Counterexamples regarding Totally Separated spaces There are known the following implications regarding totally separated spaces:


*

*Every totally separated space is totally disconnected;

*Every totally separated space is Urysohn;

*Every zerodimensional and $T_0$ is totally separated space and

*Every extremally disconnected is totally separated space.


I need counterexamples for opposite implications i.e.:


*

*there is a totally disconnected space that is not totally separated;

*there is Urysohn space that is not totally separated;

*there is a totally separated space that is not zerodimensional or $T_0$ and

*there is a totally separated space that is not externally disconnected.


I suppose the book "Counterexamples in Topology" could help here.
 A: Some hints:


*

*There are some counterexamples in the book. Also see the idea below.

*Surely, there is even a metrizable space that is not totally separated.

*Take your favourite example of a $T_2$ space that is not $T_3$ and try to modify it so it is totally separated, but not $T_3$. Most likely, it won't be extremally disconnected, and so you get 4. as well.

*Extremal disconnectedness is quite special property. Take any metrizable zero-dimensional space and most likely, it won't be extremally disconnected.


Added:
For 1. I got the following idea inspired by the counterexample 72 from the book. For every topological space $X$ we may form a “totally disconnected duplicated”: take $X × \{0, 1\}$, let $X × \{0\}$ be open discrete and let the basic neighborhood of $(x, 1)$ be $\{(x, 1)\} ∪ U × \{0\} \setminus \{(x, 0)\}$ for $U$ a neighborhood of $x$ in $X$. Denote this space by $X'$.
$X'$ is a duplicate of $X$ in the sense that $X'/\{(x, 0) \sim (x, 1)\}$ is canonically homeomorphic to $X$. $X'$ is scattered with two levels. 
If $X$ is $T_1$, then so is $X'$, and hence $X'$ is totally disconnected as any $T_1$ scattered space. On the other hand, the quasi-components of $X'$ look like this in the $T_1$ case: the singletons $\{(x, 0)\}$ are clopen quasi-components, while the quasi-components in $X × \{1\}$ correspond to quasi-components of $X$. In particular, if $X$ was connected and nondegenerate, than $X'$ is not totally separated. (I haven't checked everything so it may not be correct, it's an idea.)
