# Connectedness

• May 4th 2009, 08:23 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, 09:47 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 $A \cup B$ is not connected. Then you can write it as the union of 2 non empty separable sets.
• May 5th 2009, 01:13 PM
h2osprey
I've tried, but I didn't manage to get anywhere.
• May 5th 2009, 01:53 PM
Plato
Suppose that $A$ is not connected. Then there is a separation of $A = H \cup K$.
Where $\begin{gathered} \emptyset \ne H \subseteq A\;\& \;\emptyset \ne K \subseteq A \hfill \\ \overline H \cap K = \emptyset \;\& \;\overline K \cap H = \emptyset \hfill \\
\end{gathered}$
.

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

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

Now you have to do the other.
• May 5th 2009, 04:01 PM
h2osprey
Quote:

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

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

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

Now you have to do the other.

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

Am I using the wrong approach?
• May 5th 2009, 04:15 PM
h2osprey
Quote:

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

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

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

Now you have to do the other.

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

Am I using the wrong approach?
• May 5th 2009, 08:03 PM
manjohn12
2. Suppose $A \cup B$ is not connected. Then $A \cup B = H \cup K$ for separable sets $H$ and $K$. Now define $A_1 = H \cap A$ and $A_2 = K \cap A$. Then $A = A_1 \cup A_2$ where $A_1$ and $A_2$ are separable. This implies that $A$ is not connected. Contradiction.
• May 5th 2009, 08:33 PM
h2osprey
Quote:

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

How would you prove that $A_1 = H \cap A$ and $A_2 = K \cap A$ are non-empty?
• May 6th 2009, 05:40 AM
manjohn12
Suppose $A \cup B$ is not connected. We know that $X-A = B \cup C$ is a separation of $X-A$. Let $A \cup B = H \cup K$ be a separation of $A \cup B$. Then $\overline{B} \subset X-C$, $\overline{C} \subset X-B$, $\overline{H} \subset X-K$ and $\overline{K} \subset X-H$. Now $A \cup B \cup C = X = C \cup H \cup K$. Now $A$ is completely contained in either $H$ or $K$. WLOG suppose $A \subset H$. Then $K = H \cup K-H \subset A \cup B-A \subset B$. Also $\overline{K} \subset \overline{B} \subset X-C$. Then $\overline{C \cup H} = \overline{C} \cup \overline{H} \subset (X-B) \cup (X-K) = (C \cup A) \cup (C \cup H) \subset C \cup H$. Also $\overline{K} \subset (X-C) \cap (X-H) = K$. So $C \cup H = \overline{C \cup H}$ and $K = \overline{K}$ are closed. Hence $X = (C \cup H) \cup K$ which is a separation of $X$. Contradiction.