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Let $A$ and $B$ two connected sets in $\mathbb{R}$. Is $A+B$ also connected in $\mathbb{R}$? Then what about if it is in any space $X$? I know that any connected set in $\mathbb{R}$ is an interval or any singleton set. It is obvious in $\mathbb{R}$. But how can I prove generally? Please help me to solve it.

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  • $\begingroup$ What do you mean by $A+B$? $\endgroup$ – ervx Jul 28 '17 at 19:49
  • $\begingroup$ What's $R$? $\Bbb R$? (Sounds like it based on your interval/singleton set statement.) What's $A+B$? $\{a + b : a \in A, \ b \in B \}$? $\endgroup$ – tilper Jul 28 '17 at 19:49
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    $\begingroup$ @ervx $$A+B = \{t : a \in A \ \land \ b \in B \ \land \ a+b=t\}$$ $\endgroup$ – MathematicsStudent1122 Jul 28 '17 at 19:50
  • $\begingroup$ $A+B$={a+b:a∈A, b∈B} $\endgroup$ – addicted.insta Jul 28 '17 at 19:57
  • $\begingroup$ Continuous functions take connected sets to connected sets. In general, your space has to have addition defined. If addition is defined and continuous then the answer is yes. $\endgroup$ – John Douma Jul 28 '17 at 20:04
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The set $A\times B$ is connected, and $A+B$ is the image of this set by the continuous map $(x,y)\mapsto x+y$.

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  • $\begingroup$ Namely, this holds for $A, B\subseteq X$, where $X$ is any topological vector space. $\endgroup$ – Michael Lee Jul 28 '17 at 22:25
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Let $u\in A$, consider a continuous function $f:A+B\rightarrow \{0,1\}$. $f(u+B)$ is constant since $u+B$ is connected. Without restricting the generality, assume that $f(u+B)=0$. For every $y\in B$, $f(A+y)$ is constant, $f(A+y)=f(u+y)=0$. Let $x\in A$, you have $x+y\in A+y$, we deduce that $f(x+y)=0$, so $f$ is constant and $A+B$ is connected.

Remark that this proof can be applied to $A,B\subset E$ where $E$ is a topological vector space.

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