Which of the following topological spaces are separable

general-topologyseparable-spacessolution-verification

This is Exercise 6 from Section 2.2 on page 25 of Topology and Groupoids, by Brown.

Exercise:

A topological space is separable if it contains a countable, dense
subset. Which of the following topological spaces are separable?

  1. $\mathbb{Q}$ with the order topology
  2. $\mathbb{R}$ with the usual topology
  3. $\mathbb{I}^2$ with the television topology
  4. an uncountable set with the indiscrete topology
  5. the following spaces:

(a) $X$ is uncountable, and $N$ is a neighborhood of $x \in X$ if $x
\in N \subseteq X$ and $X \setminus N$ is finite.

(b) $X$ is uncountable, and $N$ is a neighborhood of $x \in X$ if $x
\in N \subseteq X$ and $X \setminus N$ is countable.

My attempt:

  1. Separable, because $\mathbb{Q}$ is a countable and dense subset.

  2. Separable, because $\mathbb{Q}$ is a countable and dense subset.

  3. Separable, because $\{ (p, q) \colon p, q \in \mathbb{Q} \cap \mathbb{I}^2 \}$ is a countable and dense subset.

  4. Separable, because any countable set $A \subseteq X$ will intersect $X$, which is the one and only neighborhood of any $x \in X$.

(a) Separable, because if we let $A$ be countable (and infinite), then for any $x \in X$ and neighborhood $N$ of $x$, we have $X \setminus N$ finite, so it is not possible that $X \setminus N$ contains $A$, which means that $N \cap A \neq \emptyset$.

(b)

Comments:

I think 1, 2, 4, and 5 (a) are probably correct. I am not sure about 3 because I read here that rational points in the plane are countable, but then I read here that the lexicographically ordered square is not separable.

For 5 (b) I am lost. I need to either find a countable, dense subset or prove that one can't exist.

Thanks for any help.

Edit:

The television topology is defined as follows:

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Best Answer

All of your answers except the third are correct. $\Bbb Q\cap\Bbb I^2=\varnothing$, so $\Bbb Q\cap\Bbb I^2$ definitely isn’t dense in $\Bbb I^2$. Neither is $\Bbb Q^2\cap\Bbb I^2$, since the sets $(0,1)\times\{y\}$ are all open in $\Bbb I^2$ with the television topology, there are uncountably many of them, and they are pairwise disjoint; any dense subset of the space must have a point in each of them and must therefore be uncountable. (This is actually just the lexicographically ordered square with the axes interchanged.)

$5(b)$ is not separable: if $D$ is any countable set, $X\setminus D$ is a non-empty open set disjoint from it, so $D$ is not dense in $X$.

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