PROBLEM: Let $Y\subset X$, suppose both $X$ and $Y$ are connected and $X-Y=A*\cup* B $ where ($*\cup*$ is the set separation). Show that $Y\cup A$ is connected.

My attempt: Let $U\subset Y \cup A$ be an open and closed subset in $Y\cup A$. Then $U\cap Y$ is clopen subset of $Y.$ Since $Y$ is connected , we get either $U\cap Y=\emptyset$ or $U\cap Y=Y$. Now assume that $U\cap Y = \emptyset,$ then $U \subset A$. Since $A, B$ is a separation (given in the problem) of $X-Y$, $A$ is an open and closed subset of $X-Y$.

Now using the fact that if any subset of $Z$ that is open in $Y\cup Z$ must be open in $X$ for $Y\subset X$ and $Z$ an open subset of $X-Y$, we have applying this fact to $X, Y, A$, for both cases (open and closed), we conclude that $U\subset A$ that is assumed to be open and closed in $Y \cup A$ must be open and closed in $X$. Since $X$ is connected and $U$ is not the whole space, we conclude that $U = \emptyset$.

Where I am stuck: Now the part where I am stuck is the part where we have to assume the case for when $U\cap Y=Y$. I can not seem to rigorously argue for this case. I would appreciate some help on writing this part.

  • $\begingroup$ I've never heard of an operation called set separation. What is it? $\endgroup$ – spaceisdarkgreen Apr 19 '18 at 0:42
  • $\begingroup$ I have solved it, and the definition of separation of sets is: $A$ and $B$ are separated in $X$ if each is disjoint from the other's closure for a topological space $X.$ $\endgroup$ – Aurora Borealis Apr 19 '18 at 10:08
  • $\begingroup$ I know what two sets being separated means. But in your question there is (what looks like) an operation $*\cup*$ on two sets, resulting in a third that is called separation, that I don’t know how to make sense of. (That’s how I read $X-Y=A*\cup *B.$) $\endgroup$ – spaceisdarkgreen Apr 19 '18 at 13:15
  • $\begingroup$ There is a symbol for set separation, but I dont know the latex form of it, it essentially looks like $\cup$, but there is no curve, but rather has rectangular edges. $\endgroup$ – Aurora Borealis Apr 19 '18 at 15:13

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