Strong separation of a topological space

Question

Let $X = U \cup V$ be a disconnected topological space. Then, generally we say that $U \cap V =\emptyset$. Sometimes it is given in the definition that -e.g, Munkres' Topology- we also have $\overline{U} \cap V = \overline{V} \cap U = \emptyset $ . The book I am following is Gamelin's Topology and it only says $U \cap V =\emptyset$. How can I show $\overline{U} \cup V = \overline{V} \cup U = \emptyset $ then?

Answer 1

Hint: $U=X\backslash V$ and $V=X\backslash U$ are closed.

Answer 2

If $U$ and $V$ are open, and disjoint, $U \subseteq (X \setminus V)$ and the latter set is closed, so $\overline{U} \subseteq (X \setminus V)$, so $\overline{U} \cap V =\emptyset$ too.

Answer 3

If $A,B$ are non-empty subsets of $X$ then:

$$\overline A\cap B=\varnothing\wedge A\cap\overline B=\varnothing\iff A\cup B\text{ is a connected subspace of }X$$

I suspect this is what you meet in Munkres.

Of course this also holds under the extra condition that $A\cup B=X$, and in that case we conclude that $X$ is a connected space.

I suspect this is what you see in Gamelin.

This because in that special case we find that the condition: $$\overline A\cap B=\varnothing\wedge A\cap\overline B=\varnothing$$ is equivalent with the seemingly weaker condition: $$A\cap B=\varnothing\wedge A\cap B=\varnothing$$

Observe that $A\cup B=X\wedge A\cap\overline B=\varnothing$ implies that $A$ is the complement of $\overline B$ hence is open, and consequently that $B$ as complement of $A$ is closed (so that $B=\overline B$). Likewise we find $A=\overline A$.