $\mathcal{E}$ is a partition on $X$ such that each $e\in\mathcal{E}$ can be recognized as a cell, i.e. a space homeomorphic with $\mathbb{E}^{n}:=\left\{ x\in\mathbb{R}^{n}\mid\left\Vert x\right\Vert <1\right\} $ for some nonnegative integer $n$. This $n$ is the dimension of $e$ and is denoted as $|e|$. Next to $\mathbb E^n$ we define $\mathbb{D}^{n}:=\left\{ x\in\mathbb{R}^{n}\mid\left\Vert x\right\Vert \leq1\right\}$ and $\mathbb{S}^{n-1}:=\left\{ x\in\mathbb{R}^{n}\mid\left\Vert x\right\Vert =1\right\} $ and $X^n\subseteq X$ is by definition the union of all cells having a dimension that does not exceed $n$.

The following axioms hold:

  • For every $e\in\mathcal{E}$ there is a continuous $\Phi_{e}:\mathbb{D}^{\left|e\right|}\rightarrow X$ such that $\Phi_{e}\left(\mathbb{E}^{\left|e\right|}\right)=e$, $\Phi_{e}\left(\mathbb{S}^{\left|e\right|-1}\right)\subseteq X^{\left|e\right|-1}$ and $\Phi_{e}\upharpoonleft\mathbb{E}^{\left|e\right|},e$ is homeomorphic.

  • For every $e\in\mathcal{E}$ its closure $\overline{e}$ meets only a finite number of cells belonging to $\mathcal{E}$.

  • $X$ is coherent with $\left\{ \overline{e}\mid e\in\mathcal{E}\right\} $.


Let $W\subset X$.

Is it true that here $\overline{W}=\bigcup_{e\in\mathcal{E}}\left(\overline{W\cap e}\right)$?

My own incomplete thinking

I understand that $W\subseteq\cup_{e\in\mathcal{E}}\left(\overline{W\cap e}\right)\subseteq\overline{W}$ and if $W$ would only meet a finite number of cells then $\cup_{e\in\mathcal{E}}\left(\overline{W\cap e}\right)$ is a finite union of closed sets, so is closed itself. So in that case the statement is true. Secondly I understand that - in order to prove the statement - it is enough to show that $\cup_{e\in\mathcal{E}}\left(\overline{W\cap e}\right)\cap\overline{f}$ is closed for every cell $f\in\mathcal{E}$.

I suspect that the statement is not true in general, but am not sure. So I am asking for a counterexample or a proof. Thank you in advance.


Consider: $$X=\bigcup_{n\ge1} \{\langle x,y\rangle\in\mathbb{R}^2:(x-\frac{1}{n})^2+y^2=\frac{1}{n^2}\}$$ with subspace topology from $\mathbb{R}^2$. Give it a CW-complex structure with one $0$-cell $e^0$, the origin and (countably) infinitely many $1$-cells $\{e_n^1\}_{n\ge1}$, all the circles. This is the Hawaiian earring.

Say, $W=\{\langle\frac{2}{n},0\rangle:n\ge{1}\}\subset X$. Then $\overline{W}=W\cup \{\langle 0,0\rangle\}$. But, $W\cap e_n^1=\{\langle\frac{2}{n},0\rangle\}$ , $\forall n$ and $W\cap e^0=\phi{}$.

So here: $$\bigcup_{e\in\mathcal{E}}\left(\overline{W\cap e}\right)=W\neq{\overline{W}}$$

  • $\begingroup$ Thank you. This confirms my intuition and makes it possible to go on without doubts on my mind. $\endgroup$ – Vera Dec 27 '14 at 9:52

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