# If M is disconnected....

• Apr 6th 2010, 12:08 PM
cp05
If M is disconnected....
Let (M,d) be a metric space. Prove the equivalence of the following:
(i)M is disconnected
(ii)There exists two nonempty, disjoint closed sets in M whose union is M
(iii)There exists two nonempty, disjoint open sets in M whose union is M
(iv)There exists a set A in M such that emptyset != A != M, and A is both open and closed.

these are basically the definitions of a disconnected set aren't they? I have no idea how to prove these =/
• Apr 6th 2010, 01:37 PM
Drexel28
Quote:

Originally Posted by cp05
Let (M,d) be a metric space. Prove the equivalence of the following:
(i)M is disconnected
(ii)There exists two nonempty, disjoint closed sets in M whose union is M
(iii)There exists two nonempty, disjoint open sets in M whose union is M
(iv)There exists a set A in M such that emptyset != A != M, and A is both open and closed.

these are basically the definitions of a disconnected set aren't they? I have no idea how to prove these =/

Well what's your definition of disconnected? I assume that $X$ can be written as the disjoint union of open sets.

$i)\implies ii)$: This is clear since if $A\amalg B=X$ (where $A,B$ are open) then $A=X-B$ and $B=X-A$ both of which are disjoint and closed. So then, $X=A\amalg B=\left(X-B\right)\amalg \left(X-A\right)$

$iii)\implies IV)$ Suppose that $A\amalg B=X$ is a disconnection of $X$ (with $A,B$ closed). Then, $B$ is closed by assumption and $B=X-A$ and since $A$ is closed it folllows that $B$ is open.

Oops...I just noticed that I supposed wrong your definition of disconnected. What is it?
• Apr 6th 2010, 01:40 PM
cp05
Disconnected if there exists sets A and B in M such that
1. A != null != B
2. D= A U B
3. clos(A) n B=null=A n clos(B) iff A n B=null
• Apr 6th 2010, 01:46 PM
Drexel28
Quote:

Originally Posted by cp05
Disconnected if there exists sets A and B in M such that
1. A != null != B
2. D= A U B
3. clos(A) n B=null=A n clos(B) iff A n B=null

Oh god, separated sets. That's so gross. I don't understand the last part though.

Usually it is " $X=A\cup B$ where $A,B$ are non-empty and $\overline{A}\cap B=B\cap\overline{A}=\varnothing$"

Is that what you mean to say?
• Apr 6th 2010, 01:47 PM
cp05
yes
• Apr 6th 2010, 01:51 PM
Drexel28
Quote:

Originally Posted by cp05
yes

Look here.