Is T discrete topology on X or not? I want to verify my argument  from  question of my assignments in topology.

For each n$\in  \mathbb{N}$ , let $X_n =\mathbb{R}$ and let $T_n$ be discrete topology on $X_n$. Let T be product topology on X=$\prod_{n \in \mathbb{N}} X_n$ . Is T discrete topology on X or not?

I think It will be discrete topology as every element of  $P(\prod_{\mathbb{N}} X_n)$, where P(X) means power set of X are in X.
Am I right?
 A: No, it is not the discrete topology. The set $\{(0, 0, 0, \dots,)\}$ is not open, as infinitely many of the pre-images under the projection maps are not the entire space.
To provide more detail: for $T$ to be the discrete topology, every subset of $X$ must be open. It will therefore suffice to show that a single subset of $X$ is not open. This set will be the singleton $A = \{(0, 0, \dots)\}$.
Let $\pi_n$ be the canonical projection function from $X_n$ to $X$. As $X$ has the product topology, a subset $V$ of $X$ is open iff every point of $V$ has a neighbourhood $U$ such that $\pi_n^{-1}(U) = X_n$ for all but finitely many $n$. However, $\pi_n^{-1}(A) = \{0\}$ for all $n$. Infinitely many preimages fail to be the entire space. The defining characteristic of open sets in $X$ has failed, and so, $A$ is not open. Thus, the topology of $T$ is different from the discrete topology.
A: The box topology would be the discrete topology; but the product topology is smaller.  One way to see it is that the product topology is the coarsest making the projections continuous.
However, every function from a discrete space is continuous.  That's it's the finest topology, period.
The product topology is in fact strictly coarser, because it doesn't include the entire power set.  It only includes finite products of non-empty open sets.
