What should be the correct definition of a normal space? When I learned topology from Munkres' book of the same name, he defined a topological space $X$ to be normal if


*

*One point sets are closed

*For every pair of disjoint closed sets $A$ and $B$, we have that there exist open sets $U \supseteq A$ and $V \supseteq B$ such that $U \cap V = \emptyset$.


Now consider the countably infinite product of $\Bbb{R}$ with itself, denoted $\Bbb{R}^\omega$ with the product topology. Then a point like
$$(1,1,1,\ldots )$$
cannot possibly be closed because $\Bbb{R}^\omega$ is equipped with the product topology. I believe that if we discard 1) above, $\Bbb{R}^\omega$ will be normal. Thus my question is, is it reasonable to assume that one point sets are closed in the definition of normality?
 A: What Munkres calls normal is what many of us call $T_4$, reserving normal for spaces that satisfy (2) but not necessarily (1).
However, you’re mistaken about $\Bbb R^\omega$: every singleton is closed. To use your example, if $\langle x_k:k\in\omega\rangle$ is any sequence different from the constant $1$ sequence $s$, there is a $k\in\omega$ such that $x_k\ne 1$. Let $U$ and $V$ be disjoint open intervals about $x_k$ and $1$ in $\Bbb R$; then
$$\left\{y\in\Bbb R^\omega:y_k\in U\right\}\quad\text{and}\quad\left\{y\in\Bbb R^\omega:y_k\in V\right\}$$
are disjoint open nbhds of $x$ and $s$. In particular, the former is an open nbhd of $x$ that does not contain $s$, showing that $x$ is not in $\operatorname{cl}\{s\}$. Since $x$ was arbitrary, it follows that $\operatorname{cl}\{x\}=\{x\}$, i.e., $\{x\}$ is closed.
The property that all singletons are closed is equivalent to the $T_1$ separation axiom property and is implied by the $T_2$ (or Hausdorff) property. In normal spaces (meaning those satisfying (2)) $T_1$ implies $T_2$, so some people use the term normal Hausdorff for what I call $T_4$ and Munkres calls normal.
