We show that $X=(\mathbb{Z}_+)^J$ , where $J$ is uncountable, is not normal.(a) $B$ is a finite subset of $J$ , $x\in X$ , $U(x,B)=\{y\in X|y(j)=x(j),j\in B\}=\pi_B(x)\times\sqcap_{j\notin B}X_j$ . Every $U(x,B)$ is open ($\mathbb{Z}_+$ is in the discrete topology). If $x\in U$ , $U$ is open in $X$ , then there is a finite set $B$ of indexes such that $j\notin B$ implies $U_j=X_j$ . Therefore, $U(x,B)\subseteq U$ and they form a basis.(b) Let $P_n=\{x\in X|x_j\neq n\neq x_k\Rightarrow x_j\neq x_k\}$ , i.e. $x$ may have many coordinates equal to $n$ but all other coordinates different. $\{P_n\}$ are closed and disjoint. Indeed, if $y\notin P_n$ then $y_j=y_k\neq n$ and $U(y,\{j,k\})$ does not intersect $P_n$ . For $z\in P_n$ let $A=\{j\in J|z_j\neq n\}$ . Since $z$ is injective on $A$ , $A$ is countable. Therefore, there is uncountable set of indexes $B$ such that $z_j=n$ for $j\in B$ , i.e. $z$ cannot be a point in any other $P_m$ .Suppose $U$ and $V$ separate $P_1$ and $P_2$ and derive a contradiction.(c) Essentially, what we do here: we want to construct a sequence of points in $P_1$ such that they "converge" eventually to a point of the form: 1 for all indexes except a countable subset of indexes for which it takes values 1, 2, 3, .... We take the first point $x^1$ equal to 1, then for the first $n_1$ indexes in a countable sequence $\alpha$ of indexes we define values from 1 to $n_1$ ($x^2$ ), then do the same for the first $n_2>n_1$ indexes ($x^3$ ), etc. Moreover, we want to define the sequence $\alpha$ of indexes and the sequence $\{n_i\}$ of "steps" such that $U(x^i,B_i)=U(x^i,\{\alpha_1, . . .,\alpha_{n_i}\})\subseteq U$ . Note that $x_i$ takes values different from 1 on the set $\{\alpha_1, . . .,\alpha_{n_{i-1}}\}$ while $B_i$ includes the next step as well. Ok, so, $x^1_j=1$ for all $j\in J$ and $B_0=\emptyset$ . Given $x^n$ and $B_{n-1}$ we first define $B_n$ such that $U(x^n,B_n)\subseteq U$ (this is possible because $x^n\in P_1\subseteq U$ and we can define $B_n\supseteq B_{n-1}$ ) and then just define $x^{n+1}$ by the expression.(d) Now, given the sequence of points $\alpha$ and the sequence of points $x^i$ , we define a point $y$ in $P_2$ equal to 2 for all points not in $\alpha$ and $j$ for $\alpha_j$ . $y\in V$ , so there is a finite set $B$ such that $U(y,B)\subseteq V$ . Since $B$ is finite, there is $i$ such that $B\cap\alpha\subseteq B_i=\{\alpha_1, . . ., \alpha_{n_i}\}$ . Let $z_j=j$ for $j\in B_i$ , $z_j=1$ for $j\in B_{i+1}-B_i$ and $z_j=2$ for all other indexes. Then for all $j\in B$ either $j\in B_i$ and $z_j=j=y_j$ or $j\notin\alpha$ and $z_j=2=y_j$ . For all $j\in B_{i+1}$ either $j\in B_i$ and $z_j=j=x^{i+1}_j$ or $j\in B_{i+1}-B_i$ and $z_j=1=x^{i+1}_j$ . Therefore, $z\in U(y,B)\cap U(x^{i+1},B_{i+1})\subseteq V\cap U$ .