$A\cup B$ is connected proof

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?

• How $\mathscr{U}, \mathscr{V} \subset A \;\text{or} \;B$ ? – Chinnapparaj R Oct 19 '18 at 14:58
• 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. – AdditIdent Oct 19 '18 at 15:00
• @ChinnapparajR I did not say subset but $\in$ – Pedro Gomes Oct 19 '18 at 15:00
• @AdditIdent Could you tell me how should I use that assumption? – Pedro Gomes Oct 19 '18 at 15:01
• 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 $\mathscr U \cap \mathscr V = \varnothing$. – MPW Oct 19 '18 at 15:06

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.
• The contradiction lies in the fact $V$ cannot be $\emptyset$, since $Y$ is disconnected, right? – Pedro Gomes Oct 19 '18 at 15:21
• Yes , if you start with the assumption that $Y$ is disconnected. – Sumanta Das Oct 19 '18 at 15:23
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 $$Y \subset D$$ ]