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if $S$ are disconnected on his subspace topology then there exists $U \cap S$ where $U$ are open on $X$ and $H\cup S$ where $H$ are open in $X$ that $S=(U\cap S)\cup (H\cap S) = (H\cap U)\cup S$ but by this how can i find the open sets in $X$ such that $S$ is his union?

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    $\begingroup$ You also need to add that $\emptyset = (U \cap S) \cap (H \cap S) = (U \cap H) \cap S$ of course, and $U \cap S \neq \emptyset$ and $H \cap S \neq \emptyset$. $\endgroup$ – Henno Brandsma May 11 '17 at 18:15
  • $\begingroup$ You cannot find two open subsets of $X$ that exactly union up to $S$ in general. That would imply $S$ is open, in $X$ as a union of sets open in $X$, you cannot do better than $S \subseteq U \cup U$. $\endgroup$ – Henno Brandsma May 12 '17 at 4:29
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$S$ disconnected on $X$ means $S$ is disconnected in its subspace topology by defination! Your question makes no sense

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  • $\begingroup$ I feel you should elaborate more on the real essence of connectedness i.e. any continuous map from S to {0,1} will be a constant map $\endgroup$ – user379195 May 11 '17 at 18:44
  • $\begingroup$ S disconnected in X $\Longleftrightarrow$ S disconnected in his subspce topology, i want the details about $\Leftarrow$ and u mean $\Rightarrow$ $\endgroup$ – Eduardo Silva May 11 '17 at 19:15
  • $\begingroup$ @EduardoSilva What $\Leftarrow$, $\Rightarrow$ Please state equivalence between what conditions..? $\endgroup$ – Henno Brandsma May 11 '17 at 19:26
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$S = (H\cup U)\cap S $ implies $S \subseteq H \cup U$. You seem to be implying there are two definitions of a disconnected subspace, but there is only one: $S$ is disconnected in its subspace topology.

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  • $\begingroup$ but in the conditions that i put + your comments thats the definition of disconnected spaces $\endgroup$ – Eduardo Silva May 11 '17 at 19:18
  • $\begingroup$ @EduardoSilva it's the translation of being disconnected in the subspace topology. $\endgroup$ – Henno Brandsma May 11 '17 at 19:20

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