Product topology - definition Can someone please give me a detailed explanation of the concept of product topologies? I just can't get it. I have looked in a number of decent textbooks(Munkres, Armstrong, Bredon,  Wiki :P, Class notes, a youtube video).
This is what it seems like to me:
We have two topological spaces $(X,\tau_1)$ and $(Y,\tau_2)$ and we take their product topology:
$$(X\times Y,\tau_1\times \tau_2)$$
Where this product topology $\tau_1\times \tau_2$ consists of unions of all elements of $\tau_1$ with all element of $\tau_2$ I.e. The first element of $\tau_1$ is taken in union with every element of $\tau_2$ and then the second element and so on, and all unions and intersections of these are taken.
Now I am confused as well since apparently the product topology is immediately $T_{3.5}$ but I have seen that the product of two hausdorff spaces is hausdorff, then what's the deal with this? Are two hausdorff spaces actually $T_{3.5}$ and then $T_2$ is absorbed?
 A: The product topology (on a product of two spaces $(X,\tau_1)$ and $(Y,\tau_2)$ consists of all unions of sets of the form $U \times V$, where $U \in \tau_1$ and $V \in \tau_2$. On easily checks that this forms a topology. 
A more general way of defining it, which works for products of any number of spaces $(X_i, \tau_i), i \in I$, is that it is the intersection of all topologies $\tau$ on $\prod_{i \in I} X_i$ that are such that for all $i$, the projection $p_i: (\prod_{i \in I} X_i, \tau) \rightarrow (X_i, \tau_i)$ is continuous. It's a small proof to show that for two spaces this coincides with the above definition, and it shows that the product topology is natural (it's the minimal topology that makes all projections continuous) and also is the category-theoretical product (if you care for such things).
Now, the product topology is quite natural for the lower separation axioms: $X \times Y$ is a $T_i$ space for $i=0,1,2,3,3{1\over 2}$ iff $X$ and $Y$ are both $T_i$ spaces. (For $T_4$ spaces this can fail.) It's certainly not true that $T_{3\frac{1}{2}}$ is automatic for products, as you seem to think. It does need the same to already hold for both composing spaces.
