I am reading about Hilton-Milnor theorem in different books and articles, and everywhere it is emphasized that we consider $\prod\limits_{i=1}^\infty X_i$ not simply as an infinite product, but as weak infinite product of topological spaces. What does that mean? How does it differ from cartesian product with product topology? I even remember seeing a warning in one article that those are not the same, though I don't manage to find relevant definition.
Thanks.
In this context, the spaces $X_i$ are pointed spaces. This means that if $S\subseteq T$ there is a natural inclusion $\prod_{i\in S}X_i\to\prod_{i\in T}X_i$ given by using the basepoint on the coordinates in $T\setminus S$. The weak infinite product is then the colimit of all the finite products with respect to these inclusion maps.
Concretely, the weak infinite product is the subset of the ordinary product consisting of points which are the basepoint on all but finitely many coordinates. (In other words, it is what would typically be called the "direct sum" in an algebraic context.) It is topologized by saying a set is open iff its intersection with each finite subproduct is open in the usual product topology.
Note that this topology is typically strictly finer than the subspace topology from the product topology on the entire infinite product space. For instance, let us consider the weak infinite product "$\mathbb{R}^\infty$" of countably infinitely many copies of $\mathbb{R}$ (with $0$ as the basepoint). Let $e_n\in\mathbb{R}^\infty$ be the point that is $1$ on the $n$th coordinate and $0$ in all others. With respect to the usual product topology, the sequence $(e_n)$ converges to the point that is $0$ on all coordinates (since each coordinate converges to $0$). However, the set $\{e_n:n\in\mathbb{N}\}$ is closed in $\mathbb{R}^\infty$, since its intersection with each finite $\mathbb{R}^n$ is finite and thus closed, so $(e_n)$ does not converge to $0$ in $\mathbb{R}^\infty$ (indeed, it does not converge at all).