$(K^n,K^{n-1})$ is a good pair

Question

Given $X$ a $CW$ complex I'd like to know wether the pair formed by two skeletons $(K^n,K^{n-1})$ are a good pair.

I think the answer is yes because I found some reference here (Stefan's answer) with the statement involving $(X,A)$ pairs. Are $(K^n,K^{n-1})$ $CW$ pairs ? Since I don't understand the inductive open-neighborhood costruction I wanted to know wether there was a simpler way in the case of consecutive skeletons. Local contractibility of $CW$ complex can be used.

Addendum : Do they need to be consecutive ?

Any help or reference would be appreciated.

Answer 1

Yes, $(K^n, K^{n-1})$ is always a CW-pair.

A subcomplex of a cell complex $X$ is a closed subspace $A \subset X$ that is a union of cells of $X$.

Now simply observe that all skeleta $K^m$ are closed and are unions of cells.

But in this special case there is a much simpler proof that $(K^n, K^{n-1})$ is a good pair. For each open $n$-cell $e^n_\alpha$ choose a point $z_\alpha \in e^n_\alpha$. Then $Z = \{ z_\alpha \mid \alpha \in J \}$ is closed and $N = K^n \setminus Z$ is an open neigborhood of $K^{n-1}$ in $K^n$ which deformation retracts to $K^{n-1}$. See Injective homomorphism between $H_n(\mathbb{D}^n,\mathbb{S}^{n-1}) \longmapsto H_n(X^{*},X)$.