We say $A \subseteq \mathbb{R}^n$ is disconnected when there exists open sets $U$ and $V$ in $\mathbb{R}^n$ such that $$U \cap A \neq \emptyset,V \cap A \neq \emptyset, U \cap V=\emptyset, \text{and} A \subseteq U \cup V$$. We say $A$ is connected if it is not disconnected.

Let $A\subseteq S \subseteq \mathbb{R}^n$. Show that if $A$ is connected in $\mathbb{R}^n$ if and only if $A$ is connected in $S$.

The forward direction is simple, but I don't know how to prove the other direction.

Here's my proof: If $A$ is disconnected in $S$,I want to show that $A$ is disconnected in $\mathbb{R}^n$. then there exist sets $U,V \subseteq S$ which are open in $S$ such that $$U \cap A \neq \emptyset, V\cap A \neq \emptyset, U \cap V=\emptyset,A \subseteq U \cup V$$

Since $U,V$ are open in $S$, there exist open sets $U_0,V_0$ in $\mathbb{R}^n$ such that $U=U_{0}\cap S$ and $V=V_{0} \cap S$. We can show that $$U_0 \cap A \neq \emptyset, V_0 \cap A \neq \emptyset, U_0 \cap V_0 = \emptyset$$ But how can i show that $A \subseteq U_0 \cup V_0$ I can only show $A \subseteq S \cap (U_0 \cup V_0)$.

  • $\begingroup$ How did you get $U_0 \cap V_0 =\emptyset$? $\endgroup$ – copper.hat Jun 25 '18 at 5:30
  • $\begingroup$ @copper.hat $\emptyset=U \cap V=U_0 \cap V_0 \cap S \Rightarrow U_0 \cap V_0 =\emptyset$ $\endgroup$ – bbw Jun 25 '18 at 5:31
  • $\begingroup$ That does not follow. How do you know that $(U_0 \cap V_0) \setminus S$ is empty? $\endgroup$ – copper.hat Jun 25 '18 at 5:33
  • $\begingroup$ @copper.hat i think you are right. What should I do then? $\endgroup$ – bbw Jun 25 '18 at 5:35
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    $\begingroup$ It's a bit confusing to talk about the "forward direction" when you've written the statement in two different directions in the title and the text. $\endgroup$ – joriki Jun 25 '18 at 5:50

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