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Let $A$ and $B$ be connected subspaces of a topological space $(X,\tau)$. If $A\cap B\neq\emptyset$, prove that the subspace $A\cup B$ is connected.

If the subspace $A\cup B$ is not connected, then there exist, $\mathscr{U},\mathscr{V}\subset X$ such that $\mathscr{U}\cup\mathscr{V}=A\cup B$ and $\mathscr{U}\cap\mathscr{V}=\emptyset$. $\mathscr{U},\mathscr{V}$ must belong either to $A$ or $B$, like $\mathscr{U}\in A$, which contradicts the fact $A$ and $B$ are connected. Therefore $A\cup B$ is connected.

Questions:

Is my proof right? If not. How should I prove the statement?

Thanks in advance!

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    $\begingroup$ How $\mathscr{U}, \mathscr{V} \subset A \;\text{or} \;B$ ? $\endgroup$ – Chinnapparaj R Oct 19 '18 at 14:58
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    $\begingroup$ That $A\cap B\neq\emptyset$ is an important assumption. Your proof has no hope to be true if you make no use of it. $\endgroup$ – AdditIdent Oct 19 '18 at 15:00
  • $\begingroup$ @ChinnapparajR I did not say subset but $\in$ $\endgroup$ – Pedro Gomes Oct 19 '18 at 15:00
  • $\begingroup$ @AdditIdent Could you tell me how should I use that assumption? $\endgroup$ – Pedro Gomes Oct 19 '18 at 15:01
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    $\begingroup$ Your proof is obviously flawed -- note that you can just take $\mathscr U = A$ and $\mathscr V = B\setminus A$. You've made no further demands on $\mathscr U$ or $\mathscr V$ other than $\mathscr U \cup \mathscr V = A\cup B$ and $\mathscr U \cap \mathscr V = \varnothing$. $\endgroup$ – MPW Oct 19 '18 at 15:06
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A subspace $Y$ of a topological is disconnected (w.r.t. subspace topology) if there exists two non- empty open sets $U,V$ in the subspace topology of $Y$ such that $U\bigcup V=Y$ and $U\bigcap V=\phi$. Representation of $Y$ as $Y=U\bigcup V$ by of sets of prescribed properties is called a disconnection.

Here $Y=A\bigcup B$ with $A\bigcap B\not=\phi$ and $A,B$ are connected subspaces of $X$. So if possible let $Y$ is disconnected w.r.t. subspace topology then we can find two sets $U,V$ having the properties of 1st paragraph. Now since $A$ is connected $A $ is contained in one of the open set , say $U$ (otherwise $A=(A\cap U)\bigcup (A\cap V)$ will be a disconnection of $A$). Similarly $B$ is also contained in one of the sets $U,V$. Now since $A\bigcap B\not=\phi$ we can say $B$ is contained in $U$. Hence $V=\phi$. Therefore $A\bigcup B$ is connected.

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  • $\begingroup$ The contradiction lies in the fact $V$ cannot be $\emptyset$, since $Y$ is disconnected, right? $\endgroup$ – Pedro Gomes Oct 19 '18 at 15:21
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    $\begingroup$ Yes , if you start with the assumption that $Y$ is disconnected. $\endgroup$ – Sumanta Das Oct 19 '18 at 15:23
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Call $p$, the point of intersection. Then $p$ lies in either $\mathscr{U}$ or $\mathscr{V}$. Suppose $p\in \mathscr{U}$. Since $A$ and $B$ are connected, it must lie completely in $\mathscr{U}$ or $\mathscr{V}$ and it cannot lie in $\mathscr{V}$, since it contains the point $p$ of $\mathscr{U}$. Consequently $A \cup B \subset \mathscr{U}$, which means $\mathscr{V}=\phi$, a contradiction!

[Here we use the fact that " if $Y$ is a connected subspace of $X=C \cup D$, then $Y \subset C$ or $Y \subset D$ ]

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