I am wondering if there exists a connected subset $A$ in the plane (which is not an atom $\{x\}$ for some $x\in \mathbb{R}^2$) with the property that every continuous map $\gamma : [0,1] \to A$ is constant.
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ I would start with the irrationals product with themselves, and see if I could add countable many points with one coordinate rational to foil a pair of separating sets. Also, arrange that the points added are far enough apart to prove no arc present. Alternatively, look at something like a fan of Knaster-Kuratowski. Gerhard "Is A Fan Of Counterexamples" Paseman, 2019.12.30. $\endgroup$– Gerhard PasemanCommented Dec 30, 2019 at 22:27
-
$\begingroup$ mathoverflow.net/questions/213208/… $\endgroup$– Gjergji ZaimiCommented Dec 30, 2019 at 22:40
-
$\begingroup$ Yes. Thank you. $\endgroup$– user150491Commented Dec 30, 2019 at 23:02
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
1
Yes, the Knaster-Kuratowski fan will do (use the notation as in the link): if there is a $t$ such that $\gamma(t)\neq p$ then there is an interval $I$ around $t$ such that $p\notin\gamma[I]$. Project $\gamma[I]$ down to the Cantor set; by continuity and connectedness of $I$ the image is one point $c$, so $\gamma[I]\subseteq X_c$. But $X_c$ is totally disconnected, so $\gamma[I]$ consists of just one point, $x$ say. This shows that $\{t:\gamma(t)=x\}$ is open; as it is also closed it is $[0,1]$. The other case has $\gamma$ constant with value $p$.
-
$\begingroup$ Thanks. This answers the question. $\endgroup$– user150491Commented Dec 30, 2019 at 23:02