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.

Re: Prove that E is disconnected

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**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!

Re: Prove that E is disconnected

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**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?

Re: Prove that E is disconnected

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**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?

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?

Re: Prove that E is disconnected

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**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.

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?

Re: Prove that E is disconnected

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**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.

Re: Prove that E is disconnected

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**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.

Re: Prove that E is disconnected

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**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?

Re: Prove that E is disconnected

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**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

Re: Prove that E is disconnected

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**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 $\displaystyle E_2$.