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Let $X$ be a topological space. $A\subseteq X$ is connected if and only if for any disjoint, open sets $U,V\subseteq X$, we have $A\subseteq U\cup V$ implies $A\subseteq U$ or $A\subseteq V$.

Forward: Suppose $A$ is connected. Suppose $A\subseteq U\cup V$ for disjoint, open sets, $U,V$. Then, $A=(U\cup V)\cap A =(U\cap A) \cup (V\cap A)$. Since $A$ is connected, $U\cap A=\varnothing$ or $V\cap A=\varnothing$ If $U\cap A = \varnothing$ then since, $U\cap V=\varnothing$ , we have $A\subseteq V$.

Backwards: I suppose $A$ is disconnected. So, $A$ can be expressed as the disjoint union, of two non-empty open sets, $U_A,V_A$ i.e. $A=U_A\cup V_A$ In particular, $U_A=U\cap A$ and $V_A=V\cap A$ where $U, V$ are open in $X$. So $A\subseteq U\cup V$.

However, the issue is that $U\cap V$ may not be empty. I'm not sure how to proceed from this stage.

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  • $\begingroup$ What is your question exactly? It seems that you are trying to prove a definition, which doesn’t make sense. $\endgroup$ – Bcpicao Apr 26 '20 at 22:05
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    $\begingroup$ HINT: Replace $V$ with $V\setminus\operatorname{cl}U$. Does that change $V\cap A$? $\endgroup$ – Brian M. Scott Apr 26 '20 at 22:06
  • $\begingroup$ @Bcpicao: No, the OP is trying to prove that two characterizations of connected subset are equivalent, has proved one direction of the equivalence, and is asking for help with the other direction. $\endgroup$ – Brian M. Scott Apr 26 '20 at 22:08
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    $\begingroup$ @Brian M. Scott, the OP is not clear on which definition he uses and which he likes to prove. The definition he uses is that it can’t be written as an union? And what he is trying to prove is the first proposition? $\endgroup$ – Bcpicao Apr 26 '20 at 22:22
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    $\begingroup$ @Bcpicao: On the contrary, it’s perfectly clear from the proof that he's attempting what definition he’s been given, and what he’s trying to prove is clearly stated at the top of the question. The definition of disconnected subset that he has is evidently that $A$ is the disjoint union of two non-empty relatively open subsets of $A$; the characterization that he’s trying to prove equivalent follows if and only if in the statement of the theorem at the top of the question. $\endgroup$ – Brian M. Scott Apr 26 '20 at 22:25
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As you wish to prove the backwards implication is true by proving its converse is true, explicitly negating the proposition may help you tackle the problem:

$A$ is disconnected $\implies$ $\exists U,V\subset X$ disjoint open sets s.t. $A\in U \cup V$ and $A\nsubseteq U$ and $A\nsubseteq V$.

Is it clearer now, how you should conclude?

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