Is $\mathbb{R}$ with the usual topology a $T_2$ space but not $T_3.$ Is $\mathbb{R}$ with the usual topology a $T_2$ space but not $T_3$ (regular + $T_1$)?
I feel that it is not even $T_2$, because of in $T_2$ definition we have for all $x,y$ not there exist, and because we cannot separate rational & irrational numbers on the real line. 
Could anyone tell me if I am right or wrong, Is $\mathbb{R}$ with the usual topology a $T_2$ space but not $T_3$? 
 A: $\mathbb{R}$ is $T_2$, for let $x,y\in \mathbb{R}$  with $x\neq y$. Without loss of generality, suppose $x < y$. Then let $U = (x-d, x+d)$ and $V = (y-d,y+d)$, where $d = \frac{y-x}{2}$. Then $U$ and $V$ are neighborhoods of $x$ and $y$ such that $U\cap V = \varnothing$. 
$\mathbb{R}$ is $T_3$, for let $C\subseteq\mathbb{R}$ be closed with $x\notin C$. Then define $C_1 = C\cap[x,\infty)$ and $C_2 = C\cap (\infty, x]$. Then both $C_1$ and $C_2$ are closed. Assume both sets are nonempty (if one is empty, then the argument is similar). Then $C_1$ has a minimum $y_1$ and $C_2$ has a maximum $y_2$. So, $y_2 < x < y_1$. Let $d_1 = (y_1-x)/2, d_2 = (x-y_2)/2$. Finally, let $U = (x-d_2, x+d_1)$, and $V=(-\infty, y_2+d_2)\cup (y_1-d_1, \infty)$. Then $V$ is a neighborhood of $C$, $U$ is a neighborhood of $X$, and $U\cap V = \varnothing$. So, $\mathbb{R}$ is $T_3$. 
A: $T_2$ means we can separate any two points by two open sets, now $\mathbb{R}$ with the usual topology is a metric space so there is a distance function and every two distinct points has a positive distance between them hence they can be separated by open balls, hence $\mathbb{R}$ is $T_2$.
In fact it's $T_4$, check the link below.
http://math.gmu.edu/~tlim/metricnormal.pdf
