This question already has an answer here:

Union of connected subsets is connected if intersection is nonempty

I don't understand why A∩F and B∩F are relatively open where Brian Scott commented.



marked as duplicate by Did, Siminore, user99914, Aloizio Macedo, drhab Oct 4 '15 at 14:40

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  • $\begingroup$ $A, B$ are open in $\bigcup F$ and $F \subset \bigcup F$. Thus $A\cap F, B\cap F$ are relatively open in $F$. $\endgroup$ – user99914 Oct 4 '15 at 9:28
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    $\begingroup$ Why not ask this on the other page? There is a tool named comments... $\endgroup$ – Did Oct 4 '15 at 9:30
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    $\begingroup$ @Did I suppose that the reason might be that the OP is below 50 reputation points, so they can only comment on their posts and answers to their questions. (This does not change the fact that this question is a bit unclear. It refers to a comment by Brian M. Scott. I guess the OP wanted to refer to answer by Brian M. Scott...?) $\endgroup$ – Martin Sleziak Oct 4 '15 at 9:59
  • $\begingroup$ I believe the OP is talking about the 2 comments after Brian M. Scott's answer. It is not clear to the OP why $A\cap F$ and $B\cap F$ are relatively open in $F$. $\endgroup$ – R_D Oct 4 '15 at 13:39

$A \cap F$ and $B \cap F$ are open in $F$ by definition of subspace topology on $F$. A subset $X$ of $F$ is open in $F$ iff there exists an open subset $U$ of $M$ such that $X = F \cap U$.

Since $A$ and $B$ are open subsets of $\bigcup \mathscr{F}$ there are open sets $U$ and $V$ in $M$ such that $A=U \bigcap (\bigcup \mathscr{F})$ and $B = V \bigcap (\bigcup \mathscr{F})$. So $A\cap F = U \cap F$ and $B \cap F = V \cap F$, making them open in $F$.


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