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Let $X$ be a topologic space. Let $U,V\subseteq X$ be open, disjoint such that $U\cup V=X$ and let $A\subseteq X$ be connected. Then $A\subseteq U$ or $A\subseteq V$

Let assume that $A\not\subseteq U$ or $A\not\subseteq V$, because $U\cup V=X$ we get $A\cap V\neq \emptyset$ and $A\cap U\neq \emptyset$ corresponding. $A\cap V$ and $A\cap U$ are open, non empty, disjoint sets such that: $$(A\cap U) \cup (A\cap V)=A\cap(U\cup V)=A\cap X = A$$

Therefore $A$ is not connected. Contradiction

Why can we assume that $A\cap V$ and $A\cap U$ are open? I understand that $U,V$ are open, but why $A$ is open?

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  • $\begingroup$ Sets aren't open. They're open in a space. In this case, $A\cap U$ need not be open in $X$, but it is open in $A$ (by definition). $\endgroup$ – Chrystomath Sep 14 '20 at 14:10
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A subspace $A$ is connected iff we cannot write it as a disjoint union of relatively open (i.e. open in $A$!) subsets that are both non-empty.

$A \cap U$ and $V \cap A$ are by definition open in $A$ and they are disjoint (as $U$ and $V$ are) and they cover $A$ (all we need for that is $A \subseteq U \cup V$ and that is certainly the case here). So as $A$ is given to be connected one of them must be empty. If $A \cap U = \emptyset$ so $A \cap V = A$ which implies $A \subseteq V$. Likewise if $A \cap V = \emptyset$, $A \subseteq U$ and we're done.

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When you speak about connectedness, you have to consider the subspace topology on $A$. Because $U,V$ are open in $X$, then by definition of this topology, $A \cap U$ and $A \cap V$ are open in $A$.

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They are open in $A$, by the definition of the subspace topology. $A$ is not necessarily open in $X$; but it is open in itself.

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