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

Which is:

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 $\leftarrow$

  • $\begingroup$ $A$ being closed is defined as $A^c\in \mathcal{T}$, i.e. a set is closed iff its complement is open (is an element of the topology). $\endgroup$ – JMoravitz May 25 '16 at 0:51

For the first proof, if you have a separation, $M = A \cup B$, then $A$ and $B$ are both open, and $A \cap B = \emptyset$. But, $A$ is also closed since $B$ is open, and $A = M \setminus B$. Same goes for $B$. So this is the contrapositive of the reverse direction of the statement.

When they say reciprocally, they are referring to the forward implication of the statement. Namely, proving that if $M$ is connected, then $M$ and $\emptyset$ are the only sets that are both open and closed. Again, they prove the contrapositive. Assume there exists some set $A \subset M$ that is both open and closed. Then $M \setminus A$ is open. Since $A \cap (M \setminus A) = \emptyset$, and $A \cup (M \setminus A) = M$, then $M$ is not connected.

Now for the second proof, they are assuming that $\partial X = \emptyset$. Thus $\partial X = \emptyset \subset X$. To see the forward implication, assume that $M$ and $\emptyset$ are the only subsets of $M$ that are both open and closed. Then $M$ and $\emptyset$ are the only sets with empty boundary. Therefore, if $X \subset M$ has empty boundary, then $X$ is either $M$ or $\emptyset$.


If $M=A \cup B$ is a separation, both $A$ and $B$ are open as you say. Then $A = B^c$ is the complement of an open set, so is closed. The recipricolly works the same way. If $A$ is open and closed, $M \setminus A=B$ must also be open and closed as the complement of a clopen set.


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