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Nov 9, 2018 at 9:36 vote accept Dominic van der Zypen
Nov 7, 2018 at 18:36 history edited Dominic van der Zypen CC BY-SA 4.0
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Nov 7, 2018 at 18:33 comment added Dominic van der Zypen Right, thanks @KPHart, will modify this!
Nov 7, 2018 at 12:42 answer added KP Hart timeline score: 3
Nov 7, 2018 at 11:31 comment added KP Hart The `obviously' is not quite true: in $\mathbb{R}$ we have $\mathbb{Q}\subseteq\mathbb{R}\setminus\{\sqrt2\}$ and $\pi+\mathbb{Q}\subseteq\mathbb{R}\setminus\{0\}$. In fact there is an injective function $f:[\mathbb{R}]^\omega\to\mathbb{R}$ such that $f(A)\notin A$ for all $A$, so $\mathbb{R}$ does admit a function as required.
Nov 7, 2018 at 10:15 history asked Dominic van der Zypen CC BY-SA 4.0