Timeline for Proofs of the uncountability of the reals
Current License: CC BY-SA 2.5
5 events
| when toggle format | what | by | license | comment | |
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| Oct 19, 2014 at 9:48 | comment | added | Avshalom | @gowers This might answer (2) negatively. Suppose $T \models (X \subset \mathbb{R}$ is isomorphic to $\mathbb{Q})$. Then $T \models (X \in H(\omega_{1}))$, since $\mathbb{Q} \in H(\omega_{1})$ which satisfies Replacement; hence $X$ is countable. | |
| Nov 28, 2010 at 23:21 | comment | added | Andrés E. Caicedo | In case somebody missed it, Aaron posted the above as a question: mathoverflow.net/questions/47185/… | |
| Nov 24, 2010 at 2:03 | comment | added | Aaron Meyerowitz | That is a really great point about the importance of having an order for countability. I can't say that I see why for every Borel function $f$ defined on countable subsets of reals there is a countable set of reals $X$ such that such that $f(X) \in X$ (if that is a fair restatement) but I don't see any reason to doubt that. | |
| Nov 23, 2010 at 11:21 | comment | added | Unknown | I'm obliged for your thoughts. | |
| Nov 22, 2010 at 21:02 | history | answered | gowers | CC BY-SA 2.5 |