TITLE: Is there a nonempty open bounded subset of plane whose boundary contains no 1 dimensional interval?
QUESTION [7 upvotes]: Someone asked a question here which hasn't received a correct answer because everyone seems to be misinterpreting the question.  I would like to ask the question again.
Does there exist a nonempty bounded open subset $\Omega$ of $\mathbb{R}^2$ such that there is no continuous injective map $[0,1]\to \partial\Omega$?  
Note that I am not asking about a continuous bijection because then the problem is trivial (for example the is no continuous bijection of $[0,1]$ to the boundary of a ball).

REPLY [3 votes]: As answered by Eric Wofsey on MathOverflow, such an open set is given by the bounded component of the complement of pseudo-circle. Pseudo-circle, a close relative of pseudo-arc, was introduced by R.H. Bing in Concerning hereditarily indecomposable continua (Pacific J. Math. Volume 1, Number 1 (1951), 43-51.). Here is a description: begin with a circular chain of disks such as the one below, then create another such chain inside, making it crooked, i.e., with a lot of back-and-forth movement (precise definition in Bing's paper). On the picture, the smaller chain is represented by a  polygonal curve; you should imagine covering this curve with small circles. The process continues indefinitely, and the intersection of all these chains is a pseudo-circle. 
 
The illustration is from the dissertation Factorwise Rigidity Involving Hereditarily Indecomposable Spaces by Kevin B. Gammon.