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.
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.
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.
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.