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Let $X$ and $Y$ be connected, and let $Y \subseteq X$. If $A$ and $B$ are two non-empty, disjoint open sets (open in the subspace $X-Y$) whose union is $X-Y$, or in other words if $A$ and $B$ form a separation of $X-Y$, then how to prove that $Y \cup A$ and $Y \cup B$ are connected?

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2 Answers 2

HINT: suppose $C$ and $D$ are open in $X$ and form a separation of $Y \cup A$. Then, since $Y$ is connected, it must lie entirely in one of $C$ or $D$, suppose it is $C$. But then $D \cap A$ and $C \cup B$ form a separation of $X$.

What's left to prove is that this works even when $B$ and $A$ are not open in $X$.

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Do you mean $C\cup B$ and $D\cap A$? Cause $C\cup A=Y\cup A$. –  Stefan Hamcke Feb 13 '13 at 14:19
    
@Stefan: indeed I switched the letters. Thank you :) –  Marek Feb 13 '13 at 14:27

Here is a fancier solution. We will use the following:

Lemma: Let $p: X \rightarrow Y$ be a quotient map. If each fiber $p^{-1}(\{y\})$ is connected, and $Y$ is connected, then $X$ is also connected.

(this lemma is discussed here : How to prove this result involving the quotient maps and connectedness?)

Collapse the subset $Y \cup A$ of $X$ to a point; call the resulting quotient space $M$. Consider the composition $p: Y \cup B \hookrightarrow X \overset{q}{\rightarrow} M $ (where $q$ is the corresponding quotient map). Now, it remains to show that the conditions of the Lemma above apply to the map $p: Y\cup B \rightarrow M$:

  • the image set $M$ is connected (as image of surjective continuous map $q$)
  • for every element $y \in M$ the fiber $p^{-1}(\{y\})$ is connected (each such fiber is either a one-point subset of $B$ or coincides with the connected space $Y$)
  • $p$ is a quotient map. (Proof: $p$ is clearly surjective; it is continuous as composition of imbedding and quotient map. Showing that it takes saturated closed subsets $C \subseteq Y \cup B$ to closed sets of $M$ takes a bit more effort and comes down to two cases:

Case 1: $C \subseteq B$, $C\cap Y=\emptyset$. Then $C$ will also be a closed subset of $X$(because $\overline C \cap (Y \cup B)=C$ and $\overline C \subseteq \overline B$ so that $\overline C \cap A =\emptyset$) and saturated with respect to $q$ so that $p(C)=q(C)$ is closed in $M$.

Case 2: $Y \cap B \supseteq C\supseteq Y$. Then, $C\cup A$ will be a closed subset of $X$ (because $C\cup A \supseteq A \cup Y$ and $\overline A \cap B=\emptyset$) which is saturated with respect to $q$ so that $p(C)=q(C\cup A)$ is closed in $M$.

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