Show that $B \setminus C$ is connected. Let A and B be connected sets and $ A\subset B$. If C is a component of $B \setminus A$, prove that $B \setminus C$ is connected.
I have tried to solve this problem but I am really having a hard time proving it.
I hope someone can help me. Thanks a lot.
 A: If $A = B$ then $C = \emptyset$ and $B \setminus C = B$ is connected. If $A = \emptyset$, then $C = B$ and $B \setminus C = \emptyset$ is connected. Assume that $A$ is a proper subset of $B$.
Let $C_i$ denote the connected components of $B \setminus A$ with $i \in I$ in some index set $I$. By definition, each connected component $C_i$ is a connected subset of $B \setminus A$ which is maximal in the inclusion order. Therefore, each $C_i$ is closed in $B \setminus A$.
Fix $k$ and assume $B \setminus C_k$ is not connected, i.e.
$$
B \setminus C_k = D_1 \cup D_2
$$
for $D_1$ and $D_2$ disjoint non-empty sets that are open in $B \setminus C_k$. We will derive a contradiction by showing that $B$ is not connected.
First, note that every $X \subset B \setminus C_k$ can be written as
$$
X = (X \cap D_1) \cup (X \cap D_2)
$$
where $X \cap D_i$ is open in $X$ for $i = 1, 2$. Therefore, every connected subset $X \subset B \setminus C_k$ is either contained in $D_1$ or in $D_2$. In particular, $A$ is contained in $D_1$ or in $D_2$ and for each $i \in I, i \ne k$, $C_i$ is contained in $D_1$ or in $D_2$.
Assume, without loss of generality, that $A \subset D_1$. Define
$$
E_1 = D_1 \cup C_k \\
E_2 = D_2
$$
and note that $E_1$ and $E_2$ are disjoint non-empty sets and $E_1 \cup E_2 = B$. We will show that both are open in $B$ by demonstrating that $E_2$ is clopen in $B$.
Recall that $E_2$ is open in $B \setminus C_k$, so there is an open set $G$ in $B$ such that $E_2 = G \cap (B \setminus C_k)$. Let $X'$ denote the closure of $X$ in $B$. Noting that $C_k' \subset C_k \cup A$ and $G \cap A = \emptyset$, we can write $E_2$ as the intersection of two sets open in $B$
$$
E_2 = G \cap (B \setminus C_k')
$$
so $E_2$ is open in $B$.
Similarly, recall that $E_2$ is closed in $B \setminus C_k$ so there is a closed set $F$ in $B$ such that $E_2 = F \cap (B \setminus C_k)$. Noting that $E_2' \subset E_2 \cup A$ and $F \cap A = \emptyset$, we can write $E_2$ as the intersection of two sets closed in $B$
$$
E_2 = F \cap E_2'
$$
so $E_2$ is closed in $B$.
Thus, $E_2$ is both open and closed in $B$ and consequently $B = E_1 \cup E_2$ is the union of two disjoint non-empty open sets contradicting the fact that $B$ is connected. Therefore, the assumption  that $B \setminus C_k$ can be written as the union of two disjoint non-empty open sets is false and thus $B \setminus C_k$ is connected.
