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Question: Is there a path connected subset of $\mathbb R^2$, without any bounded path connected subset (aside from singletons)?

Motivation: If we replace "path connected" by "connected", then the answer is positive. It was proven by Stefan Mazurkiewicz, see the original article.

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  • $\begingroup$ this may be of interest: mathoverflow.net/questions/36587/… $\endgroup$
    – erz
    Commented Sep 2, 2018 at 14:50
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    $\begingroup$ If the set $E$ is path connected, then it contains a path (barring trivialities), and that path is itself a bounded path connected subset of $E$. Or am I missing something? $\endgroup$
    – Nate Eldredge
    Commented Sep 2, 2018 at 15:22
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    $\begingroup$ Or a single point is a path connected subset of any set. $\endgroup$
    – Christian Remling
    Commented Sep 2, 2018 at 16:17
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    $\begingroup$ @ForeverMozart: The set $\mathbb{R} \times \{0\}$, i.e. the $x$-axis, certainly doesn't. Any set with nonempty interior certainly does, because you can put a tiny copy of the topologist's sine curve in there. $\endgroup$
    – Nate Eldredge
    Commented Sep 2, 2018 at 21:57
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    $\begingroup$ Let $E$ be a path-connected subset of $\mathbb{R}^2$ which is not a singleton, and let $x,y \in E$ be distinct points. Then there is a path from $x$ to $y$ in $E$, i.e. there is a continuous function $\gamma: [0,1] \rightarrow E$ such that $\gamma(0)=x$ and $\gamma(1)=y$. The unit interval $[0,1]$ is compact and connected, hence its image $\gamma([0,1])$ under the continuous function $\gamma$ must also be compact (hence bounded in $\mathbb{R}^2$) and connected. The subset $\gamma([0,1])\subseteq E$ is your desired bounded path-connected subset of $E$, which is not a singleton since $x\neq y$. $\endgroup$
    – Stefano Gogioso
    Commented Sep 2, 2018 at 23:25

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