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I'm studying Intro to Topology by Mendelson.

The problem statement is,

Let $A,B\subset X$, $X$ a topological space. If $A$ is connected, $B$ open and closed, and $A\cap B\neq\emptyset$ then $A\subset B$.

My proof is,

By way of contradiction, suppose that $A$ is not a subset of $B$. Then there exists an $a\in A$ such that $a\in C(B)$. Consider the sets $P=A\cap B$ and $Q=A\cap C(B)$. Note that both $P$ and $Q$ are nonempty and open. Thus, we have that $A\subset A\cap B\cup A\cap C(B)$ and $P\cap Q=A\cap B\cap C(B)=A\cap\emptyset=\emptyset\subset C(A)$. Also, $P\cap A\neq\emptyset$ and $Q\cap A\neq\emptyset$. Therefore, $A$ is disconnected, which is a contradiction, since $A$ is assumed connected.

I'm not sure if this was even the right approach, but it's my best shot so far.

Thanks for any hints or feedback!

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  • $\begingroup$ $A\cap B$ is open and closed as a subset of $A$ with the relative topology. So either $A\cap B=\emptyset$ or $A\cap B=A$. $\endgroup$ – egreg Jul 24 '13 at 9:21
  • $\begingroup$ This could surely made less indirect and also you could take $P=B$ and $Q=C(B)$ without intersecting them with $A$ first. $\endgroup$ – Hagen von Eitzen Jul 24 '13 at 9:23
  • $\begingroup$ @HagenvonEitzen, I didn't think of approaching it that way. I'll try that and see where I get. Thanks for the response. $\endgroup$ – Shant Danielian Jul 24 '13 at 9:29
  • $\begingroup$ @Hagen: Whether you can take $P=B$ and $Q=X\setminus B$ depends on how disconnected has been defined and on what’s been proved to be equivalent. $\endgroup$ – Brian M. Scott Jul 24 '13 at 9:29
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Your argument is essentially correct, but the wording needs a bit more care in one place. When you say that the sets $P=A\cap B$ and $Q=A\setminus B$ are open, the default reading is ‘open in $X$’, which isn’t necessarily the case. You want to say that $P$ and $A$ are open in $A$. Specifically, $P$ and $Q$ are non-empty disjoint relatively open subsets of $A$ whose union is $A$, and therefore $A$ is not connected.

Added: If you’ve not already done so, you might find it useful for future reference to prove that a set $A$ in a space $X$ is disconnected iff it has a non-empty, proper subset that is clopen in the relative topology on $A$. Here you could have applied that result immediately: $B\cap A$ is clearly non-empty and relatively clopen in $A$, so if $A$ is connected ...

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  • $\begingroup$ Would I already know by default that $P$ and $Q$ are open in $A$ or would I show that? It's not absolutely clear how that's the case, but I would have to think about it first. $\endgroup$ – Shant Danielian Jul 24 '13 at 9:35
  • $\begingroup$ @Shant: $B$ is open in $X$, so by the definition of the relative topology $B\cap A$ is open in $A$. Similarly, $X\setminus B$ is open in $X$, so ... ? $\endgroup$ – Brian M. Scott Jul 24 '13 at 9:37
  • $\begingroup$ Which would mean that $(X-B)\cap A$ is also relatively open in $A$. Thank you for helping connect the dots. $\endgroup$ – Shant Danielian Jul 24 '13 at 9:42
  • $\begingroup$ @Shant: Yep, you’ve got it. You’re welcome. $\endgroup$ – Brian M. Scott Jul 24 '13 at 9:45
  • $\begingroup$ @BrianM.Scott Since the set B is open and closed in X, I can conclude that the set B∩A is open and closed in A. But A is connected, then B∩A should be either empty or the whole set A. Is that right? $\endgroup$ – Melissa Oct 24 '17 at 17:01
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It's simpler if you consider that $A\cap B$ is open and closed in the subspace $A$. Thus, since $A$ is connected, either $$ A\cap B=\emptyset\quad\text{or}\quad A\cap B=A $$

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  • $\begingroup$ How do you know that $A\cap B$ is open and closed in the subspace $A$? Are you saying that both $A$ and $B$ are open and closed in the subspace $A$? Yet, $B$ is open and closed in $X$. Is it necessarily open in $A$ as well? $\endgroup$ – Shant Danielian Jul 24 '13 at 9:32
  • $\begingroup$ @ShantDanielian $A\cap B$ is open in $A$ by definition of relative topology. If $C$ is closed in $X$, then it's easy to see that $A\cap C$ is closed in $A$ (with respect to the relative topology). $\endgroup$ – egreg Jul 24 '13 at 9:38
  • $\begingroup$ Thank you egreg for your response, it makes more sense now. $\endgroup$ – Shant Danielian Jul 24 '13 at 9:43

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