$Y=\prod_{n\in\mathbb{N}}X$.

Question

Let X be a compact CW-complex. The infinite cartesian product $Y=\prod_{n\in\mathbb{N}}X$ is a compact topological space, and as CW-complex It should have a finite number of cells. But in the CW-decomposition of Y given in Cartesian product of two CW-complexes infinite cells appear. What is happening?

Answer 1

It is not a CW-complex unless $X$ has only a single point. To see this, let $X$ be a CW-complex which has more than one point.

Case 1. $\dim(X) = 0$. Then $X$ is a finite discrete space and $Y$ is an infinite space. If it were a compact CW-complex, then it could have only finitely many $0$-cells. Hence it must have at least one (open) cell $e$ of dimension $n > 0$ which is a homeomorphic copy of $\mathring{D}^n$. We conclude that $Y$ has a connected component with more than one point. On the other hand each component of $Y$ is the product of components of $X$, i.e. has only a single point. This is a contradiction.

Case 2. $\dim(X) > 0$. Then $X$ has an open cell $e$ of dimension $n > 0$ and $Y$ contains a subspace homeomorphic to $E = \prod_{n \in \mathbb{N}} e$. If $Y$ were a CW-complex (which is compact!), it would embed into some $\mathbb{R}^m$. See Hatcher "Algebraic Topology" Corollary A.10. Hence also $E$ would embed into $\mathbb{R}^m$. But now $E$ contains a copy of $\mathbb{R}^{m+1}$, hence $\mathbb{R}^{m+1}$ would embed into $\mathbb{R}^m$. But this is impossible which is the consequence of topological dimension theory. See for example Chapter 7 of Engelking "General Topology" (in particular Theorems 7.1.1, 7.3.3, 7.3.19).