is the vector space $\mathbb{R}^\mathbb{N}$ locally compact? is the vector space $\mathbb{R}^\mathbb{N}$ locally compact?
for example, let $x=(x_1,x_2,....)$ any point of $\mathbb{R}^\mathbb{N}$ and let $V=[x_1-\epsilon,x_1+\epsilon] \times [x_2-\epsilon,x_2+\epsilon]\times...$
is $V$ a compact neighborhood of $x$? or is there a mistake?
 A: A neighborhood of $0\in\mathbb{R}^{\mathbb{N}}$ must contain a subset of the form
$$
U_1\times U_2\times \dots \times U_n \times \dotsm
$$
where all but a finite number of the $U_i$ are equal to $\mathbb{R}$ and the others are open intervals containing $0\in\mathbb{R}$.
If this subset is contained in a compact subset of $\mathbb{R}^{\mathbb{N}}$, then also
$$
C_1\times C_2\times \dots \times C_n \times \dotsm
$$
would be compact, where $C_i$ is the closure of $U_i$. Obviously this is a contradiction.

The product topology is the least topology making all projections into continuous functions. Let's talk generally for $X=\prod_{i\in I}X_i$ where $X_i$ is a topological space (a metric space, if you prefer; or just $\mathbb{R}$). We can define, for every $i\in I$, the projection map $\pi_i\colon X\to X_i$.
Our choice for the topology on $X$ means that, in order to check whether a map $f\colon Z\to X$ is continuous, we just need to check that $\pi_i\circ f\colon Z\to X_i$ is continuous for all $i\in I$.
Now, what subsets of $X$ are neighborhoods of $x\in X$? Only those that need to be as a consequence of the definition. Remember that the family of neighborhoods of $x$ is closed under finite intersections and you'll know why, in my argument above, only a finite number of the subsets $U_i$ can be a proper subset of $\mathbb{R}$.
What you had in mind is the box topology; but this is, in general, strictly finer than the product topology when considering the product of an infinite family. You can try your hand at showing when the two topologies are equal.
Why isn't the box topology used? Well, an infinite product of compact spaces is not compact when given the box topology (again, apart for trivial cases). Tykhonov's theorem, one of the pillars of topology, tells us that the product of any family of compact spaces is compact.
Local compactness (which is preserved for the box topology) is not as relevant as compactness; moreover there are many other reasons for using the product topology rather than the box topology: basically, it is a “functorial” construction.
