Problem of the Week #73 - October 21st, 2013

[Chris L T521]

Here's this week's problem.

-----

Problem: Let $X$ and $Y$ be connected topological spaces and let $Y\subset X$. Show that if $A$ and $B$ form a separation of $X-Y$, then $Y\cup A$ and $Y\cup B$ are connected.

-----

 

[Chris L T521]

No one answered this week's question. You can find the solution below.

[sp]Since $A$ and $B$ are non-empty sets in $X-Y$, there exist non-empty open sets $U_A$ and $U_B$ of $X$ such that $A=U_A\cap(X-Y)$ and $B= U_B\cap(X-Y)$, so $X-Y$ can be written as a disjoint union of $A$ and $B$, i.e.\[X-Y = A\sqcup B = [U_A\cap(X-Y)]\sqcup [U_B\cap(X-Y)]\]
Now, suppose that $Y\cup A$ is not connected. Then $Y\cup A$ has a separation; thus there exist non-empty open sets $U_C$ and $U_D$ of $X$ such that
\[Y\cup A = [U_C\cap(Y\cup A)] \sqcup [U_D\cap(Y\cup A)]\]
where both $U_C\cap(Y\cup A)$ and $U_D\cap(Y\cup A)$ are non empty. Hence, we have that
\[Y = Y \cap (Y\cup A) = (U_C\cap Y) \sqcup (U_D\cap Y).\]
Since $Y$ is connected, we can assume that $U_C\cap Y=Y$, i.e. $Y\subset U_C$ and $U_D\cap Y=\emptyset$. In this case, $A\cap U_D\neq \emptyset$ and thus
\[A=(U_C\cap A)\sqcup (U_D\cap A).\]
This, together with the fact that $Y\subset U_C$, $A\subset U_A$ and $B\subset U_B$ implies that
\[\begin{aligned}X=Y\sqcup A\sqcup B &\subset (U_B\cup U_C)\cup A\\ &\subset (U_B\cup U_C)\cup (U_D\cap A) \\ &\subset (U_B\cup U_C)\cup(U_A\cap U_D)\\ &\subset X\end{aligned}\]
Now, note that $\emptyset \neq A\cap U_D \subset U_A\cap U_D$ and $(U_B\cup U_C)\cap (U_A\cap U_D) = \emptyset$ since
\[(U_A\cap U_D)\cap U_B \subset U_A\cap U_B=\emptyset\quad\text{and}\quad (U_A\cap U_D)\cap U_C\subset U_D\cap U_C = \emptyset\]
With this, we see that $Y$, $A$ and $B$ form a separation on $X$, which contradicts the fact that $X$ is connected.[/sp]