1. ## Connected sets

Let A be connected subset of X and let AB ⊂ cl(A). Show that B is connected and hence, in particuar, cl(A) is connected.

Hint: (Use) Let G∪H be a disconnection of A and let B be a connected subset of A then we see that either BH= or B∩G=∅, and so either BG or B⊂H.

2. Originally Posted by AcCeylan
Let A be connected subset of X and let AB ⊂ cl(A). Show that B is connected and hence, in particuar, cl(A) is connected.
Suppose that $G\cup H=B$ is a disconnection of $B$.

That means that $G\cap B\ne\emptyset~\&~H\cap B\ne\emptyset$.

Also $G\cap\overline{H}=\emptyset~\&~H\cap\overline{G}=\ emptyset$

Using the fact that $B\subseteq\overline{A}$ prove that contradicts the given that $A$ is connected.

3. Do you mean that I must show cl(A) is disconnected and this will imply A is also disconnected. hence,the result will give a contradiction..? am I right?
if not how can I do this , can you explain, please thank you

4. Originally Posted by AcCeylan
Do you mean that I must show cl(A) is disconnected and this will imply A is also disconnected. hence,the result will give a contradiction..? am I right?
THEOREM The closure of a connected set is a connect set.
That is what you are really asked to prove.

As a lemma (previous theorem) you should have proved that if $A$ is a connected set and $A\subset G\cup H$ where $G~\&~H$ are separated sets then $A\subset G$ or $A\subset H$.
Two set are said to be separated if neither is empty and neither contains a point nor a limit of the other.

Now it is quite easy to prove the lemma. If the conclusion is false then $(A\cap G)\cup(A\cap H)=A$ is a separation of $A$.
But $A$ is connected. So that is a contradiction.

5. Thanks for everything but I have one more question..you said that
"Using the fact that B ⊂ cl(A) prove that contradicts the given that A is connected"
How can I show this?

6. Originally Posted by AcCeylan
Thanks for everything but I have one more question..you said that
"Using the fact that B ⊂ cl(A) prove that contradicts the given that A is connected"
How can I show this?
Suppose that $G\cup H=B$ is a separation of $B$.
Because $A\subseteq B$ and $A$ is connected we have $A\subseteq G$ or $A\subseteq B$.
Say $A\subseteq G$. Now is it possible for $H$ to be nonempty?
If the answer is no, then you have a contradiction. And that means that $B$ is connected.