I am trying to prove that if $X$ is a CW-complex and $e_\alpha^n$ is an open $n$-cell, then $e_\alpha^n$ must be open in $X$. For do so, lets recall the following definitions:
Definition 1. Let $\{I_n\}_{n=0}^\infty$ be a sequence of disjoint sets such that $I_0\ne\emptyset$. For $n=0$ define $X^0=I_0$ with the discrete topology. Given $X^{n-1}$, put $X^n=X^{n-1}$ if $I_n=\emptyset$. If $I_n\ne \emptyset$ we assume that we have a familily of maps $\{\varphi_\alpha:\partial D^n\to X^{n-1}\}$ and define $X^n$ as the quotient space: $$X^n=\frac{X^{n-1}\bigsqcup_{\alpha\in I_n}D^n}{x\sim \varphi_\alpha(x), \,\forall\alpha\in I_n}$$ Then a CW-complex is the colimit $X$ of the sequence of closed embeddings: $$X^0\hookrightarrow X^1\hookrightarrow\ldots\hookrightarrow X^n\hookrightarrow\ldots$$
Definition 2 Let $\Phi_\alpha^n$ be the composition of the quotient map $X^{n-1}\bigsqcup_\alpha D^n\to X^n$ with the inclusion $X^n\hookrightarrow X$. Then we call $e_\alpha^n=\Phi_\alpha^n(\mathring{D}^n)$ an open $n$-cell.
Any help?
In general it is not true. An open $n$-cell $e^n_\alpha$ is always open in the $n$-skeleton $X^n$ of $X$, but not necessarily open in $X$. In fact, if there is any attaching map $\varphi : S^{m-1} \to X^{m-1}$ of an $m$-cell with $m > n$ such that $\varphi(S^{m-1}) \cap e^n_\alpha \ne \emptyset$, then $e^n_\alpha$ is not open in $X$.