Let $(X, \tau) $ be a topological space and $A, B\subset X$ connected.
We know that $A\cap B\neq \emptyset$ implies $A\cup B$ is connected.
We also know that $A\cup B$ connected doesn't imply $A\cap B\neq \emptyset$
For an example $(\Bbb{R}, \tau_{\textrm{cofinite}})$ with $A=\Bbb{Q}$ and $B=\Bbb{R}\setminus \Bbb{Q}$
Now consider the following situation:
A topological space $(X, \tau)$ have the property $\text{SG}$ if $\forall A, B\subset X$ (with $A,B$ more than $1$ point) connected be such that $A\cup B$ connected implies $A\cap B\neq\emptyset$
The property $\text{SG}$ is vacuously satisfied by any totally disconnected space.
Is the classification of the $\text{SG}$ property already done?
If yes then list some interesting property of the space satisfying $\text{SG}$ property.
Best Answer
Recall that a topological space is biconnected, if it is connected but is not the union of two disjoint connected subsets, which both have more than one point.
This is a standard notation, see for instance this paper of M.E. Rudin. (In this paper, by definition, connected spaces have more than one point. So, this is true for biconnected spaces as well. But for our purpose it is better that spaces with at most one point are biconnected as well.)
In the following, $X$ is always a topological space.
The following lemma, which is certainly folklore, is crucial. For the easy proof see for instance here, lemma 9 on p. 10 (T1 is not needed.)
Lemma. Let $X$ be connected, $C$ a connected subset of $X$ and $T$ a component of $X \setminus C$. Then $X \setminus T$ is connected.
This implies immediately connectness of $B^\prime$ in the following (the other properties are obvious):
Corollary. Let $X$ be connected, $A, B$ disjoint connected subsets of $X$ and $a \in A$. Define $A^\prime := C_{X \setminus B}(a) =$ the component of $a$ in $X \setminus B$ and $B^\prime := X \setminus A^\prime$. Then $A \subset A^\prime$, $B \subset B^\prime$, $A^\prime$ and $B^\prime$ are connected and $X$ is the disjoint union of $A^\prime$ and $B^\prime$.
For the next theorem it is convenient to define:
$X$ is strongly SG, if for all connected subsets $A, B$ of $X$, each with more than one point, it is $A \cap B \neq \emptyset$.
So, in contrast to SG, here it is not assumed that $A \cup B$ is connected. It is easy to see (but will not be used here) that a space is strongly SG, iff it is SG and has at most one component with more than one point.
Theorem. The following are equivalent:
PROOF. $1 \Rightarrow 2$: Let $A, B$ be connected subsets, each with more than one point. Assume $A \cap B = \emptyset$. Pick $a \in A$ and define $A^\prime, B^\prime$ as in the corollary, which implies a contradiction to biconnectness of $X$.
$2 \Rightarrow 3$ is trivial.
$3 \Rightarrow 1$ is straightforward, since $X$ is connected.
Remark. The crucial step $1 \Rightarrow 2$ is also mentioned in M.E. Rudin's paper (see above). But I couldn't find a proof of it. Therefore I included it here.
Corollary. The following are equivalent:
PROOF. $1 \Rightarrow 2$: Let $C$ be a connected subset of $X$. Trivially, $C$ is SG as well. Hence biconnected.
$2 \Rightarrow 3$ is trivial.
$3 \Rightarrow 1$: Let $A, B$ be subsets of $X$ with more than one point each and $A, B, A \cup B$ be connected. Hence, $A \cup B$ is contained in a component $C$ of $X$. Since $C$ is biconnected, it is SG. Hence, $A \cap B \neq \emptyset$.
$1 \Leftrightarrow 4$ is straightfoward (or, formally, $3 \Leftrightarrow 4$ follows immediately by the theorem).
Notes.