1. ## [SOLVED] Disconnected Set

Problem:
Suppose that A is a subset of $\mathbb{R}^n$ that fails to be connected(i.e. disconnected), and let U, V be open subset of $\mathbb{R}^n$ that separate A.
Suppose that B is a subset of A that is connected. Prove that either $B \subseteq U$ or $B \subseteq V$

Proof:
Let U, V be disjoint open subsets that separate A, then by definition we know three things:
(i) $A \cap U = \emptyset$ and $A \cap V = \emptyset$
(ii) $(A \cap U ) \cap (A \cap V) = \emptyset$
(iii) $(A \cup U ) \cap ( A \cup V) = A$

Let $B \subset A$ that is connected.

This is what I believe it to be, but don't understand why it's true.
$B \cap U \neq \emptyset \implies B \subseteq U$
$B \cap V \neq \emptyset \implies B \subseteq V$

If B is a subset of A, then shouldn't U and V disconnect B as well? Thus, B can not be contained in U and B can not be contained in V.

Thank you so much for reading. Any help is greatly appreciated.

2. Originally Posted by Paperwings
Problem:
Suppose that A is a subset of $\mathbb{R}^n$ that fails to be connected(i.e. disconnected), and let U, V be open subset of $\mathbb{R}^n$ that separate A.
Suppose that B is a subset of A that is connected. Prove that either $B \subseteq U$ or $B \subseteq V$
Try this.
If $B\subseteq A$ then $(B \cap U) \cup (B\cap V) = B$.
Now $(B\cap U)\cap (B\cap V) = \emptyset$.
Therefore $B\cap U = B$ or $B\cap V = B$.
Thus, $B\subseteq U$ or $B\subseteq V$.