Skip to main content
8 events
when toggle format what by license comment
May 4, 2017 at 14:18 comment added erz @JeremyBrazas why cannot we just remove a point from each of the circles?
May 4, 2017 at 8:11 answer added Francesco Polizzi timeline score: 3
May 3, 2017 at 17:59 answer added Will Brian timeline score: 3
May 3, 2017 at 17:50 comment added Pietro Majer @erz: any non-empty open subset of the Sierpinski carpet $S$ contains the boundary of some square $Q$ disjoint from $S$ (one of the squares removed in the construction). Of course $\partial Q$ is a loop which is not contractible in $\mathbb{R}^2\setminus Q$, and a fortiori in $S$ .
May 3, 2017 at 17:22 comment added Christian Remling The infinite-dimensional torus $S^{\mathbb N}$ with product topology is a more obvious example perhaps of a space with no simply connected open sets.
May 3, 2017 at 16:01 comment added erz @PietroMajer Although it looks very intuitive, how exactly do you prove it?
May 3, 2017 at 15:52 comment added Pietro Majer A negative example for the stronger property is the Sierpinski carpet en.wikipedia.org/wiki/Sierpinski_carpet , a connected compact plane set, whose non-empty open subsets are not simply connected.
May 3, 2017 at 14:26 history asked erz CC BY-SA 3.0