Equivalent Definition of disconnectedness of a Metric Space X

Question

I'm looking for a proof of this theorem, which states the equivalent

definition for disconnectedness of a metric space X.

Especially, I'm looking for a proof of (1) <=> (2)!

does anyone can prove this or have a proof of this theorem?

Answer 1

First, we prove that the statements (2),(3) and (4) are equivalent.

$(2) \Rightarrow (3)$: Since $X = A_1 \cup A_2$ and $A_1 \cap A_2 = \emptyset$, it follows that $A_1 = X\setminus A_2$ and $A_2 = X\setminus A_1$. As complements of open sets $A_1$ and $A_2$ must be closed.

$(3) \Rightarrow (4)$: We see that $A_1 = X \setminus A_2$ and therefore, as a complement of a closed set, $A_1$ has to be open. But $A_1$ is closed by assumption and nonempty, so we have found a proper subset of $X$ which is both closed and open.

$(4) \Rightarrow (2)$: If $A$ is a proper open and closed subset of $X$, then $B = X\setminus A$ is open and $X = A \cup B$.

Let's move on to $(1) \Leftrightarrow (2)$. We'll do this by proving the equivalence of $(1)$ and $(3)$.

$(1) \Rightarrow (3)$: Since we have $A_1 \cap \bar{A_2} = \emptyset$ and $A_2 \subseteq \bar{A_2}$ it follows that $A_1$ and $A_2$ are disjoint. Then we have $A_1 = X \setminus A_2$, because $X = A_1 \cup A_2$. But this implies $A_2 = \bar{A_2}$ (if not, then $\bar{A_2}$ contains elements from $A_1$, contradicting our assumption) and hence $A_2$ is closed. By the same argument we have that $A_1$ is closed. Therefore, $A_1$ and $A_2$ satisfy the conditions of $(3)$.

$(3) \Rightarrow (1)$: Since $A_1$ and $A_2$ are closed, it follows that $A_1 = \bar{A_1}$ and $A_2 = \bar{A_2}$. But then $A_1$ and $A_2$ already satisfy the conditions of (1), because they're disjoint.