When $J$ is uncountable $R^{J}$ is not Normal So this is an exercise of Munkres,
Let $X=\mathbb{N}^J$ and the first step is to show that if $x \in X$ and $B $ is a finite subset let $U(x,B)$ consist of the element $y$ of $X$ such that $x(\alpha) = y(\alpha)$ for $\alpha \in B$ then $U(x,B)$ is a basis for $X$. I can see that in the subspace topology this is going to be an open set, my problem is that when trying to prove that this is basis for $X$ and when thinking of elements of $X$ as functions from $J$ to $\mathbb{N}$ basically what we are saying is that there is at least an index $\alpha$ where any two functions agree, so after thinking about this i think that this might be possible because $J$ is uncountable and $\mathbb{N}$ is countable, is this the case? Thanks in advance.
 A: It's quite easy to see that these sets form a base; you just need to realise that 
$\Bbb N$ is discrete, it holds for any power of a discrete space:
Suppose $O$ is product open in $X$ and $x : J \to \Bbb N$ is a point of $O$.
Then there is a standard basic open subset $\prod_{\alpha \in J} O_\alpha$
with $x \in \prod_{\alpha \in J} O_\alpha \subseteq O$, so (by definition of the standard base) there is a finite subset $F \subseteq J$ such that $O_\alpha = \Bbb N$ for all $\alpha \notin F$ and all $O_\alpha$ are open.
Now $x(\alpha) \in O_\alpha$ for all $\alpha \in J$ so 
$$U(x,F) \subseteq \prod_{\alpha \in J} O_\alpha$$
(suppose $y \in U(x,F)$ then for all $\alpha \in F$, $y(\alpha) = x(\alpha) \in O_\alpha$, and for all $\alpha \notin F$, $y(\alpha) \in O_\alpha=\Bbb N$ is obvious, so $y \in \prod_{\alpha \in J} O_\alpha$.)
Finally remark that a set of the form $U(x,B)$ is open in the product because it is exactly of the basic open form $\prod_{\alpha \in J} O_\alpha$ where $O_\alpha = \{x_\alpha\}$ for $x \in B$ and $O_\alpha = \Bbb N$ for $\alpha \notin B$, where we use that all singletons are open in $\Bbb N$ (the discreteness), and $B$ is finite (as required for a basic open set).
