Is the following set connected given that the union and intersection is connected Suppose $U_1, U_2$ are open sets in a space $X$. Suppose $U_1 \cap U_2$ and $U_1 \cup U_2$ are connected. Can we conclude that $U_1$ must be connected??
I am trying to find a counterexample, but I failed. PErhaps it is true? Can someone help me find a counter example? thanks
 A: Suppose $A$ and $B$ form a separation of $U_1$, i.e., $A$ and $B$ are disjoint nonempty open sets such that $A\cup B = U_1$. Because $U_1 \cap U_2$ is a connected subset of $U_1$, it must be entirely contained in either $A$ or $B$ (else we would get a separation of $U_1 \cap U_2$ by intersecting $A$ and $B$ with $U_1 \cap U_2$); WLOG let $U_1 \cap U_2 \subset A$. Then, $A\cup U_2$ and $B$ forms a separation of $U_1 \cup U_2$, a contradiction.
A: Suppose $U_1$ is disconnected. Then $U_1 = V \cup W$ where $V$ and $W$ are non-empty, disjoint, open subsets of $X$. Since $U_1 \cap U_2$ is connected, one of $V \cap U_2$ and $W \cap U_2$ must be empty. Suppose without loss of generality that $V \cap U_2 = \varnothing$. We have $U_1 \cup U_2 = V \cup W \cup U_2$. Note that $V$ and $W \cup U_2$ are non-empty and open. Furthermore, $V \cap (W \cup U_2) = \varnothing$. It follows that $U_1 \cup U_2$ is disconnected.
A: We assume that both $U_1$ and $U_2$ are nonempty (otherwise the result is trivial). If $U_1=A\cup B$, being $A$ and $B$ disjoint open sets, then $U_1\cap U_2$ is the disjoint union of the open sets $A\cap U_2$ and $B\cap U_2$, so by connectedness one of the sets is empty, say $A\cap U_2=\emptyset$. Therefore $U_1\cup U_2$ is the disjoint union of the open disjoint sets $A$ and $B\cup U_2$, so again by connectedness one of these sets is empty, namely $A$ (because $U_2\ne\emptyset$ by assumption). This proves that $U_1$ is connected.
A: I'm assuming all of the above are non-empty.
If we can't separate $U_1 \cap U_2$ into disjoint sets, how could we possible separate $U_1$ into disjoint sets?
That is, suppose for contradiction that $U_1 = S_1 \sqcup S_2$ (which is to say $U_1 = S_1 \cup S_2$ with $S_1 \cap S_2 = \emptyset$). Then we have 
$$U_1 \cap U_2 = (S_1 \sqcup S_2) \cap U_2 = (S_1 \cap U_2) \sqcup (S_2 \cap U_2).$$ But then we've just written $U_1 \cap U_2$ as a disjoint union, contradicting our assumption that $U_1 \cap U_2$ is connected.
