Timeline for How "wild" can hereditarily decomposable continua be?
Current License: CC BY-SA 4.0
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| Oct 26 at 11:46 | comment | added | Alessandro Codenotti | The drawings are a bit confusing, but the description of the projection $X_2\to X_1$ helps, you want $X_1\setminus\{x^1_1\}$ to have a path component homeomorphic to $(0,1]$ on which to map the pathcomponent of the added continuum which is also homeomorphic to an half-open arc. In other words the component of $X_1\setminus\{x^1_1\}$ has been twisted into the wiggly part of the new continuum | |
| Oct 26 at 10:36 | comment | added | Daron | @AlessandroCodenotti I was hoping someone would mention Nadler 2.27 because I don't understand what the heck is going on! The text says the first arc must be "inserted" at any point other than the three extreme points of the sin continuum. But the images show only the arc being inserted at the extreme points! Do you know how the example is supposed to work? Are the extra arcs supposed to be inserted perpendicular to the existing space? | |
| Oct 24 at 11:54 | comment | added | Alessandro Codenotti | Very nice question! I do not know the answer to the main question so let me just give two examples of somewhat wild hereditarily decomposable continua, namely those presented in Exercise 2.27 and 2.32 of Nadler's Continuum Theory. Both continua are planar, hereditarily decomposable and totally path disconnected (they contain no nontrivial arcs). Moreover the second one has the property that all of its subcontinua separate the plane | |
| Oct 24 at 9:57 | history | edited | Daron | CC BY-SA 4.0 | deleted 2 characters in body |
| Oct 24 at 8:19 | history | asked | Daron | CC BY-SA 4.0 |