Given.
$$D \text{ – some dense set in ℝ} $$
$$P(D) – \text{power set over D} $$
$$T_{d_E} – \text{canonical topology on ℝ} $$
$$T_D = T_{d_E}\cup P(D) $$
$$T – \text{ topology generated from the subbasis } T_D$$
$$\underline X = (ℝ, T) $$
Prove.
$\underline X$ is totally-disconnected.
Attempt.
If $A \subseteq R, \ card(A) > 1$ and $A$ is open, then it contains some $d \in D$ by the definition of density.
Since $\underline X$ is a $T_1$ space, then $\{d\}$ is closed but it's also open in $A$ due to the way $T_D$ is defined. So $\{d\} \neq A$ is "clopen" set in $A$ and hence $A$ is disconnected.
Same goes by the above logic for the case if $A$ is not open but is a neighbourhood containing an open set.
If the above is right then one can handle the cases: $[ a, b], $[a, b), $(a, b]$ and $(a,b)$.
But how can one show that e.g. non-interval set is disconnected: $B = (a,b)\cup(c,d)$?
If the topology defined on $\underline X$ were $T(\le)$ Topology, then one could argue that since $B$ is not an interval then it can't be connected. But $T_D$ is not an order topology, so how can I prove that $B$ type sets are disconnected too?
Best Answer
$T_D$ is not a topology but it is a subbase for a topology, which is probably what's meant here. Taking $D= \Bbb P$ gives us the Michael line, and this question is meant to generalise this situation, I suppose.
To see that $\underline{X}$ is totally disconnected, take a subspace $A$ having $2$ or more points. If $d \in D \cap A \neq \emptyset$ then $\{d\}$ is a clopen subset of $A$, so $A$ is not connected. Otherwise $A$ has empty interior in the Euclidean topology and $A \subseteq \Bbb R\setminus D$. In the usual topology any $A$ that has empty interior and has two or more points is disconnected (as it's not an interval of any type) and so the same holds for $A$ (which has the Euclidean topology, been disjoint from $D$). QED