# Prove that E is disconnected

• Nov 25th 2011, 04:41 PM
wopashui
Prove that E is disconnected
Prove that a closed subset E of a metric space (M,d) is disconnected iff there exists disjoint nonempty closed sets E1, E2 such that E=E1UE2.

for the --> direction, we have E is disconnected, then by definition there exists nonempty disjoint open sets A,B s.t E=AUB, how do I turn this into a union of 2 closed sets, we can't just say since E is closed, then its subsets is closed, right?

also need some hint for the other direction.
• Nov 25th 2011, 07:59 PM
Drexel28
Re: Prove that E is disconnected
Quote:

Originally Posted by wopashui
Prove that a closed subset E of a metric space (M,d) is disconnected iff there exists disjoint nonempty closed sets E1, E2 such that E=E1UE2.

for the --> direction, we have E is disconnected, then by definition there exists nonempty disjoint open sets A,B s.t E=AUB, how do I turn this into a union of 2 closed sets, we can't just say since E is closed, then its subsets is closed, right?

also need some hint for the other direction.

Note that $\displaystyle E-A=B$ and so $\displaystyle E-A$ is open, so $\displaystyle A$ is closed. Similarly, $\displaystyle B$ is closed. Ta-da!
• Nov 27th 2011, 07:14 PM
wopashui
Re: Prove that E is disconnected
Quote:

Originally Posted by Drexel28
Note that $\displaystyle E-A=B$ and so $\displaystyle E-A$ is open, so $\displaystyle A$ is closed. Similarly, $\displaystyle B$ is closed. Ta-da!

so i need to let E1=A and E2=B, then why is A,B are disconnection of E?
• Nov 27th 2011, 09:42 PM
Drexel28
Re: Prove that E is disconnected
Quote:

Originally Posted by wopashui
so i need to let E1=A and E2=B, then why is A,B are disconnection of E?

I'm sorry, what do you mean there? The idea is that if $\displaystyle \{A,B\}$ form a disconnection of $\displaystyle X$ (they are disjoint, open, non-empty, and union to $\displaystyle X$) then $\displaystyle A,B$ are both closed, and so can clearly be taken to be your $\displaystyle E_1,E_2$, no?
• Nov 30th 2011, 12:32 PM
wopashui
Re: Prove that E is disconnected
ok, I have done one direction, I need to do the --> direction now, supposing E is disconnected, there exists disjoint E1,E2 subset of E, such that E=E1UE2, and there exist open sets A,B such that E1 subset of A and E2 subset of B.
We need to show that E1, E2 are closed, since we know E is closed, what can we say about E1 and E2? E=E1UE2 implies E1, E2 are closed?
• Nov 30th 2011, 12:42 PM
Plato
Re: Prove that E is disconnected
Quote:

Originally Posted by wopashui
ok, I have done one direction, I need to do the --> direction now, supposing E is disconnected, there exists disjoint E1,E2 subset of E, such that E=E1UE2, and there exist open sets A,B such that E1 subset of A and E2 subset of B.
We need to show that E1, E2 are closed, since we know E is closed, what can we say about E1 and E2? E=E1UE2 implies E1, E2 are closed?

It is really difficult to follow where you are from the above.
Please state in an "if...then..." format which half you need to prove now.
• Nov 30th 2011, 06:23 PM
wopashui
Re: Prove that E is disconnected
sorry for the confusion, now i need to prove if a closed subset E of a metric space (M,d) is disconnected , then there exists disjoint nonempty closed sets E1, E2 such that E=E1UE2.

so supposing E is disconnected, there exists disjoint E1,E2 subset of E, such that E=E1UE2, and there exist open sets A,B such that E1 subset of A and E2 subset of B.
We need to show that E1, E2 are closed, since we know E is closed, what can we say about E1 and E2? E=E1UE2 implies E1, E2 are closed?
• Dec 1st 2011, 02:59 AM
Plato
Re: Prove that E is disconnected
Quote:

Originally Posted by wopashui
sorry for the confusion, now i need to prove if a closed subset E of a metric space (M,d) is disconnected , then there exists disjoint nonempty closed sets E1, E2 such that E=E1UE2.

I think I know what definition of connect set you are using.
If $\displaystyle E$ is not connected the there exists two disjoint open sets $\displaystyle A~\&~B$ each having non-empty intersection with $\displaystyle E$ such that $\displaystyle E\subset A\cup B$.
Let $\displaystyle E_1=E\cap A~\&~E_2=E\cap B$. Clearly $\displaystyle E_1\cup E_2=E.$.
Now if $\displaystyle x$ is a limit point of $\displaystyle E_i$ then $\displaystyle x$ is a limit poiint of $\displaystyle E$.
Therefore, $\displaystyle x\in E_i$ because of disjoint open sets.
• Dec 1st 2011, 07:55 AM
wopashui
Re: Prove that E is disconnected
Quote:

Originally Posted by Plato
I think I know what definition of connect set you are using.
If $\displaystyle E$ is not connected the there exists two disjoint open sets $\displaystyle A~\&~B$ each having non-empty intersection with $\displaystyle E$ such that $\displaystyle E\subset A\cup B$.
Let $\displaystyle E_1=E\cap A~\&~E_2=E\cap B$. Clearly $\displaystyle E_1\cup E_2=E.$.
Now if $\displaystyle x$ is a limit point of $\displaystyle E_i$ then $\displaystyle x$ is a limit poiint of $\displaystyle E$.
Therefore, $\displaystyle x\in E_i$ because of disjoint open sets.

sorry, are you just restating the definition, or this is the proof? I need to show that given E is closed and disconnected, then exists disjoint nonempty closed sets E1, E2 such that E=E1UE2.
• Dec 1st 2011, 08:02 AM
Plato
Re: Prove that E is disconnected
Quote:

Originally Posted by wopashui
I need to show that given E is closed and disconnected, then exists disjoint nonempty closed sets E1, E2 such that E=E1UE2.

I showed exactly that. $\displaystyle E_1~\&~E_2$ are disjoint closed non-empty sets the union of which is $\displaystyle E$.
Now what do you not understand?
• Dec 1st 2011, 08:21 AM
wopashui
Re: Prove that E is disconnected
Quote:

Originally Posted by Plato
I showed exactly that. $\displaystyle E_1~\&~E_2$ are disjoint closed non-empty sets the union of which is $\displaystyle E$.
Now what do you not understand?

for the last two lines, if the limit point is in the set, the set is closed, but what you are stating is a bit confused me
• Dec 1st 2011, 08:44 AM
Plato
Re: Prove that E is disconnected
Quote:

Originally Posted by wopashui
for the last two lines, if the limit point is in the set, the set is closed, but what you are stating is a bit confused me

We want to show that $\displaystyle E_1$ is closed.
A closed set contains all of its limit points.
So if $\displaystyle t$ is a limit point of $\displaystyle E_1$ it is also a limit point of $\displaystyle E$ and because $\displaystyle E_1\subset A$ we know that $\displaystyle t$ cannot be in $\displaystyle E_2$ so $\displaystyle t\in E_1$. Thus $\displaystyle E_1$ contains all of its limit points so it is closed.
Likewise for $\displaystyle E_2$.