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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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I think you have a typo somewhere. Do you mean $x \in \mathbb{R}^n$ in your second formula? – Zhen Lin Aug 17 '11 at 5:37
Yes, Im sorry. Of course is $x \in \mathbb{R}^n$. Thanks – Hiperion Aug 17 '11 at 15:42
up vote 1 down vote accepted

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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Typo: You want $x+K=A\cup B$ and $(A-x)\cup (B-x)$. If you do it this way, you can only claim that $A$ and $B$ are relatively open in $x+K$, and then you have to work just a little to make sure that $A-x$ and $B-x$ are relatively open in $K$. – Brian M. Scott Aug 17 '11 at 5:51
@Brian: you're certainly right. I thought that relative openness is also preserved under the continuous maps, isn't it? – Ilya Aug 17 '11 at 6:54
There are several details to check, and I’m guessing that the OP isn’t very familiar with the subject. What you really have is open sets $U,V$ s.t. $x+K\subseteq U\cup V$, $U\cap (x+K)\neq\varnothing\neq V\cap (x+K)$, and $(x+k)\cap U\cap V=\varnothing$. I agree that it’s easy enough to check that $U-x$ and $V-x$ ‘work’ for $K$ (or more generally that $f^{-1}[U]$ and $f^{-1}[V]$ ‘work’ for $f^{-1}[x+K]$), but a beginner sometimes needs to be reminded just what has to be checked. – Brian M. Scott Aug 17 '11 at 7:09
@Brian: ok, thanks. Should I leave my answer? Your is more beautiful and general, though maybe OP is looking for the proof without homeomorphisms? – Ilya Aug 17 '11 at 7:29
Sure, why not? You never know what someone will find helpful. Just change the intersections to unions to fix the typos; any residual confusion on the OP’s part ought to be cleared up by the comments. – Brian M. Scott Aug 17 '11 at 7:35

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.

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thank you very much! – Hiperion Aug 17 '11 at 16:01

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