# Connectedness

• May 4th 2009, 07:23 PM
h2osprey
Connectedness
1) Let (X, d) be a metric space and let A, B be two closed subsets of X such that the union and intersection of A and B are connected. Prove that A is connected.

2) Let (X, d) be a connected metric space and let A be a connected subset of X. Assume that the complement of A is the union of two separated sets B and C. Prove that the union of A and B are connected.

• May 5th 2009, 08:47 AM
manjohn12
1. Suppose for contradiction that A is not connected. Then you can write it as the union of 2 non-empty separated sets.

2. So A^C is not connected. Now suppose $\displaystyle A \cup B$ is not connected. Then you can write it as the union of 2 non empty separable sets.
• May 5th 2009, 12:13 PM
h2osprey
I've tried, but I didn't manage to get anywhere.
• May 5th 2009, 12:53 PM
Plato
Suppose that $\displaystyle A$ is not connected. Then there is a separation of $\displaystyle A = H \cup K$.
Where $\displaystyle \begin{gathered} \emptyset \ne H \subseteq A\;\& \;\emptyset \ne K \subseteq A \hfill \\ \overline H \cap K = \emptyset \;\& \;\overline K \cap H = \emptyset \hfill \\ \end{gathered}$.

If $\displaystyle B \cap H = \emptyset$ then $\displaystyle H \cup \left( {B \cup K} \right)$ is a separation of $\displaystyle A\cup B$, contradiction.

So $\displaystyle \left( {B \cap H} \right) \cup \left( {B \cap K} \right) = B \cap A$ is also a separation of $\displaystyle A\cap B$.

Now you have to do the other.
• May 5th 2009, 03:01 PM
h2osprey
Quote:

Originally Posted by Plato
Suppose that $\displaystyle A$ is not connected. Then there is a separation of $\displaystyle A = H \cup K$.
Where $\displaystyle \begin{gathered} \emptyset \ne H \subseteq A\;\& \;\emptyset \ne K \subseteq A \hfill \\ \overline H \cap K = \emptyset \;\& \;\overline K \cap H = \emptyset \hfill \\ \end{gathered}$.

If $\displaystyle B \cap H = \emptyset$ then $\displaystyle H \cup \left( {B \cup K} \right)$ is a separation of $\displaystyle A\cup B$, contradiction.

So $\displaystyle \left( {B \cap H} \right) \cup \left( {B \cap K} \right) = B \cap A$ is also a separation of $\displaystyle A\cap B$.

Now you have to do the other.

For 2) I managed to get $\displaystyle A = ((X\setminus B) \cap D1) \cup ((X\setminus B) \cap D2)$ and also, since $\displaystyle B \cap C = \emptyset, (X\setminus B) \cup (X\setminus C) = X$
Hence $\displaystyle X = (D1 \cup (X\setminus B)) \cup D2$, but I got stuck since assuming (without loss of generality) $\displaystyle ((X\setminus B) \cap D1)$ is empty doesn't guarantee that $\displaystyle (D1 \cup (X\setminus B)) \cup D2$ are separated (thus drawing the contradiction since X is connected).

Am I using the wrong approach?
• May 5th 2009, 03:15 PM
h2osprey
Quote:

Originally Posted by Plato
Suppose that $\displaystyle A$ is not connected. Then there is a separation of $\displaystyle A = H \cup K$.
Where $\displaystyle \begin{gathered} \emptyset \ne H \subseteq A\;\& \;\emptyset \ne K \subseteq A \hfill \\ \overline H \cap K = \emptyset \;\& \;\overline K \cap H = \emptyset \hfill \\ \end{gathered}$.

If $\displaystyle B \cap H = \emptyset$ then $\displaystyle H \cup \left( {B \cup K} \right)$ is a separation of $\displaystyle A\cup B$, contradiction.

So $\displaystyle \left( {B \cap H} \right) \cup \left( {B \cap K} \right) = B \cap A$ is also a separation of $\displaystyle A\cap B$.

Now you have to do the other.

For 2) I managed to get $\displaystyle A = ((X\setminus B) \cap D1) \cup ((X\setminus B) \cap D2)$ and also, since $\displaystyle B \cap C = \emptyset, (X\setminus B) \cup (X\setminus C) = X$
Hence $\displaystyle X = (D1 \cup (X\setminus B)) \cup D2$, but I got stuck since assuming (without loss of generality) $\displaystyle ((X\setminus B) \cap D1)$ is empty doesn't guarantee that $\displaystyle (D1 \cup (X\setminus B)) \cup D2$ are separated (thus drawing the contradiction since X is connected).

Am I using the wrong approach?
• May 5th 2009, 07:03 PM
manjohn12
2. Suppose $\displaystyle A \cup B$ is not connected. Then $\displaystyle A \cup B = H \cup K$ for separable sets $\displaystyle H$ and $\displaystyle K$. Now define $\displaystyle A_1 = H \cap A$ and $\displaystyle A_2 = K \cap A$. Then $\displaystyle A = A_1 \cup A_2$ where $\displaystyle A_1$ and $\displaystyle A_2$ are separable. This implies that $\displaystyle A$ is not connected. Contradiction.
• May 5th 2009, 07:33 PM
h2osprey
Quote:

Originally Posted by manjohn12
2. Suppose $\displaystyle A \cup B$ is not connected. Then $\displaystyle A \cup B = H \cup K$ for separable sets $\displaystyle H$ and $\displaystyle K$. Now define $\displaystyle A_1 = H \cap A$ and $\displaystyle A_2 = K \cap A$. Then $\displaystyle A = A_1 \cup A_2$ where $\displaystyle A_1$ and $\displaystyle A_2$ are separable. This implies that $\displaystyle A$ is not connected. Contradiction.

How would you prove that $\displaystyle A_1 = H \cap A$ and $\displaystyle A_2 = K \cap A$ are non-empty?
• May 6th 2009, 04:40 AM
manjohn12
Suppose $\displaystyle A \cup B$ is not connected. We know that $\displaystyle X-A = B \cup C$ is a separation of $\displaystyle X-A$. Let $\displaystyle A \cup B = H \cup K$ be a separation of $\displaystyle A \cup B$. Then $\displaystyle \overline{B} \subset X-C$, $\displaystyle \overline{C} \subset X-B$, $\displaystyle \overline{H} \subset X-K$ and $\displaystyle \overline{K} \subset X-H$. Now $\displaystyle A \cup B \cup C = X = C \cup H \cup K$. Now $\displaystyle A$ is completely contained in either $\displaystyle H$ or $\displaystyle K$. WLOG suppose $\displaystyle A \subset H$. Then $\displaystyle K = H \cup K-H \subset A \cup B-A \subset B$. Also $\displaystyle \overline{K} \subset \overline{B} \subset X-C$. Then $\displaystyle \overline{C \cup H} = \overline{C} \cup \overline{H} \subset (X-B) \cup (X-K) = (C \cup A) \cup (C \cup H) \subset C \cup H$. Also $\displaystyle \overline{K} \subset (X-C) \cap (X-H) = K$. So $\displaystyle C \cup H = \overline{C \cup H}$ and $\displaystyle K = \overline{K}$ are closed. Hence $\displaystyle X = (C \cup H) \cup K$ which is a separation of $\displaystyle X$. Contradiction.