A topological space $(X,\tau)$ is said to be seperable if it has a dense subset which is countable.Determine which of the following spaces are seperable.
(i) the set $\mathbb{R}$ with the usual topology:
$\mathbb{Q} \subset \mathbb{R}$ is a dense subset and is countable,So $\mathbb{R}$ is seperable.
(ii)a countable set with discrete topology :
All proper subset of $X$ will be close ,so its closure is itself.So they aren't dense.$X \in \tau$ So $X$ is open and $\phi \in \tau$ so $X$ is close also.$X$ is closed,so its closure is $X$ itself.So is dense and $X$ is countable.So $(X,\tau)$ is seperable.
(iii)a countable set with the finite closed topology:
Similarly here $X$ is dense and being countable,the topological space is seperable.
(iv)$(X,\tau)$ where $X$ is finite.
Similarly here $X$ is dense in itself and being finite is countable.So Topological space is seperable.
(v)$(X,\tau)$ where $\tau$ is finite.
Taking $\tau$ as indiscrete topology.So only open and close sets are $X$ and $\phi$. $X$ is dense but unless it is countable we cannot say $\tau$ is seperable.
(vi)a uncountable set with the discrete topology.
only dense set is $X$ but $X$ is not countable so Topology is not seperable.
(vii)an uncountable set with the finite closed topology
Here $X$ is dense set but not countable.I am unable to conclude whether any countable subset of $X$ can be dense or not?
Please check i believe my arguments are correct for first six topology.And give me a hint for seventh one.
Best Answer
For v : pick one element from each $O \in \tau\setminus\{\emptyset\}$ and convince yourself all such points together are dense.
So in fact any space where $\tau$ is at most countable is separable.
(vii): any countably infinite subset cannot be closed (it's not finite) so the only closed superset is $X$ and so any countably infinite subset (which exists) is dense.
The other arguments are OK.