I came across some tables in "counterexamples for topology" that contradict some of the properties I have determined in the following table I have created.It says that none of the properties on the chart I created are satisfied by the indiscrete topology. I thought the indiscrete topology was regular, normal, and separable, the chart says it is not? What is going on here? It says $4$ is for the indiscrete.
$$\begin{array}{cc|c}
& \mbox{T1} & \mbox{Hausdorff}&\mbox{Regular} & \mbox{Normal}&\mbox{Separable}\\
\hline
\mathbb{R}&Y & Y & Y&Y&Y\\
\mathbb{R}^n&Y & Y & Y&Y&Y\\
\mbox{indiscrete}&N & N & Y&Y&Y\\
\mbox{discrete}&Y & Y & Y&Y&Y\\
\mbox{Cofinite}& Y& N&N &N&Y \\
\mbox{Cocountable}&Y&N&N&N&Y\\
\mathbb{R}_l&Y &Y &Y &Y&Y\\
\mbox{line w 2 origins} &Y &N &N &N&Y\\
\mbox{ordered square} & Y&Y &Y &Y&N \\
\mathbb{R}_k &Y &Y &N &N&Y\\
\{0,1\}^A &Y &Y &Y &Y&Y
\end{array}$$
Best Answer
Some people, unfortunately, including Steen and Seebach, fail to recognize that the $T_k$ designations are supposed to form a hierarchy, so that $T_k$ implies $T_\ell$ whenever $k\ge\ell$. For you and me $T_3$ means regular plus $T_1$, and $T_4$ means normal plus $T_1$. For people like Steen and Seebach who follow the other convention, regular and normal are our $T_3$ and $T_4$, respectively, and $T_3$ and $T_4$ are our regular and normal, respectively. Thus, for them a regular or normal space must be $T_1$, but a $T_3$ or $T_4$ space need not be $T_1$. They also switch $T_{3½}$ and completely regular.