I'm trying to understand this proof that:
$M$ connected $\iff$ $M$ and $\emptyset$ are the only subsets of $M$ open and closed at the same time
If $M=A\cup B$ is a separation, then $A$ and $B$ are open and closed. Recriprocally, if $A\subset M$ is open and closed, then $M = A\cup(M-A)$. What? I know that if $M=A\cup B$ is a separation, $A$ and $B$ are both open. But why closed? Also, the 'recriprocally' part is totally nonsense to me. Anybody could help?
Also, there's another proof, which states: $M$ and $\emptyset$ are the only subsets of $M$ at the same time closed and open $\iff$ if $X\subset M$ has empty boundary, theb $X=M$ or $X=\emptyset$
which is proved as the following:
given $X\subset M$, we know the condition $X\cap \partial X = \emptyset$ implies $X$ is open, while the condition $\partial X \subset X$ implies $X$ is closed. Then, $X$ is open and closed $\iff$ $\partial X = X\cap \partial X = \emptyset$, this show $\iff$ for the theorem above.
First of all, I think that the condition $X$ has boundary empty implies that $X\cap \partial X$ is empty, but who said anything about $\partial X\subset X$? Also, where's the $\rightarrow$ of this proof? I can only see $\leftarrow$