Finding a CW decomposition of $X$ having a prescribed point $x\in X$ as a 0-cell. Let $(X,E)$ be a CW complex and let $x\in X$. Does there exist a CW decomposition of $X$ having $x$ as a 0-cell?
The book I'm reading says that subdividing the cell containing $x$ gives the desired CW decomposition, with no further details. (that's the whole content of the proof!) Of course anyone can see that subdivision is the key point here so I feel that such an explanation is not really a proof but just some advice.
Some chapters earlier, I solved a similar exercise concerning finite simplicial complexes, i.e. polyhedra. I used an argument similar to the barycentric subdivision of a simpmlicial complex there. However, when we deal with a CW complex $X$, it seems that there is no uniform way to handle the subdivision for all cells, because unlike polyhedra, closure of cells need not meet nicely anymore for CW complexes. I have no idea how to subdivide the cells and give them appropriate characteristic maps.
An detailed proof of the qusetion would be very nice. Please enlighten me.
 A: Here's one simple way to do it. The proof is by induction on the dimension of the cell containing $x$.
Let $e$ be the cell containing $x$, let $n$ be the dimension of $e$, and let $\chi : B^n \to X$ be the characteristic map of $e$. The boundary image $\chi(\partial B^n)$ is contained in the $n-1$ dimensional skeleton $X^{(n-1)}$. Choosing any point $y \in \chi(\partial B^n)$, if $y$ is not already a $0$-cell, we may by induction change the CW structure on $X^{(n-1)}$ so that $y$ is a $0$-cell.
Let $\xi = \chi^{-1}(x)$, a point in the interior of $B^n$. Choose $\eta \in \partial B^n$ to be a point such that $\chi(\eta)=y$. Let $(\xi,\eta) \subset B^n$ be the open line segment in $B^n$ with endpoints $\xi,\eta$. 
Remove $e$ from the set of $n$-cells of $X$, add $x$ to the set of 0-cells, add $\alpha = \chi(\xi,\eta)$ to the set of $1$-cells, and add $\chi(\text{int}(B) - (\{\xi\} \cup (\xi,\eta)) = e - (\{x\} \cup \alpha)$ to the set of $n$-cells. 
It should be evident that $e - (\{x\} \cup \alpha)$ is homeomorphic to the open $n$-ball, and with a little work you should be able to write down a formula for a characteristic map.
