Product topology of $\Re^X$ et similia I have a problem with visualizing how exactly the product topology of something like $\Re^X$ looks like.  
Just a quick summary of my line of reasoning.  
Given a family of topological spaces $\{(X_i,\tau_i)\}_{i \in I}$, with $X = \prod_{i \in I} X_i$, and defining for each $j \in I$ the projection $P_j : X_j \to X$ as $P_j(x) = x_j$,  the product topology $\tau$, denoted $\prod_{i \in I} \tau_i$, is the weakest topology on $X$ that makes each projection $P_i$ continuous.
So far so good.   
Thus, I try to apply these concept to something $\Re^X$, but I do not get much out of it. In particular, my problem is that I don't see any family of topological spaces associated with $\Re^X$, but only one topological space, i.e. $\Re^X$ itself!
[Thus, no product topology?!]
Indeed, this is how I see that object:
$$\Re^X := \{ f \ | \ f: X \to \Re \}.$$
So, if I think about a product topology $\tau$ on $\Re^X$, I think about a tricky idea that does not really deal with a family of topologies but - at the same time - is a convenient way to get each functional to be continuous. Thus, basically the projection should be
$$ P : \Re^X \to \Re^X,$$
more or less as an identity function that - thanks to the product topology - ensures us that any function in $\Re^X$ is continuous.
I am not sure if this does make sense, in particular in the light of the fact that I "know" that the product topology on $\Re^X$ is called the topology of pointwise convergence on $X$, because a net $\{f_\alpha\}$ in $\Re^X$ satisfies $f_\alpha \to f$ if and only if $f_\alpha (x) \to f(x)$ in $\Re$ for each $x$ in $X$.
[Of course, this does not make my life easier at all...]
Just to add one last point, let's forget what happens when we move to $\Re^{X^\prime}$, where $X^\prime$ is the topological dual of $X$... there I have no clue at all.
So, in the end, is my intuition correct?
If it is so, how should I translate it to the definition of pointwise convergence on $X$?  
Any help is much appreciated, considering that I am basically self-thaught.
Thanks a lot in advance for any help or feedback!
 A: The space $\mathbb{R}^X$ is indeed a product. It is the product of $X$-many copies of $\mathbb{R}$. In other words, $X$ just takes the role of the index set, and then $$\mathbb{R}^X=\prod_{x\in X}\mathbb{R}_x,$$where for every $x\in X$ we have $\mathbb{R}_x=\mathbb{R}$. Think about it for a minute to see that an element in this product is nothing else than a function $X\to\mathbb{R}$. Any function, no conditions at all.
Next note that by the (universal) property of the product topology (which you seem to know well), the topology of $\mathbb{R}^X$ is generated by the basis whose elements are products $$\prod_{x\in X}U_x\subset\prod_{x\in X}\mathbb{R}_x,$$ where $U_x$ is open in $\mathbb{R}$ for every $x$, and $U_x=\mathbb{R}$ for every $x$ except for finitely many $x$s.
Then just go step by step, following the definitions, to see that convergence in this topology is indeed equivalent to pointwise convergence.
And by the way, for each $x\in X$ the projection $\pi_x:\mathbb{R}^X\to\mathbb{R}$ is given by $\pi_x(f)=f(x)$.
