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Let $A,B \subseteq \mathbb{R}^n$ be connected subsets of $\mathbb{R}^n$ with $A \cap B \neq \emptyset$. Prove that $A \cup B$ is connected

Heres what I have so far:

Assume for the sake of contradiction that $A \cup B$ is not connected. Then we have by definition that $U$ and $V$ are non-empty relatively open sets in $A \cup B$ with $U \cap V = \emptyset$ and $A \cup B = U \cup V$. We also know that as $U$ and $V$ are non-empty relatively open sets in $A \cup B$, $U$ and $V$ such that $U \subseteq A \cup B$ and for some open set $C$ that $U = (A \cup B) \cap C$ and similarly $V \subseteq A \cup B$ and for some open set $D$ we have $V = (A \cup B)\cap D$.

This is about all I have from the givens, we have to come to some contradiction, but I'm not sure where it'll come from.

If anyone has any hints, I don't want the solution to this problem just a place to head towards.

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    $\begingroup$ You titled this "Prove A∪B is open" but in the text you say "prove A∪B is connected". Is the second what you want? $\endgroup$
    – user247327
    Oct 21, 2020 at 21:39
  • $\begingroup$ You are correct I want to prove its connected I'll fix that right now $\endgroup$
    – Joey
    Oct 21, 2020 at 21:42

2 Answers 2

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Hint: Pick a point $p\in A\cap B$. If $A\cup B$ is a union of two disjoint open sets, say $U,V$, then $p$ must lie in one of them, say, $U$. Prove that $U\cap A$ is open and closed in $A$.

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I will use this characterisation of connected space : a topological space $X$ is connected if and only if every continuous function $f : X \to \{0,1\}$ is constant.

Suppose $A$ and $B$ are connected topological spaces and that $A\cap B \neq \varnothing$. Let $x\in A\cap B$.

Let $f : A\cup B \to \{0,1\}$ be continuous. Then its restrictions $f_A$ to $A$ and $f_B$ to $B$ are continuous too. As $A$ is connected and $x \in A$, $f_A$ is constant and takes only $f(x)$ as a value. Similarly, $f_B$ is constant and takes only $f(x)$ as a value. Thus, $f$ is constant, and $A\cup B$ is connected.

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