Timeline for Path connected without bounded path connected subset?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 3, 2018 at 18:10 | review | Close votes | |||
| Sep 9, 2018 at 18:08 | |||||
| Sep 2, 2018 at 23:25 | comment | added | Stefano Gogioso | 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 $x\neq y$. | |
| S Sep 2, 2018 at 21:57 | history | suggested | Stefano Gogioso | CC BY-SA 4.0 |
I have added a clarification excluding singletons from the question statement, because the reference provided excludes singletons as well (see page 100, point 17)
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| Sep 2, 2018 at 21:57 | comment | added | Nate Eldredge | @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. | |
| Sep 2, 2018 at 21:46 | review | Suggested edits | |||
| S Sep 2, 2018 at 21:57 | |||||
| Sep 2, 2018 at 19:34 | comment | added | Forever Mozart | Perhaps a better question would be: Does every path connected subset of the plane have a non-degenerate bounded connected subset which is not path connected? | |
| Sep 2, 2018 at 16:17 | comment | added | Christian Remling | Or a single point is a path connected subset of any set. | |
| Sep 2, 2018 at 15:22 | comment | added | Nate Eldredge | 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? | |
| Sep 2, 2018 at 14:51 | comment | added | erz | i think the answer works for path connectedness as well, although this is not what you ask | |
| Sep 2, 2018 at 14:50 | comment | added | erz | this may be of interest: mathoverflow.net/questions/36587/… | |
| Sep 2, 2018 at 14:37 | history | asked | Portland | CC BY-SA 4.0 |