1

Is the cocountable complement topology always separable? Also are there Really any other issues with this table I created?

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

I am going to need to know these.Any help will be greatly appreciated.

Here is a link to all the discrepancies that were cleared up.

Table of separation properties of various topological spaces

Arturo Magidin
  • 398,050

1 Answers1

1

As said before: "Cocountable" means nothing. What set ? If the set is uncountable then the cocountable topology is not separable, because all countable sets are closed and so the closure is never $X$.

If $X$ is countable, then any topology on it is separable. So you do need to specify the underlying set in order to be able to fill in the table at all.

For "discrete" the same holds: not separable if $X$ is uncountable, and separable otherwise. Specify the set! Also for "indiscrete" you need to be precise. And what is the size of $A$ in $\{0,1\}^A$?

The table is incomplete and partially false as it stands, and the spaces are underspecified.

Henno Brandsma
  • 242,131
  • @DanielSchmidt What is the definition of "discrete" they use for this excercise? If they don't specify the set, the problem is ill-defined. – Henno Brandsma Mar 04 '21 at 22:51
  • For the discrete topology on any set they specify that only the underlying set and empty set are open.Or sorry that is for indiscrete.I meant all subsets are open. –  Mar 04 '21 at 22:52
  • @DanielSchmidt That's the indiscrete topology. If we take set $X={p}$ then $X$ in the indiscrete topology is $T_1$ and Hausdorff, if $X={p,q}$ it's neither. So it should have specified the set too. The answers depend on it. $\Bbb Z$ in the discrete topology is separable, $\Bbb R$ is not, so it should have been specified. – Henno Brandsma Mar 04 '21 at 22:55
  • 1
    @DanielSchmidt My advice: take $A,X=\Bbb R$ for all unspecified spaces. Then give the correct answers. – Henno Brandsma Mar 04 '21 at 23:03