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I'm trying to show:

If $K\subset \mathbb{R}^n$ is connected and $x\in \mathbb{R}^n$ then $x+K=\{x+y\in \mathbb{R}^n: y\in K\}$ is connected.

Thanks for your help.

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    Yes, Im sorry. Of course is $x \in \mathbb{R}^n$. Thanks2011-08-17

2 Answers 2

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Suppose that $x+K = A\cup B$ where $A,B$ are open (relatively) in $x+K$ and disjoint. Then $A-x$ and $B-x$ are still open (relatively in $K$) and disjoint but $\{A-x\}\cup \{B-x\} = K$ which contradicts with the fact that $K$ is connected.

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    Notational nitpick: It might be better to write $-x+A$ rather than $A-x$, since the latter could be confused for $A-\{x\}$.2011-08-17
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Fix $y \in \mathbb{R}^n$ and let $f:\mathbb{R}^n \to \mathbb{R}^n:x \mapsto x+y$; it’s easy to check that $f$ is continuous and that $f[K] = y + K$. Continuous functions preserve connectedness, so $y + K$ is connected. (For that matter, it’s easy enough to check that $f$ is a homeomorphism, which makes the conclusion even more readily apparent.)

Gortaur’s proof is a special case of the usual proof that continuous preserve connectedness.