Let $Y\subset X$ be such that both $X$ and $Y$ are connected. Show that if $A$ and $B$ is a separation of $X-Y$, then $Y\cup A$ and $Y\cup B$ are connected.
I found a proof for this problem in this post. Here it is:
Here is my attempt for QN2 I think to prove it by contradiction, assume $Y\cup A$ and $Y\cup B$ are not connected then for $P$ and $Q$ disjoint $Y\cup A=P\coprod Q$ and for $M$ and $N$ disjoint $Y\cup B=M\coprod N$ $(Y\cup A)\cup (Y\cup B)=(P\coprod Q)\cup (M\coprod N)$. The left side will give $X$, and the right side can be written as a disjoint union, this contradicts the fact that $X$ is connected, so $Y\cup A$ and $Y\cup B$ must be connected
Now, I find that there is a logical gap in the proof which is. The proof shows us that: it can't happen that $Y\cup A$ and $Y\cup B$ are both not connected. So at least one of the two subspaces are connected. But it doesn't show that both are. How to overcome this obstacle?
Any ideas?
You may show
by assuming
and then derive a contradiction.
You do not show above statement by assuming
and deriving a contradiction.
An attempt for a proof could be:
Say $X-Y=A+B$ where $A$ and $B$ are $(X-Y)$-open (i.e. $A$ and $B$ are a separation of $X-Y$) and say $Y\cup A$ NOT connected (this should lead to a contradiction).
Say $Y\cup A = P + Q$ is a separation (i.e. $Q$ and $P$ $(Y\cup A)$-open)
Then $P\cap Y=\emptyset$ or $Q\cap Y=\emptyset$ (since $Y$ is connected)
Say $P\cap Y=\emptyset$
Then $Y\subset Q$ and $$\begin{aligned} X &=A\cup Y\cup B\\ &=P + (Q \cup B) \end{aligned}$$
But still $P$, $Q$ and $B$ are not said to be open in $X$ like $$X=P' + Z$$ where $P'$ and $Z$ are $X$-open and disjoint - what is $P'$ and $Z$?