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

Attempt at an answer: since $A,B$ form a separation of $X-Y$ then $X-Y = A \cup B$ and $A \cap B = \emptyset$. Then $X = A \cup B \cup Y = (A\cup Y) \cup (B \cup Y)$, but $X$ is connected hence it must be that $(A \cup Y) \cap ( B \cup Y) \neq \emptyset$, but this must imply that $A \cup Y$ and $B \cup Y$ are connected otherwise $X$ could possibly be disconnected if they are not connected. Is this enough, I feel like I a missing some details ( $Y$ is not necessarily an open set in $X$).

Thank you.

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    $\begingroup$ Yes, it is true (assuming that $U,V$ and $W$ are sets). $\endgroup$ – Lord Soth Nov 3 '13 at 2:05
  • $\begingroup$ okay, thanks but I am going to update the question with something more advanced and show you my problem $\endgroup$ – kolonel Nov 3 '13 at 2:08
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    $\begingroup$ Sounds great! If your question is totally different than this one, please note that you may want to consider asking a new question. $\endgroup$ – Lord Soth Nov 3 '13 at 2:09
  • $\begingroup$ okay thank you lord soth :) , i just did $\endgroup$ – kolonel Nov 3 '13 at 2:18
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Your first question:

$x\in U$ if and only if $x\in U\vee x\in U$.

Your second question:

It's the exercise 12 in section 23 Munkres.

I think your proof isn't quite right.

$Proof:$

Suppose $Y\cup A$ disconnected (The other one is similar, just swap $A$ and $B$ through out), let $U,V$ be a separation. Then $\overline{U}\cap V=\emptyset$ and $U\cap\overline{V}=\emptyset$ and $Y\cup A=U\cup V$. Similarly, since $A,B$ separates $X\setminus Y$, we have $\overline{A}\cap B=\emptyset$ and $A\cap\overline{B}=\emptyset$ and $X\setminus Y=A\cup B$.

Since $Y$ is connected, either $Y\subseteq U$ or $Y\subseteq V$. For simplicity, assume $Y\subseteq U$. Since $Y\cup A=U\cup V$ and both of which are disjoint unions, we have $V\subseteq A$, $\overline{V}\subseteq\overline{A}$ as well.

Now we can infer $\overline{V}\cap B=\emptyset$ and $V\cap\overline{B}=\emptyset$.

Since $(U\cup V)\cup B=(A\cup B)\cup Y=(X\setminus Y)\cup Y=X$. We will show $U\cup B$ and $V$ form a separation for $X$. Note $(\overline{U\cup B})\cap V=(\overline{U}\cup\overline{B})\cap V=(\overline{U}\cap V)\cup(\overline{B}\cap V)=\emptyset$. And $(U\cup B)\cap\overline{V}=(U\cap\overline{V})\cup(B\cap\overline{V})=\emptyset$. Hence they are indeed a separation for $X$, which is a contradiction.

$\mathrm{QED}$

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  • $\begingroup$ okay, thanks but I am going to update the question with something more advanced and show you my problem $\endgroup$ – kolonel Nov 3 '13 at 2:09
  • $\begingroup$ Here's a complete proof. :) $\endgroup$ – Kaa1el Nov 3 '13 at 4:45
  • $\begingroup$ thank you kaalel, i would vote u up, but i dont currently have enough credits $\endgroup$ – kolonel Nov 3 '13 at 5:11

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