Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
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
add comment

2 Answers

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.

share|improve this answer
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
show 2 more comments

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.

share|improve this answer
thank you very much! –  Hiperion Aug 17 '11 at 16:01
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.