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I was attempting to make a collection of some of the separation properties of various topological spaces, can anyone tell me which of the properties I got incorrect? Also does anyone know of any source with a collection like this with more topological spaces and more separation properties?

$$\begin{array}{cc|c} & \mbox{T1} & \mbox{Hausdorff}&\mbox{Regular} & \mbox{Normal}&\mbox{Separable}\\ \hline \mathbb{R}&Y & Y & Y&Y&Y\\ \mathbb{R}^n&Y & Y & Y&Y&Y\\ \mbox{indiscrete}&N & N & Y&Y&Y\\ \mbox{discrete}&Y & Y & Y&Y&Y\\ \mbox{Cofinite}& Y& N&N &N&Y \\ \mbox{Cocountable}&Y&N&N&N&Y\\ \mathbb{R}_l&Y &Y &Y &Y&Y\\ \mbox{line w 2 origins} &Y &N &N &N&Y\\ \mbox{ordered square} & Y&Y &Y &Y&N \\ \mathbb{R}_k &Y &Y &N &N&Y\\ \{0,1\}^A &Y &Y &Y &Y&Y \end{array}$$

Comment:$\{0,1\}^A$ is the set of all functions from a set $A$ to $\{0,1\}$. $\mathbb{R}$ and $\mathbb{R}^n$ have the standard topology.$\mathbb{R}_l$ is $\mathbb{R}$ with the lower limit topology and $\mathbb{R}_k$ is $\mathbb{R}$ with the $k$ topology. I am not assuming regular/normal necessarily implies $T_1$.

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I assume that by ordered square you mean the lexicographically ordered square.

Separability is not a separation property. An uncountable space with the discrete or the co-countable topology is not separable, and $\{0,1\}^A$ is separable if and only if $|A|\le\mathfrak{c}=|\Bbb R|$. (That last result follows from the Hewitt-Marczewski-Pondiczery theorem, mentioned (with links) in this answer.)

Finite spaces with the cofinite topology and countable spaces with the co-countable topology are discrete and therefore regular and normal.

Brian M. Scott
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As a source of more of such tables the classic book “Counterexamples in Topology” is still very nice (it has such tables in the back for many more such properties), and a modern adaptation (taking that and other sources and making it searchable in a modern way) is $\pi$-base where you can search for a space and find its properties or look for a combination of properties, and find a space for it.

As to your table: separable is not a separation property, despite the name. It’s more of a notion of size IMO.

You cannot say that the cocountable topology is not Hausdorff, because that’s only the case if the underlying set is uncountable. And on a countable or finite set, any topology is separable, so you should specify the exact underlying set in the case of cofinite, discrete, cocountable topologies. Maybe fix the set on $\Bbb R$ for definiteness and contrastive value.

$\{0,1\}^A$ is separable iff $|A|\le |\Bbb R|$, so that needs an extra specification too.

PatrickR
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Henno Brandsma
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