Timeline for Is it possible to show that an infinite set has a countable (infinite) subset, without using the Axiom of Choice?
Current License: CC BY-SA 2.5
5 events
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| Jan 7, 2011 at 18:19 | comment | added | Andrés E. Caicedo |
@Asaf: Amorphous sets are Dedekind finite. The point is that if you remove a point $a$ from a set $X$, and the result has the same size as $X$, you can iterate, and get an injection of ${\mathbb N}$: Keep track of the orbit of $a$ as you iterate the bijection $X\to X\setminus\{a\}$.
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| Jan 7, 2011 at 18:09 | comment | added | Asaf Karagila♦ | @Peter: If $X$ is amorphous (as defined by Stefan in his answer) is it not Dedekind-infinite with some bijection to a cofinite subset? | |
| Jan 7, 2011 at 17:45 | comment | added | Peter LeFanu Lumsdaine | It’s perhaps worth noting that this may not be the most familiar definition of Dedekind-infinite (“X is D-infinite if it is bijective to some proper subset of itself”), but that these two definitions are equivalent in ZF. | |
| Jan 7, 2011 at 17:37 | vote | accept | tibet | ||
| Jan 7, 2011 at 17:38 | |||||
| Jan 7, 2011 at 17:31 | history | answered | Chris Eagle | CC BY-SA 2.5 |